I am a new programmer and I am trying to sort a vector of integers by their parities - put even numbers in front of odds. The order inside of the odd or even numbers themselves doesn't matter. For example, given an input [3,1,2,4], the output can be [2,4,3,1] or [4,2,1,3], etc. Below is my c++ code, sometimes I got luck that the vector gets sorted properly, sometimes it doesn't. I exported the odd and even vectors and they look correct, but when I tried to combine them together it is just messed up. Can someone please help me debug?
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
unordered_multiset<int> even;
unordered_multiset<int> odd;
vector<int> result(A.size());
for(int C:A)
{
if(C%2 == 0)
even.insert(C);
else
odd.insert(C);
}
merge(even.begin(),even.end(),odd.begin(),odd.end(),result.begin());
return result;
}
};
If you just need even values before odds and not a complete sort I suggest you use std::partition. You give it two iterators and a predicate. The elements where the predicate returns true will appear before the others. It works in-place and should be very fast.
Something like this:
std::vector<int> sortArrayByParity(std::vector<int>& A)
{
std::partition(A.begin(), A.end(), [](int value) { return value % 2 == 0; });
return A;
}
Because the merge function assumes that the two ranges are sorted, which is used as in merge sort. Instead, you should just use the insert function of vector:
result.insert(result.end(), even.begin(), even.end());
result.insert(result.end(), odd.begin(), odd.end());
return result;
There is no need to create three separate vectors. As you have allocated enough space in the result vector, that vector can be used as the final vector also to store your sub vectors, storing the separated odd and even numbers.
The value of using a vector, which under the covers is an array, is to avoid inserts and moves. Arrays/Vectors are fast because they allow immediate access to memory as an offset from the beginning. Take advantage of this!
The code simply keeps an index to the next odd and even indices and then assigns the correct cell accordingly.
class Solution {
public:
// As this function does not access any members, it can be made static
static std::vector<int> sortArrayByParity(std::vector<int>& A) {
std::vector<int> result(A.size());
uint even_index = 0;
uint odd_index = A.size()-1;
for(int element: A)
{
if(element%2 == 0)
result[even_index++] = element;
else
result[odd_index--] = element;
}
return result;
}
};
Taking advantage of the fact that you don't care about the order among the even or odd numbers themselves, you could use a very simple algorithm to sort the array in-place:
// Assume helper function is_even() and is_odd() are defined.
void sortArrayByParity(std::vector<int>& A)
{
int i = 0; // scanning from beginning
int j = A.size()-1; // scanning from end
do {
while (i < j && is_even(A[i])) ++i; // A[i] is an even at the front
while (i < j && is_odd(A[j])) --j; // A[j] is an odd at the back
if (i >= j) break;
// Now A[i] must be an odd number in front of an even number A[j]
std::swap(A[i], A[j]);
++i;
--j;
} while (true);
}
Note that the function above returns void, since the vector is sorted in-place. If you do want to return a sorted copy of input vector, you'd need to define a new vector inside the function, and copy the elements right before every ++i and --j above (and of course do not use std::swap but copy the elements cross-way instead; also, pass A as const std::vector<int>& A).
// Assume helper function is_even() and is_odd() are defined.
std::vector<int> sortArrayByParity(const std::vector<int>& A)
{
std::vector<int> B(A.size());
int i = 0; // scanning from beginning
int j = A.size()-1; // scanning from end
do {
while (i < j && is_even(A[i])) {
B[i] = A[i];
++i;
}
while (i < j && is_odd(A[j])) {
B[j] = A[j];
--j;
}
if (i >= j) break;
// Now A[i] must be an odd number in front of an even number A[j]
B[i] = A[j];
B[j] = A[i];
++i;
--j;
} while (true);
return B;
}
In both cases (in-place or out-of-place) above, the function has complexity O(N), N being number of elements in A, much better than the general O(N log N) for sorting N elements. This is because the problem doesn't actually sort much -- it only separates even from odd. There's therefore no need to invoke a full-fledged sorting algorithm.
Related
I have to write a function which accepts an int array parameter and checks to see if it is a
permutation.
I tried this so far:
bool permutationChecker(int arr[], int n){
for (int i = 0; i < n; i++){
//Check if the array is the size of n
if (i == n){
return true;
}
if (i == arr[n]){
return true;
}
}
return false;
}
but the output says some arrays are permutations even though they are not.
When you write i == arr[n], that doesn't check whether i is in the array; that checks whether the element at position n is i. Now, that's even worse here, as the array size is n, so there's no valid element at position n: it's UB, array is overindexed.
If you'd like to check whether i is in the array, you need to scan each element of the array. You can do this using std::find(). Either that, or you might sort (a copy of) the array, then check if i is at position i:
bool isPermutation(int arr[], int n){
int* arr2 = new int[n]; // consider using std::array<> / std::vector<> if allowed
std::copy(arr, arr + n, arr2);
std::sort(arr2, arr2 + n);
for (int i = 0; i < n; i++){
if (i != arr2[i]){
delete[] arr2;
return false;
}
}
delete[] arr2;
return true;
}
One approach to checking that the input contains one of each value is to create an array of flags (acting like a set), and for each value in your input you set the flag to true for the corresponding index. If that flag is already set, then it's not unique. And if the value is out of range then you instantly know it's not a permutation.
