I have this problem: Given a vector with n numbers, sort the numbers so that the even ones will be on odd positions and the odd numbers will be on even positions. E.g. If I have the vector 2 6 7 8 9 3 5 1, the output should be 2 7 6 9 8 3 5 1 . The count should start from 1. So on position 1 which is actually index 0 should be an even number, on position 2 which is actually index 1 should be an odd number and so on. Now this is easy if the odd and even numbers are the same, let's say 4 even number and 4 odd numbers in the vector, but what if the number of odd numbers differs from the number of even numbers like in the above example? How do I solve that. I attached the code with one of the tries I did, but it doesn't work. Can I get some help please. I ask you to keep it simple that means only with vectors and such. No weird methods or anything cause I'm a beginner and I only know the basics. Thanks in advance!
I have to mention that n initial is globally declared and is the number of vector elements and v_initial is the initial vector with the elements that need to be rearranged.
The task says to add the remaining numbers to the end of the vector. Like if there are 3 odd and 5 even numbers, The 2 extra even numbers should be thrown at the end of the vector
void vector_pozitii_pare_impare(int v_initial[])
{
int v_pozitie[50],c1=0,c2=1;
for (i = 0; i < n_initial; i++)
{
if (v_initial[i] % 2 == 0)
{
bool isTrue = 1;
for (int k = i + 1; k < n_initial; k++)
{
if (v_initial[k] % 2 != 0)
isTrue = 0;
}
if (isTrue)
{
v_pozitie[c1] = v_initial[i];
c1++;
}
else
{
v_pozitie[c1] = v_initial[i];
c1 += 2;
}
}
else
{
bool isTrue = 1;
for (int j = i + 1; j < n_initial; j++)
{
if (v_initial[j] % 2 == 0)
{
isTrue = 0;
}
if (isTrue)
{
v_pozitie[c2] = v_initial[i];
c2++;
}
else
{
v_pozitie[c2] = v_initial[i];
c2 += 2;
}
}
}
}
This may not be a perfect solution and it just popped out right off my mind without being tested or verified, but it's just to give you an idea.
(Let A,B,C,D be odd numbers and 0,1,2 even numbers correspondingly)
Given:
A 0 B C D 1 2 (random ordered list of odd/even numbers)
Wanted:
A 0 B 1 C 2 D (input sequence altered to match the wanted odd/even criteria)
Next, we invent the steps required to get from given to wanted:
// look at 'A' -> match, next
// Result: A 0 B C D 1 2
// look at '0' -> match, next
// Result: A 0 B C D 1 2
// look at 'B' -> match, next
// Result: A 0 B C D 1 2
// look at 'C' -> mismatch, remember index and find first match starting from index+1
// Result: A 0 B C D ->1<- 2
// now swap the numbers found at the remembered index and the found one.
// Result: A 0 B 1 D C 2
// continue until the whole list has been consumed.
As I said, this algorithm may not be perfect, but my intention is to give you an example on how to solve these kinds of problems. It's not good to always think in code first, especially not with a problem like this. So you should first think about where you start, what you want to achieve and then carefully think of how to get there step by step.
I feel I have to mention that I did not provide an example in real code, because once you got the idea, the execution should be pretty much straight forward.
Oh, and just a small remark: Almost nothing about your code is C++.
A simple solution, that is not very efficient would be to split the vector into 2 vectors, that contain even and uneven numbers and then always take one from the even, one from the uneven and then the remainder, from the one that is not completely entered.
some c++ (that actually uses vectors, but you can use an array the same way, but need to change the pointer arithmetic)
I did not test it, but the principle should be clear; it is not very efficient though
EDIT: The answer below by #AAAAAAAAARGH outlines a better algorithmic idea, that is inplace and more efficient.
void change_vector_even_uneven(std::vector<unsigned>& in_vec){
std::vector<unsigned> even;
std::vector<unsigned> uneven;
for (auto it = in_vec.begin(); it != in_vec.end(); it++){
if ((*it) % 2 == 0)) even.push_back(*it);
else uneven.push_back(*it);
}
auto even_it = even.begin();
auto uneven_it = uneven.begin();
for (auto it = in_vec.begin(); it != in_vec.end(); it++){
if (even_it == even.end()){
(*it) = (*uneven_it);
uneven_it++;
continue;
}
if (uneven_it == uneven.end()){
(*it) = (*even_it);
even_it++;
continue;
}
if ((it - in_vec.begin()) % 2 == 0){
(*it) = (*even_it);
even_it++;
}
else{
(*it) = (*uneven_it);
uneven_it++;
}
}
}
The solutions is simple. We sort the even and odd values into a data structure. In a loop, we iterate over all source values. If they are even (val & 2 == 0) we add them at the end of a std::deque for evens and if odd, we add them to a std::deque for odds.
Later, we we will extract the the values from the front of the std::deque.
So, we have a first in first out principle.
The std::deque is optimized for such purposes.
Later, we make a loop with an alternating branch in it. We, alternatively extract data from the even queue and then from the odd queue. If a queue is empty, we do not extract data.
We do not need an additional std::vector and can reuse the old one.
With that, we do not need to take care for the same number of evens and odds. It will of course always work.
