Related
I'm trying to sort the elements in a vector from smallest to biggest.
My algorithm:
Find the smallest number in the vector using least()
Find all occurrences of the smallest number in the vector and copy them into another vector named result
Delete those occurrences from the original vector
Repeat until all elements from the original vector have been deleted
As the title suggests, I'm getting vector subscript out of range error. I tried debugging by printing out the indexes, but they seem correct.
An example you could input could be:
5
5 5 4 3 2 1 2
The program should output: 1 2 2 3 4 5 5
Here's my code:
#include <iostream>
#include <vector>
#include <string>
using namespace std;
int least(vector<int> groups) {
int lowest = groups[0];
for (int i = 0; i < groups.size(); i++) {
//cout << i << endl;
if (groups[i] < lowest) {
lowest = groups[i];
}
}
return lowest;
}
vector<int> sort(vector<int> groups) {
vector<int> result;
while (groups.size() != 0) {
int smallest = least(groups);
//cout << smallest << endl;
for (int i = 0; i < groups.size(); i++) {
if (groups[i] == smallest) {
result.push_back(smallest);
}
//cout << i + " " + groups.size();
}
groups.erase(remove(groups.begin(), groups.end(), smallest), groups.end());
}
return result;
}
int main() //sort from least to greatest
{
int t, n, taxiCount = 0, numOfChildren = 0;
cin >> t;
vector<int> groups, result;
for (int i = 0; i < t; i++) {
cin >> n;
groups.push_back(n);
}
result = sort(groups);
//cout << result.size();
for (int i : result) {
cout << i + " ";
}
}
I know there are better ways to sort through a vector, but I'd like to stick to my way for now.
I am doing this particular exercise where I have to implement the Binary Search algorithm which returns the index of the first occurence of an element in a sorted array, if it contains duplicates. Since I am primarily working on my algorithmic skills in C++, I am only trying to do it in C++. Here is my code:
#include <iostream>
#include <cassert>
#include <vector>
using std::vector;
int binary_search(const vector<int> &a, int x, int n) {
int left = 0, right = n-1;
while(left<= right){
int mid = left + (right-left)/2;
if(x== a[mid]){
return mid;
}else if(a[mid]>x){
right = mid-1;
}else{
left = mid+1;
}
}
return -1;
}
int first_occurence(const vector<int>&a, int x, int n) {
int out = binary_search(a, x, n);
if(out !=-1){
for(int i = out;i>0&& a[i]==x;--i ){
out = i;
}
}
return out;
}
int main() {
int n;
std::cin >> n;
vector<int> a(n);
for (size_t i = 0; i < a.size(); i++) {
std::cin >> a[i];
}
int m;
std::cin >> m;
vector<int> b(m);
for (int i = 0; i < m; ++i) {
std::cin >> b[i];
}
for (int i = 0; i < m; ++i) {
std::cout << first_occurence(a, b[i], n) << ' ';
}
}
The first input to the program tells how many items the array should contain, the second is the enumeration of these elements, third line tells how many keys to search for and the final line are the individual keys. The output is the indices for the key or -1 when no such key is found.
My strategy is to use a function to find the index of a key. If it is found, then in case of duplicates, the first occurrence has to have a lower index. That is what the first_occurence() method does; keep looping back till the first occurence is found.
For the following input:
10
1 5 4 4 7 7 7 3 2 2
5
4 7 2 0 6
The output is:
-1 4 -1 -1 -1
Which is only correct for the key 7. I have been trying to debug this for quite some time but I can not figure out the problem.
returns the index of the first occurence of an element in a sorted array,
Your binary search algorithm requires that the data is sorted before you call it.
Example:
#include <algorithm>
#include <sstream>
int main() {
std::istringstream in(R"aw(10
1 5 4 4 7 7 7 3 2 2
5
4 7 2 0 6
)aw");
int n;
in >> n;
vector<int> a(n);
for (auto& v : a) {
in >> v;
}
std::sort(a.begin(), a.end()); // <- add this
// display the sorted result:
for (auto v : a) std::cout << v << ' ';
std::cout << '\n';
int m;
in >> m;
vector<int> b(m);
for (auto& v : b) {
in >> v;
}
for (auto v : b) {
std::cout << v << ' ' << first_occurence(a, v, n) << '\n';
}
}
I need some help, I know this question was asked before but I don't get it and I cant solve it, so I need help. I need to move the elements of my array to a position to left. So if the input will be 1,2,3,4,5 then the output will be 2,3,4,5,1. I have done the same to right but to left I cant figure it out, please also explain the logic , thanks.
#include <iostream>
using namespace std;
int a[100],n,i,tempr,templ;
int main()
{
cin>>n;
for(i=1;i<=n;i++) cin >> a[i];
for(i=1;i<=n;i++)
{
tempr = a[n];
a[n] = a[i];
a[i] = tempr;
cout<<"Right: "<<a[i]<<endl;
}
for(i=1;i<=n;i++)
{
templ = a[2];
a[2] = a[i];
a[i] = templ;
cout<<"Left: "<<a[i]<<endl;
}
return 0;
}
Please help!