Now, you would normally expect to allocate additional data for this temporary set. But, since your function accepts the input as non-constant data, we can use a trick where you use the same array but store extra information by making values negative.
It will even work for all positive int values, since as of C++20 the standard now guarantees 2's complement representation. That means for every positive integer, a negative integer exists (but not the other way around).
bool isPermutation(int arr[], int n)
{
// Ensure everything is within the valid range.
for (int i = 0; i < n; i++)
{
if (arr[i] < 1 || arr[i] > n) return false;
}
// Check for uniqueness. For each value, use it to index back into the array and then
// negate the value stored there. If already negative, the value is not unique.
int count = 0;
while (count < n)
{
int index = std::abs(arr[count]) - 1;
if (arr[index] < 0)
{
break;
}
arr[index] = -arr[index];
count++;
}
// Undo any negations done by the step above
for (int i = 0; i < count; i++)
{
int index = std::abs(arr[i]) - 1;
arr[index] = std::abs(arr[index]);
}
return count == n;
}
Let me be clear that using tricky magic is usually not the kind of solution you should go for because it's inevitably harder to understand and maintain code like this. That should be evident simply by looking at the code above. But let's say, hypothetically, you want to avoid any additional memory allocation, your data type is signed, and you want to do the operation in linear time... Well, then this might be useful.
A permutation p of id=[0,...,n-1] a bijection into id. Therefore, no value in p may repeat and no value may be >=n. To check for permutations you somehow have to verify these properties. One option is to sort p and compare it for equality to id. Another is to count the number of individual values set.
Your approach would almost work if you checked (i == arr[i]) instead of (i == arr[n]) but then you would need to sort the array beforehand, otherwise only id will pass your check. Furthermore the check (i == arr[n]) exhibits undefined behaviour because it accesses one element past the end of the array. Lastly the check (i == n) doesn't do anything because i goes from 0 to n-1 so it will never be == n.
With this information you can repair your code, but beware that this approach will destroy the original input.
If you are forced to play with arrays, perhaps your array has fixed size. For example: int arr[3] = {0,1,2};
If this were the case you could use the fact that the size is known at compile time and use an std::bitset. [If not use your approach or one of the others given here.]
template <std::size_t N>
bool isPermutation(int const (&arr)[N]) {
std::bitset<N> bits;
for (int a: arr) {
if (static_cast<std::size_t>(a) < N)
bits.set(a);
}
return bits.all();
}
(live demo)
You don't have to pass the size because C++ can infer it at compile time. This solution also does not allocate additional dynamic memory but it will get into problems for large arrays (sat > 1 million entries) because std::bitset lives on automatic memory and therefore on the stack.
i am trying to write a code that will delete all the elements if an array has same element at different index . it works fine for one element deletion or elements at odd index i.e 1,3,5 etc but it neglects one element if the consecutive index have same element.
i have just tried this to get my hands on arrays
for(int i=0;i<n;i++) //for deletion
{
if(arr[i]==_delete)
{
arr[i]=arr[i+1];
--n;
}
}
I suggest you use std::vector as a container for your objects.
std::vector<TYPE> vec ;
// initialise vector
You can use
vec.erase(std::remove_if(vec.begin(), vec.end(),
[](const auto & item){return item == _delete;}), vec.end());
Alternatively, you can use std::list. Its list::erase has linear time complexity.
As an additional solution, if you want to deal with built-in C++ arrays, the standard std::remove algorithm can be rewritten like this:
void remove(int _delete) {
int j = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] != _delete) {
arr[j++] = arr[i];
}
}
// update the size!
n = j;
}
It's quite pretty:
We keep in the array the elements we only need, and override the ones in which we are not interested (they can be either equal or not to _delete and start at position j till the end)
The question ask for a function that takes an array of integers pre-filled with random elements. The function goes through the array linearly and if any element is equal to any of the preceding elements, regenerate that element. This goes on until the whole array is made of unique elements.
I know that this is very inefficient way of generating an array with unique elements but I wanted to give it a try. I wrote this code which falls into an infinite loop.
void makeUnique(int* const objArr, const int& size)
{
int i = 1,j = 0;
do {
j = 0;
while (j < i) {
if (objArr[j] == objArr[i]) {
objArr[i] = rand();
--i;
break;
}
++j;
}
++i;
} while (i<size);
}
Where am I going wrong?
I have an array of structs where I keep track of how many times each unique word was seen in a given text:
struct List {
char word[20];
int repeat;
};
Now I need to sort this:
as 6
a 1
appetite 1
angry 1
are 2
and 4
...
To this:
a 1
as 6
and 4
are 2
angry 1
appetite 1
...