Please see below one of millions of possible solutions:
#include <iostream>
#include <vector>
#include <deque>
int main() {
std::vector testData{ 2, 6, 7, 8, 9, 3, 5, 1 };
// Show initial data
std::cout << "\nInitial data: ";
for (const int i : testData) std::cout << i << ' ';
std::cout << '\n';
// We will use a deques to store odd and even numbers
// With that we can efficiently push back and pop front
std::deque<int> evenNumbers{};
std::deque<int> oddNumbers{};
// Sort the original data into the specific container
for (const int number : testData)
if (number % 2 == 0)
evenNumbers.push_back(number);
else
oddNumbers.push_back(number);
// Take alternating the data from the even and the odd values
bool takeEven{ true };
for (size_t i{}; !evenNumbers.empty() && !oddNumbers.empty(); ) {
if (takeEven) { // Take even numbers
if (not evenNumbers.empty()) { // As long as there are even values
testData[i] = evenNumbers.front(); // Get the value from the front
evenNumbers.pop_front(); // Remove first value
++i;
}
}
else { // Now we take odd numbers
if (not oddNumbers.empty()) { // As long as there are odd values
testData[i] = oddNumbers.front(); // Get the value from the front
oddNumbers.pop_front(); // Remove first value
++i;
}
}
// Next take the other container
takeEven = not takeEven;
}
// Show result
std::cout << "\nResult: ";
for (const int i : testData) std::cout << i << ' ';
std::cout << '\n';
return 0;
}
Here is yet another solution (using STL), in case you want a stable result (that is, the order of your values is preserved).
#include <algorithm>
#include <vector>
auto ints = std::vector<int>{ 2, 6, 7, 8, 9, 3, 5, 1 };
// split list to even/odd sections -> [2, 6, 8, 7, 9, 3, 5, 1]
const auto it = std::stable_partition(
ints.begin(), ints.end(), [](auto value) { return value % 2 == 0; });
auto results = std::vector<int>{};
results.reserve(ints.size());
// merge both parts with equal size
auto a = ints.begin(), b = it;
while (a != it && b != ints.end()) {
results.push_back(*a++);
results.push_back(*b++);
}
// copy remaining values to end of list
std::copy(a, it, std::back_inserter(results));
std::copy(b, ints.end(), std::back_inserter(results));
The result ist [2, 7, 6, 9, 8, 3, 5, 1]. The complexity is O(n).
This answer, like some of the others, divides the data and then reassembles the result. The standard library std::partition_copy is used to separate the even and odd numbers into two containers. Then the interleave function assembles the result by alternately copying from two input ranges.
#include <algorithm>
#include <iostream>
#include <vector>
template <typename InIt1, typename InIt2, typename OutIt>
OutIt interleave(InIt1 first1, InIt1 last1, InIt2 first2, InIt2 last2, OutIt dest)
{
for (;;) {
if (first1 == last1) {
return std::copy(first2, last2, dest);
}
*dest++ = *first1++;
if (first2 == last2) {
return std::copy(first1, last1, dest);
}
*dest++ = *first2++;
}
}
void reorder_even_odd(std::vector<int> &data)
{
auto is_even = [](int value) { return (value & 1) == 0; };
// split
std::vector<int> even, odd;
std::partition_copy(begin(data), end(data), back_inserter(even), back_inserter(odd), is_even);
// merge
interleave(begin(even), end(even), begin(odd), end(odd), begin(data));
}
int main()
{
std::vector<int> data{ 2, 6, 7, 8, 9, 3, 5, 1 };
reorder_even_odd(data);
for (int value : data) {
std::cout << value << ' ';
}
std::cout << '\n';
}
Demo on Compiler Explorer
As suggested, I am using vectors and STL.
No need to be a great mathematician to understand v_pozitie will start with pairs of odd and even and terminate with the integers not in the initial pairs.
I am then updating three iterators in v_positie (no need of temporary containers to calculate the result) : even, odd and end,(avoiding push_back) and would code this way :
#include <vector>
#include <algorithm>
void vector_pozitii_pare_impare(std::vector<int>& v_initial, std::vector<int>& v_pozitie) {
int nodd (0), neven (0);
std::for_each (v_initial.begin (), v_initial.end (), [&nodd] (const int& n) {
nodd += n%2;
});
neven = v_initial.size () - nodd;
int npair (neven < nodd ?neven:nodd);
npair *=2;
std::vector<int>::iterator iend (&v_pozitie [npair]), ieven (v_pozitie.begin ()), iodd (&v_pozitie [1]);
std::for_each (v_initial.begin (), v_initial.end (), [&iend, &ieven, &iodd, &npair] (const int& s) {
if (npair) {
switch (s%2) {
case 0 :
*ieven++ = s;
++ieven;
break;
case 1 :
*iodd++ = s;
++iodd;
break;
}
--npair;
}
else *iend++ = s;
});
}
int main (int argc, char* argv []) {
const int N = 8;
int tab [N] = {2, 6, 7, 8, 9, 3, 5, 1};
std::vector<int> v_initial (tab, (int*)&tab [N]);
std::cout << "\tv_initial == ";
std::for_each (v_initial.begin (), v_initial.end (), [] (const int& s) {std::cout << s << " ";});
std::cout << std::endl;
std::vector<int> v_pozitie (v_initial.size (), -1);
vector_pozitii_pare_impare (v_initial, v_pozitie);
std::cout << "\tv_pozitie == ";
std::for_each (v_pozitie.begin (), v_pozitie.end (), [] (const int& s) {std::cout << s << " ";});
std::cout << std::endl;
}
Related
Suppose I have a given sum, say sum = 4. I am also given a vector = {2,4}. There are two ways to generate the given sum from the given vector (elements may be reused).
One way is just {4} cause 4 = 4.
Second way is {2,2} cause 2 + 2 = 4.
I have to find the shortest possible combination, therefore in this particular case the answer is {4}.
Here is my approach - I go through the tree, and when on the leaf I get a 0, we hit the base case, return {} vector, and fill up the vector while traversing the tree. When I get to a node, I choose the smaller of the two (or more) vectors. This way when I reach the root node, I should get a vector of the shortest combination that can yield me the target sum.
As of yet, I do not care about time constraints as such, I know there's a lot of repetitive computing going on so I will have to memoize it once I can get the basic version correct.
I have been trying to figure why this code is not working. Any insight would be appreciated.
#include <vector>
#include <algorithm>
#include <iostream>
using namespace std;
vector<int> findBestSum(int targetSum, const vector<int> &elements, vector<vector<int>> &temp) {
if (targetSum == 0)
return {};
else if (targetSum < 0)
return {-1};
else {
vector<int> small;
for (auto &i : elements) {
int remainder = targetSum - i;
vector<int> returnedVector = findBestSum(remainder, elements, temp);
if ((!returnedVector.empty() && find(returnedVector.begin(), returnedVector.end(), -1) == returnedVector.end()) || returnedVector.empty()) {
returnedVector.push_back(i);
temp.push_back(returnedVector);
}
int smallestLength = temp[0].size();
for (auto &j : temp)
if (smallestLength >= j.size())
small = j;
}
return small;
}
}
int main() {
int targetSum = 6;
const vector<int> elements{2, 3, 5}; // answer should be [3,3] however I just get a 3...
vector<vector<int>> temp;
vector<int> bestSumVector = findBestSum(targetSum, elements, temp);
for (auto i : bestSumVector)
cout << i << " ";
}
Update (14th of July, 2021):
After a few busy months I have tried to lock horns with this problem and this time my code looks like this:
#include <iostream>
#include <vector>
#include <map>
#include <numeric>
using namespace std;
bool howSum(int &targetSum, vector<int> &elementVector, vector<int> &howSumVector, vector<vector<int>> &allSums) {
static int originaltargetsum = targetSum;
if (targetSum == 0)
return true;
else if (targetSum < 0)
return false;
else {
for (auto i : elementVector) {
int remainder = targetSum - i;
bool flag = howSum(remainder, elementVector, howSumVector, allSums);
if (flag) {
howSumVector.push_back(i);
if (targetSum == originaltargetsum ||
accumulate(howSumVector.begin(), howSumVector.end(), 0) == originaltargetsum) {
allSums.push_back(howSumVector);
howSumVector.clear();
}
return true;
}
}
return false;
}
}
int main() {
int sum = 8;
vector<int> elements = {1, 4, 5};
vector<vector<int>> allSums = {};
vector<int> workingBench = {};
howSum(sum, elements, workingBench, allSums);
for (auto &i : allSums) {
for (auto &j : i) {
cout << j << " ";
}
cout << endl;
}
}
For this I have sum as 8 and elements as {1, 4, 5}.
Also I'm storing and displaying all possible solutions right now (once that is correctly done, finding shortest vector and memoization should be easy). Possible solutions in this case are:
[1, 1, 1, 1, 1, 1, 1, 1]
[4, 4]
[5, 1, 1, 1]
[4, 1, 1, 1, 1]
Currently my code only shows the first possible combination. I'm pretty sure I'm returning true and false incorrectly, please help me out here.
I took a stab at this. I do have a working solution, hopefully it is what you want:
#include <iostream>
#include <vector>
#include <algorithm>
void howSum(int targetSum, const std::vector<int> & elementVector, const std::vector<int> & howSumVector, std::vector<std::vector<int>> & allSums)
{
static int originaltargetsum = targetSum;
if (targetSum == 0)
{
allSums.push_back(howSumVector);
return;
}
else if (targetSum < 0)
{
return;
}
else
{
for (const auto i : elementVector)
{
// an element less than or equal to 0 would cause an infinite loop
if (i <= 0)
continue;
std::vector<int> newSumVector = howSumVector;
newSumVector.push_back(i);
std::vector<int> newElementVector;
std::copy_if(std::begin(elementVector), std::end(elementVector), std::back_inserter(newElementVector), [i](int element){ return element >= i; });
howSum(targetSum - i, newElementVector, newSumVector, allSums);
}
}
}
int main()
{
int sum = 8;
std::vector<int> elements = { 1, 4, 5 };
std::vector<std::vector<int>> allSums = {};
std::vector<int> workingBench = {};
howSum(sum, elements, workingBench, allSums);
for (const auto & i : allSums)
{
for (const auto & j : i)
{
std::cout << j << " ";
}
std::cout << std::endl;
}
return 0;
}
I think, in general, you were over-thinking or over-engineering the problem. Like others have mentioned, your current code is returning true too early, and nothing besides the first element/combination is tested. With recursion, it is important to take care in your return cases - really, you only want a base case or two, and otherwise you want to recur.