(By alphabetically I mean only by first letter)
So far, I have come up with this:
for (i = 0; i < length - 1; i++) {
min_pos = i;
for (j = i + 1; j < length; j++) // find min
if (array[j].word[0] < array[min_pos].word[0]) {
min_pos = j;
}
swap = array[min_pos]; // swap
array[min_pos] = array[i];
array[i] = swap;
}
This code works perfectly for sorting alphabetically, but I just can't write proper code to sort BOTH alphabetically and by length.
Make a comparator function.
Add an operator< to your List:
bool operator<(const List &lhs) const {
if(word[0] != lhs.word[0]) {
return word[0] < lhs.word[0];
}
return strlen(word) < strlen(lhs.word);
}
And now use this operator to sort, using whichever algorithm strikes your fancy.
Others have pointed out that there are faster and cleaner ways to sort. But if you want to use your own selection sort, as you've written, then you just need to make a few changes to your code.
Separate the "do I need to swap" logic from the swapping logic itself. Then the code becomes much cleaner and it's more clear where to add the extra check.
I've only copied the inner loop here. You'd want to replace your existing inner loop with this one. I'm not clear on why you need swap_pos and min_pos, so I've left the semantics alone.
for (j = i + 1; j < length; j++) { // find min
// first, determine whether you need to swap
// You want to swap if the first character of the new word is
// smaller, or if the letters are equal and the length is smaller.
bool doSwap = false;
if (array[j].word[0] < array[min_pos].word[0]) {
doSwap = true;
}
else if (array[j].word[0] == array[min_pos].word[0] &&
strlen(array[j].word) < array[min_pos].word) {
doSwap = true;
}
// do the swap if necessary
if (doSwap) {
swap_pos = j;
swap = array[min_pos]; // swap
array[min_pos] = array[i];
array[i] = swap;
}
}
To more clearly illustrate the necessary logic changes, I've purposely avoided making major style changes or simple optimizations.
You can pass a lambda to sort to do this:
sort(begin(array), end(array), [](const auto& lhs, const auto& rhs){ return *lhs.word < *rhs.word || *lhs.word == *rhs.word && (strlen(lhs.word) < strlen(rhs.word) || strlen(lhs.word) == strlen(rhs.word) && strcmp(lhs.word, rhs.word) < 0); });
Live Example
Use tuple lexicographical compare operators
An easy way to not write this condition is to
#include <tuple>
Then std::tie can be used:
std::tie(array[j].word[0], array[j].repeat) < std::tie(array[min_pos].word[0], array[min_pos].repeat)
This works because std::tie creates a tuple of lvalue references to its arguments. (Which means std::tie requires variables. If You want to compare results from functions std::make_tuple or std::forward_as_tuple would be better)
And std::tuple has operators which
Compares lhs and rhs lexicographically, that is, compares the first elements, if they are equivalent, compares the second elements, if those are equivalent, compares the third elements, and so on.
And the above description is also the idea how to make a comparison of more than value.
I am trying to program the Sieve of Eratosthenes, but I am not sure how to delete elements from the vector I made given a specific condition. Does anyone know how to achieve this? Here is my code:
#include <iostream>
#include <vector>
using namespace std;
int prime(int n);
int prime(int n)
{
vector<int> primes;
for(int i = 2; i <= n; i++)
{
primes.push_back(i);
int t = i % (i + 1);
if(t == 0)
{
delete t; // is there a way of deleting the elements from
// the primes vector that follow this condition t?
}
cout << primes[i] << endl;
}
}
int main()
{
int n;
cout << "Enter a maximum numbers of primes you wish to find: " << endl;
cin >> n;
prime(n);
return 0;
}
Your algorithm is wrong:
t = i % (i + 1);
is
i
which is always != 0 because i is larger than 1.
By the way if you absolutely want to remove the t-th element you have to be sure that the vector is not empty and then you do:
primes.erase(primes.begin()+t);
Even if you fix the algorithm your approach is inefficient: erasing an element in the middle of a vector means copying back of one position all the ones following the erased element.
You don't usually want to delete elements in the middle of a Sieve of Eratosthenes, but when you do want to, you usually want to use the remove/erase idiom:
x.erase(std::remove_if(x.begin(), x.end(), condition), x.end());
std::remove basically just partitions the collection into those that don't meet the specified condition, followed by objects that may have been used as the source of either a copy or a move, so you can't count on their value, but they are in some stable state so erasing them will work fine.
The condition can be either a function or a functor. It receives (a reference to a const) object that it examines and determines whether it lives or dies (so to speak).
Find here a c++ pseudocode for the sieve algorithm. Once you've understood the algorithm you can start working on this.
primes(vector& primes, size_t max){
vector primesFlag(1,max);
i=1
while(i*i<max){
++i;
for(j=i*i; j < max; j+= i){
primesFlag[j] = 0;
}
}
primes.clear()
primes.reserve(...);
for(j >= 2;
if primesFlag[j] = 1
primes.push_back(j);
}