With the solution I have here, the main thing I have added is copying the current combination of elements for each element you need to test. That solves your main issue of not testing every combination of numbers. In addition to that, it seemed better to append to allSums when the targetSum was reached. With those changes, I was able to do away with the bool return value and simplify the code a bit. Running the code above gives these solutions:
1 1 1 1 1 1 1 1
1 1 1 1 4
1 1 1 4 1
1 1 1 5
1 1 4 1 1
1 1 5 1
1 4 1 1 1
1 5 1 1
4 1 1 1 1
4 4
5 1 1 1
This does have some duplicates (because of the order things are tested) but I felt like it is good enough since you only want the smallest solution, 4 4. To find this, you would just need to sort the allSums vector by inner vector size and then take the first entry.
I think you need to change the implementation to correctly process elements of the vector.
In your implementation it doesn't go over all vector items, just the first one.
This is one way to do it if you use vector elements as the first parameter in your function.
vector<int> findBestSum(int element, int targetSum, const vector<int>& elements,
vector<vector<int>>& temp) {
if (targetSum == 0)
return {};
else if (targetSum < 0)
return { -1 };
else {
int remainder = targetSum - element;
vector<int> returnedVector = findBestSum(element, remainder, elements, temp);
if ((!returnedVector.empty() && find(returnedVector.begin(), returnedVector.end(), -1) == returnedVector.end()) || returnedVector.empty()) {
returnedVector.push_back(element);
return returnedVector;
}
return returnedVector;
}
}
int main() {
const int targetSum = 6;
const vector<int> elements{ 2, 3, 5 }; // answer should be [3,3] however I just get a 3...
vector<vector<int>> temp;
for (auto i : elements) {
vector<int> returnedVector = findBestSum(i, targetSum, elements, temp);
if ((!returnedVector.empty() && find(returnedVector.begin(), returnedVector.end(), -1) == returnedVector.end()) || returnedVector.empty())
temp.push_back(returnedVector);
}
if (temp.size() > 0) {
vector<int> bestSum = {};
size_t small = 0;
size_t smallestLength = temp[0].size();
for (auto& j : temp)
if (smallestLength >= j.size()) {
small = j.size();
bestSum = j;
}
for (auto i : bestSum)
cout << i << " ";
}
else
cout << " sum not found" << endl;
}
I tried to solve this exercise
I got 66 percent
I can not understand why
can you help?
The exercise is:
Write a function:
int solution(vector &A);
that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.
For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.
Given A = [1, 2, 3], the function should return 4.
Given A = [−1, −3], the function should return 1.
The solution I wrote is:
#include <algorithm>
#include<cmath>
using namespace std;
int solution(vector<int> &A) {
if (A.size() == 0 || (A.size() == 1 && A.at(0) <= 0))
return 1;
if (A.size() == 1)
return A.at(0) + 1;
sort(A.begin(), A.end());
if (A.at(A.size() - 1) <= 0)
return 1;
auto ip = std::unique(A.begin(), A.end());
A.resize(distance(A.begin(), ip));
A.erase(remove_if(A.begin(), A.end(), [](const int i) { return i < 0; }),A.end());
if (A.at(0) != 1)
return 1;
if (A.size() == 1)
return (A.at(0) != 1 ? 1 : 2);
int i = 0;
for (; i < A.size(); ++i) {
if (A.at(i) != i + 1)
return A.at(i - 1) + 1;
}
return A.at(A.size()) + 1;
}
The following algorithm has a complexity O(n). No need to sort or to erase.
We know that the first missing value is less or equal to n+1, if n is the array size.
Then we simply have to use an array of size n+2, present[n+2], initialised to 0, and then to look at all values A[i]:
if (A[i] <= 1+n && A[i] > 0) present[A[i]] = 1;
Finally, in a second step we simply have to examine the array present[.], and search for the first index k such that present[k]==0.
#include <iostream>
#include <vector>
int find_missing (const std::vector<int> &A) {
int n = A.size();
std::vector<int> present (n+2, 0);
int vmax = n+1;
for (int i = 0; i < n; ++i) {
if (A[i] <= vmax && A[i] > 0) {
present[A[i]] = 1;
}
}
for (int k = 1; k <= vmax; ++k) {
if (present[k] == 0) return k;
}
return -1;
}
int main() {
std::vector<int> A = {1, 2, 0, 3, -3, 5, 6, 8};
int missing = find_missing (A);
std::cout << "First missing element = " << missing << std::endl;
return 0;
}
Well this is wrong
if(A.size()==1)
return A.at(0)+1;
If A is {2} that code will return 3 when the correct answer is 1
Also
A.erase(remove_if(A.begin(), A.end(),[](const int i) {return i < 0; }),A.end());
should be
A.erase(remove_if(A.begin(), A.end(),[](const int i) {return i <= 0; }),A.end());
Also
return A.at(A.size()) + 1;
is a guaranteed vector out of bounds error.
Even a small amount of testing and debugging would have caught these errors. It's a habit you should get into.
I think there are far too many special cases in the code, which only serve to complicate the code and increase the chance of bugs.
This answer is the implementation of the proposal given in the comment by PaulMcKenzie.
So, all credits go to PaulMcKenzie
It is not the fastest solution, but compact. The idea is basically.
Sort the data
Then compare the adjacent values, if the next value is equal to the previous value+1.
If not, then we found a gap. This can be implemented by using the function std::adjacent_find. Description can be found here.
We put all the side conditions into the lambda. If std::adjacent_find cannot find a value, then we take the next possible positive value.
I am not sure, what I could describe more. Please see the below example:
#include <iostream>
#include <vector>
#include <algorithm>
int solution(std::vector<int>& data) {
// Sort
std::sort(data.begin(), data.end());
// Check if there is a gap in the positive values
const auto gap = std::adjacent_find(data.begin(), data.end(), [](const int p, const int n) { return (n !=p) && (n != (p + 1) && p>0); });
// If there is no gap, the take the next positive value
return (gap == data.end()) ? (data.back() > 0 ? data.back() + 1 : 1) : *gap + 1;
}
int main() {
//Some test cases
std::vector<std::vector<int>> testCases{
{1,3,6,4,1,2},
{1,2,3},
{-1,-3}
};
for (auto& testCase : testCases)
std::cout << solution(testCase) << '\n';
return 0;
}
others have already pointed out what are the main errors, but I would like to invite you to try a different solution instead of trying to fix all the bugs and spend much time on debugging, because your solution seems a little overcomplicated.
Here I propose a way you can think about the problem:
What is the minimum number the function can return? Since it returns a positive integer, it is 1, in the case 1 is not in the array. Since that we can use any number <=0 to see if we found our result scanning the vector (see next);
In case one is not in the array, how do I find the wanted number? Your intuition is correct, if your vector is sorted it is easier: you can iterate over your data, and when you find an "hole" between two subsequent elements, then the value of the first element of the hole + 1 is you result
What do I do if the array contains 1 and has no holes? Well, you return the smallest element that is not in the array, so the last element + 1. You may notice that by checking if your "candidate" value (that is a number that shouldn't be returned, so <=0) has changed during the scanning;
Let's go to the code:
int solution(std::vector<int>& v){
int retVal=0;
std::sort(v.begin(), v.end());
for(int i=0; i<v.size()-1; i++){
if(v[i]>0 && v[i+1]>v[i]+1){
retVal=v[i]+1;
break;
}
}
if(retVal==0) {
if (v.back() > 0)
retVal = v.back() + 1;
else
retVal = 1;
}
return retVal;
}
As suggested you can use the standard library a little bit more, but I think this is reasonably simple and efficient.
Other note:
I think your assignment does not bother you with this, but I mention just for completeness. Most of the times you don't want a function to modify your parameters: you can pass the vector "by value" meaning that actually you pass a complete copy of your data, without touching the original one, or you can pass a const reference and create a copy inside the function.
What I'm trying to do
Return an integer, which would represent the number of numbers that I would need to add in order to make my array consecutive (once its sorted, first).
Sample input and output
Input: [6, 2, 3, 8].
Output: 3.
Reason: After sorting the vector to [2, 3, 6, 8], we would need 3 numbers in order to make the whole vector consecutive (4, 5, 7 need to be added to the array, hence returning 3).
Where am I at?
Took care of a base-case of 1 integer in the vector (return 0; nothing to make consecutive).
Created a counter, set to 0.
Sorted the vector, ascending.
Loop through the vector, to check:
If the difference between the value on the right and the current value != 1, then add 1 to the counter. This would mean that the next number is not consecutive, and therefore we need a number.
#include <iostream>
#include <algorithm>
#include <vector>
#include <cstdlib>
int makeArrayConsecutive2(std::vector<int> statues) {
if (statues.size() == 1) {
return 0;
}
int counter = 0;
std::sort(statues.begin(), statues.end());
for (int i = 0; i < statues.size(); i++) {
if (statues[i+1] - statues[i] != 1) {
counter += 1;
}
}
return counter;
}
So what's the issue?
4/5 of my test cases passed, except for this one:
Input: [5, 4, 6]
Expected: 0
Actual: 1
I'm having trouble understanding why this is not working for that one case. If everything else is working, how come its just this one case? I suspect its an index issue, but I tried playing around with -1 from my indexes, and still not working.
Thank you for your help.
The standard library already provides std::adjacent_difference, which can do the vast majority of the work for you:
#include <numeric>
#include <iostream>
#include <vector>
#include <algorithm>
int makeArrayConsecutive(std::vector<int> input)
{
std::sort(input.begin(), input.end());
std::vector<int> diffs;
std::adjacent_difference(input.begin(), input.end(), std::back_inserter(diffs));
std::transform(diffs.begin(), diffs.end(), diffs.begin(), [](int a) { return a - 1; });
int result = std::accumulate(diffs.begin() + 1, diffs.end(), 0);
return result;
}
int main()
{
std::cout << makeArrayConsecutive({ 6, 2, 3, 8 }) << "\n";
std::cout << makeArrayConsecutive({ 5, 4, 6 });
}
Result (as expected):
3
0
Note: I've left each operation separate for clarity. You could (for one example) pretty easily eliminate the std::transform by using a functor with std::accumulate instead:
int makeArrayConsecutive(std::vector<int> input)
{
std::sort(input.begin(), input.end());
std::vector<int> diffs;
std::adjacent_difference(input.begin(), input.end(), std::back_inserter(diffs));
int result = std::accumulate(diffs.begin() + 1, diffs.end(),
0,
[](int total, int val) { return total + val - 1; });
return result;
}
Likewise, you could eliminate the extra storage for the differences by writing the differences back into the input array, which could give a substantial reduction in storage requirements if the input was very big:
int makeArrayConsecutive(std::vector<int> input)
{
std::sort(input.begin(), input.end());
std::adjacent_difference(input.begin(), input.end(), input.begin());
int result = std::accumulate(input.begin() + 1, input.end(),
0,
[](int total, int val) { return total + val - 1; });
return result;
}
...but if you're just trying to figure out what's going on, I think the first version is probably the simplest to follow.
These are the corrected lines:
for (int i = 0; i < statues.size()-1; i++) {
if (statues[i+1] - statues[i] != 1) {
counter += statues[i+1] - statues[i] -1;
}
}
I am currently struggling with a homework problem for my Algorithms Class. A summary of the instruction:
The user enters an integer 'n' to determine the number of test cases.
The user individually enters another integer 'num' to determine the # of elements in each test case.
The user enters the elements of the individual array.
The algorithm has to process the array and determine whether it can be partitioned into two subsequences, each of which is in strictly increasing order. If the result is positive, the program prints "Yes", otherwise it prints "No".
I have 24 hours to complete this assignment but am struggling with the primary problem - I cannot properly process the user input. (come up with an algorithm to split the two subsequences)
update: I got to this solution. It passes 4/5 tests but fails the time constraint in the last test.
#include<iostream>
#include<string>
using namespace std;
bool run(){
int numbers;
int *arr;
cin >> numbers;
arr = new int[numbers];
for (int i = 0; i < numbers; i++)
cin >> arr[i];
long long int MAX = 0;
long long int MAX2 = 0;
string stra = "";
string strb = "";
string result = "";
string total = "";
long long int sum = 0;
for (int i = 0; i < numbers; i++){
if (arr[i] >= MAX && arr[i] != arr[i - 1]){
stra += to_string(arr[i]);
MAX = arr[i];
}
else
if (arr[i] >= MAX2 && MAX2 != MAX){
strb += to_string(arr[i]);
MAX2 = arr[i];
}
}
for (int i = 0; i < numbers; i++){
result = to_string(arr[i]);
total += result;
}
long long int len1 = stra.length();
long long int len2 = strb.length();
sum += len1 + len2;
delete[] arr;
if (sum != total.length())
return false;
else
return true;
}
int main()
{
int test;
cin >> test;
while (test > 0)
{
if (run())
cout << "Yes\n";
else
cout << "No\n";
test--;
}
system("pause");
}
Example input:
2
5
3 1 5 2 4
5
4 8 1 5 3
Example output:
Yes
No
Explanation: For the array 3 1 5 2 4, the two strictly increasing subsequences are: 3 5 and 1 2 4.
It seems that the existence of any equal or decreasing subsequence of at least three elements means the array cannot be partitioned into two subsequences, each with strictly increasing order, since once we've placed the first element in one part and the second element in the other part, we have no place to place the third.
This seems to indicate that finding the longest decreasing or equal subsequence is a sure solution. Since we only need one of length 3, we can record in O(n) for each element if it has a greater or equal element to the left. Then perform the reverse. If any element has both a greater or equal partner on the left and a smaller or equal partner on the right, the answer is "no."
We can visualise the O(n) time, O(1) space method by plotting along value and position:
A choosing list B here
A x would be wrong
x
value B z
^ B x
| x
| A
| x
|
| B
| x
- - - - - - - -> position
We notice that as soon as a second list is established (with the first decrease), any element higher than the absolute max so far must be assigned to the list that contains it, and any element lower than it can, in any case, only be placed in the second list if at all.
If we were to assign an element higher than the absolute max so far to the second list (that does not contain it), we could arbitrarily construct a false negative by making the next element lower than both the element we just inserted into the second list and the previous absolute max, but greater than the previous max of the second list (z in the diagram). If we had correctly inserted the element higher than the previous absolute max into that first list, we'd still have room to insert the new, arbitrary element into the second list.
(The JavaScript code below technically uses O(n) space in order to show the partition but notice that we only rely on the last element of each part.)
function f(A){
let partA = [A[0]];
let partB = [];
for (let i=1; i<A.length; i++){
if (A[i] > partA[partA.length-1])
partA.push(A[i]);
else if (partB.length && A[i] <= partB[partB.length-1])
return false;
else
partB.push(A[i]);
}
return [partA, partB];
}
let str = '';
let examples = [
[30, 10, 50, 25, 26],
[3, 1, 5, 2, 4],
[4, 8, 1, 5, 3],
[3, 1, 1, 2, 4],
[3, 4, 5, 1, 2],
[3, 4, 1],
[4, 1, 2, 7, 3]
];
for (e of examples)
str += JSON.stringify(e) + '\n' + JSON.stringify(f(e)) + '\n\n';
console.log(str);
I would go over the entire array once and check two maximal values. If the actual array value is smaller than both maxima, it is not possible, otherwise the proper maximum is increased.
The algorithm does not have to traverse the whole array, if the split condition is violated before.
Here is my code
#include <algorithm>
#include <iostream>
#include <vector>
bool isAddable(const int item, int &max1, int &max2) {
if (max2 > item) {
return false;
}
else {
if (max1 > item) {
max2 = item;
}
else {
max1 = item;
}
return true;
}
}
void setStartValue(int &max1, int &max2, const std::vector<int> &vec) {
max1 = *std::min_element(vec.begin(), vec.begin() + 3);
max2 = *std::max_element(vec.begin(), vec.begin() + 3);
}
bool isDiviableIntoTwoIncreasingArrays(const std::vector<int> &vec) {
if (vec.size() < 3) {
return true;
}
int max1, max2;
setStartValue(max1, max2, vec);
for (int i = 2; i < vec.size(); ++i) {
if (max1 > max2) {
if (!isAddable(vec[i], max1, max2)) {
return false;
}
}
else {
if (!isAddable(vec[i], max2, max1)) {
return false;
}
}
}
return true;
}
int main() {
std::vector<int> userVec;
int tmp1;
while (std::cin >> tmp1) {
userVec.emplace_back(tmp1);
}
const std::vector<int> v1{3, 1, 5, 2, 4};
const std::vector<int> v2{4, 8, 1, 5, 3};
const std::vector<int> v3{3, 4, 1};
for (const std::vector<int> &vec : {userVec, v1, v2, v3}) {
if (isDiviableIntoTwoIncreasingArrays(vec)) {
std::cout << "Yes\n";
}
else {
std::cout << "No\n";
}
}
}
I think you could resort to using a brute force solution. Notice here I use vectors(I think you should as well) to store the data and I use recursion to exhaust out all possible combinations. Keep the problem in mind, solve it and then focus on trivial tasks like parsing the input and matching the way your coursework expects you to enter data. I have added inline comments to make this understandable.
bool canPartition(vector<int>& nums) {
if(nums.empty()) return false;
vector<int> part1 = {}, part2 = {}; // two partitions
auto ans = canPart(nums, part1, part2, 0); // pass this to our recursive function
return ans;
}
bool canPart(vector<int>& nums, vector<int>& part1, vector<int>& part2, int i)
{
if(i >= nums.size()) // we are at the end of the array is this a solution?
{
if(!part1.empty() && !part2.empty()) // only if the partitions are not empty
{
//if you want you could print part1 and part2 here
//to see what the partition looks like
return true;
}
return false;
}
bool resp1empty = false, resp2empty = false, resp1 = false, resp2 = false;
if(part1.empty()) // first partition is empty? lets add something
{
part1.push_back(nums[i]);
resp1empty = canPart(nums, part1, part2, i + 1);
part1.pop_back(); // well we need to remove this element and try another one
}
else if(nums[i] > part1.back()) // first partition is not empty lets check if the sequence is increasing
{
part1.push_back(nums[i]);
resp1 = canPart(nums, part1, part2, i + 1);
part1.pop_back();
}
if(part2.empty()) // is partition two empty? lets add something
{
part2.push_back(nums[i]);
resp2empty = canPart(nums, part1, part2, i + 1);
part2.pop_back();
}
else if(nums[i] > part2.back()) // check if sequence is increasing
{
part2.push_back(nums[i]);
resp2 = canPart(nums, part1, part2, i + 1);
part2.pop_back();
}
//if any of the recursive paths returns a true we have an answer
return resp1empty || resp2empty || resp1 || resp2;
}
You can now try this out with a main function:
vector<int> v = {3,1,5,2,4};
cout << canPartition(v);
The key take away is make a small test case, add a few more non trivial test cases, solve the problem and then look into parsing inputs for other test cases
I think this comes down to whether you have an option for a number to appear in the first list or second list or not.
So, we will keep adding numbers to list 1 and if we can't add any element, we will make it as the start of the new list.
Let's say, we have both the lists going. If we come across an element to whom we can't add to any of the lists, we return false.
There does arise a situation where we could add an element to any of the 2 lists. In this scenario, we adopt a greedy approach as to add to which list.
We prepare an array of minimum values from the right. For example, for [30,10,50,25,26], we will have an array of minimums as [10,25,25,26,(empty here since last)].
Now, let's trace how we could divide them into 2 lists properly.
30 => List A.
10 => List B. (since you can't add it first list, so make a new one from here)
50 => List A.
Here, 50 applies to come after either 30 or 10. If we choose 10, then we won't be able to accommodate the next 25 in either of the 2 lists and our program would fail here itself, since our lists would look like [30] and [10,50]. However, we could continue further if we add 50 to 30 by checking for the minimum stored for it in our minimums array, which is 25.
25 => List B.
26 => List B.
So, our final lists are [30,50] and [10,25,26].
Time complexity: O(n), Space complexity: O(n) and you can print the 2 lists as well.
If we come across a sorted array which is strictly increasing, we return true for them anyway.
This question has been asked before but I cannot find it for C++.
If I have a vector and I have a starting number, does std::algorithm provide me a way to find the next highest missing number?
I can obviously write this in a nested loop, I just cant shake the feeling that I'm reinventing the wheel.
For example, given: vector foo{13,8,3,6,10,1,7,0};
The starting number 0 should find 2.
The starting number 6 should find 9.
The starting number -2 should find -1.
EDIT:
Thus far all the solutions require sorting. This may in fact be required, but a temporary sorted vector would have to be created to accommodate this, as foo must remain unchanged.
At least as far as I know, there's no standard algorithm that directly implements exactly what you're asking for.
If you wanted to do it with something like O(N log N) complexity, you could start by sorting the input. Then use std::upper_bound to find the (last instance of) the number you've asked for (if present). From there, you'd find a number that differs from the previous by more than one. From there you'd scan for a difference greater than 1 between the consecutive numbers in the collection.
One way to do this in real code would be something like this:
#include <iostream>
#include <algorithm>
#include <vector>
#include <numeric>
#include <iterator>
int find_missing(std::vector<int> x, int number) {
std::sort(x.begin(), x.end());
auto pos = std::upper_bound(x.begin(), x.end(), number);
if (*pos - number > 1)
return number + 1;
else {
std::vector<int> diffs;
std::adjacent_difference(pos, x.end(), std::back_inserter(diffs));
auto pos2 = std::find_if(diffs.begin() + 1, diffs.end(), [](int x) { return x > 1; });
return *(pos + (pos2 - diffs.begin() - 1)) + 1;
}
}
int main() {
std::vector<int> x{ 13, 8, 3, 6, 10, 1,7, 0};
std::cout << find_missing(x, 0) << "\n";
std::cout << find_missing(x, 6) << "\n";
}
This is somewhat less than what you'd normally think of as optimal to provide the external appearance of a vector that can/does remain un-sorted (and unmodified in any way). I've done that by creating a copy of the vector, and sorting the copy inside the find_missing function. Thus, the original vector remains unmodified. The disadvantage is obvious: if the vector is large, copying it can/will be expensive. Furthermore, this ends up sorting the vector for every query instead of sorting once, then carrying out as many queries as desired on it.
So I thought I'd post an answer. I don't know anything in std::algorithm that accomplishes this directly, but in combination with vector<bool> you can do this in O(2N).
template <typename T>
T find_missing(const vector<T>& v, T elem){
vector<bool> range(v.size());
elem++;
for_each(v.begin(), v.end(), [&](const T& i){if((i >= elem && i - elem < range.size())range[i - elem] = true;});
auto result = distance(range.begin(), find(range.begin(), range.end(), false));
return result + elem;
}
First you need to sort the vector. Use std::sort for that.
std::lower_bound finds the first element that is greater or equal with a given element. (the elements have to be at least partially ordered)
From there you iterate while you have consecutive elements.
Dealing with duplicates: One way is the way I went: consider consecutive and equal elements when iterating. Another approach is to add a prerequisite that the vector / range contains unique elements. I chose the former because it avoids erasing elements.
Here is how you eliminate duplicates from a sorted vector:
v.erase(std::unique(v.begin(), v.end()), v.end());
My implementation:
// finds the first missing element in the vector v
// prerequisite: v must be sorted
auto firstMissing(std::vector<int> const &v, int elem) -> int {
auto low = std::lower_bound(std::begin(v), std::end(v), elem);
if (low == std::end(v) || *low != elem) {
return elem;
}
while (low + 1 != std::end(v) &&
(*low == *(low + 1) || *low + 1 == *(low + 1))) {
++low;
}
return *low + 1;
}
And a generalized version:
// finds the first missing element in the range [first, last)
// prerequisite: the range must be sorted
template <class It, class T = decltype(*std::declval<It>())>
auto firstMissing(It first, It last, T elem) -> T {
auto low = std::lower_bound(first, last, elem);
if (low == last || *low != elem) {
return elem;
}
while (std::next(low) != last &&
(*low == *std::next(low) || *low + 1 == *std::next(low))) {
std::advance(low, 1);
}
return *low + 1;
}
Test case:
int main() {
auto v = std::vector<int>{13, 8, 3, 6, 10, 1, 7, 7, 7, 0};
std::sort(v.begin(), v.end());
for (auto n : {-2, 0, 5, 6, 20}) {
cout << n << ": " << firstMissing(v, n) << endl;
}
return 0;
}
Result:
-2: -2
0: 2
5: 5
6: 9
20: 20
A note about sorting: From the OP's comments he was searching for a solution that wouldn't modify the vector.
You have to sort the vector for an efficient solution. If modifying the vector is not an option you could create a copy and work on it.
If you are hell-bent on not sorting, there is a brute force solution (very very inefficient - O(n^2)):
auto max = std::max_element(std::begin(v), std::end(v));
if (elem > *max) {
return elem;
}
auto i = elem;
while (std::find(std::begin(v), std::end(v), i) != std::end(v)) {
++i;
}
return i;
First solution:
Sort the vector. Find the starting number and see what number is next.
This will take O(NlogN) where N is the size of vector.
Second solution:
If the range of numbers is small e.g. (0,M) you can create boolean vector of size M. For each number of initial vector make the boolean of that index true. Later you can see next missing number by checking the boolean vector. This will take O(N) time and O(M) auxiliary memory.