First problem is bad indexing:
for(i=1;i<=n;i++) cin >> a[i]; //wrong logic, C++ indexing start from 0
Correct approach:
for(i=0;i<n;i++) //all your loops
Second problem is wrong logic for shifting elements:
Corrected version:
//input example: 1 2 3 4 5
//to the left
int temp = a[0]; //remember first element
for(i=0;i<n-1;i++)
{
a[i] = a[i+1]; //move all element to the left except first one
}
a[n-1] = temp; //assign remembered value to last element
//output: 2 3 4 5 1
cout << "To left: " << endl;
for(i=0;i<n;i++)
cout << a[i] << endl;
//to the right
temp = a[n-1]; //remember last element
for(i=n-1;i>=0;i--)
{
a[i+1] = a[i]; //move all element to the right except last one
}
a[0] = temp; //assign remembered value to first element
//output: 1 2 3 4 5 because elements are shifted back by right shift
cout << "To right: " << endl;
for(i=0;i<n;i++)
cout << a[i] << endl;
EDIT:
How to display both shifts:
#include <iostream>
using namespace std;
int to_left[5], to_right[5],n,i,tempr,templ;
int main()
{
cout << "Input array size: ";
cin >> n;
for(i=0;i<n;i++)
{
cin >> to_left[i]; //read values to first array
to_right[i]=to_left[i]; //then copy values to second one
}
//shift first array to left
int temp = to_left[0];
for(i=0;i<n-1;i++)
{
to_left[i] = to_left[i+1]; //move all element to the left except first one
}
to_left[n-1] = temp; //assign remembered value to last element
//output: 2 3 4 5 1
cout << "To left: " << endl;
for(i=0;i<n;i++)
cout << to_left[i] << endl;
//shift second array to right
temp = to_right[n-1]; //remember last element
for(i=n-1;i>=0;i--)
{
to_right[i+1] = to_right[i]; //move all element to the right except last one
}
to_right[0] = temp; //assign remembered value to first element
//output: 1 2 3 4 5 because elements are shifted back by right shift
cout << "To right: " << endl;
for(i=0;i<n;i++)
cout << to_right[i] << endl;
return 0;
}
Note that your code look very much like C code. In C++, you can declare variables in any segment of code, not just at the beginning. In C++, you can declare variable in for loop like this: for(int i=0; i<...) - no need for global variable i
For reference, this would be good C++ code example that satisfies problem you are facing:
#include <iostream>
#include <vector>
int main()
{
std::size_t n; //size_t is unsiged type used for various sizes of containers or types
std::cout << "Input array size: ";
std::cin >> n;
std::vector<int> to_left(n), to_right(n); //two dynamic arrays containing integers, takin n as their size
for(std::size_t i=0;i<to_left.size();++i) //use vector size(), instead of n, also ++i in considered better for loops that i++ (may be faster)
{
std::cin >> to_left[i];
to_right[i]=to_left[i];
}
int temp = to_left[0]; //declare temp here, not at the begining of code
for(std::size_t i=0;i<n-1;++i)
to_left[i] = to_left[i+1];
to_left[n-1] = temp;
std::cout << "To left: " << std::endl;
for(std::size_t i=0;i<n;++i)
std::cout << to_left[i] << std::endl;
temp = to_right[n-1]; //reuse temp
for(int i=to_right.size()-1;i>=0;--i) //note int, not std::size_t, because size_t is always >=0, loop would never end.
to_right[i+1] = to_right[i];
to_right[0] = temp;
std::cout << "To right: " << std::endl;
for(std::size_t i=0;i<n;i++)
std::cout << to_right[i] << std::endl;
return 0;
}
And here would be ideal C++ code:
#include <iostream>
#include <vector>
#include <algorithm>
int main()
{
std::size_t n;
std::cout << "Input array size: ";
std::cin >> n;
std::vector<int> to_left(n), to_right(n);
for(std::size_t i=0;i<to_left.size();++i)
{
std::cin >> to_left[i];
to_right[i]=to_left[i];
}
// rotate first array to the left
std::rotate(to_left.begin(), to_left.begin() + 1, to_left.end());
// rotate second array to right
std::rotate(to_right.rbegin(), to_right.rbegin() + 1, to_right.rend());
std::cout << "To left:" << std::endl;
for(auto x : to_left) //C++11 feature, x iterates through container
std::cout << x << std::endl;
std::cout << "To right:" << std::endl;
for(auto x : to_right)
std::cout << x << std::endl;
return 0;
}
Or you can use memmove(...) projected exactly for those purpose, here your sample:
#include <iostream>
#include <cstring>
using namespace std;
//rotate Left
void r_left(int *a,int n)
{
int tmp=a[0];
memmove(a,a+1,sizeof(int)*(n-1));
a[n-1]=tmp;
}
//rotate right
void r_right(int *a,int n)
{
int tmp=a[n-1];
memmove(a+1,a,sizeof(int)*(n-1));
a[0]=tmp;
}
void show(int *a,int n)
{
while(n--)
cout<<*a++<<' ';
cout<<endl;
}
int main()
{
int ar[]={1,2,3,4,5};
int n=sizeof(ar)/sizeof(ar[0]);
r_left(ar,n);
show(ar,n);
r_right(ar,n);
show(ar,n);
return 0;
}
easiest way to swap elements in C++ is to use std::iter_swap()
so for an array of 4 elements to swap elements 1 and 4 you would do the following
int a[4];
std::iter_swap(a, a+3);
note that you also need to #include <algorithm> for this to work
the basic logic of the function is that you give the location in memory of the 2 elements, so as the first element of an array is also its location in memory, you can pass a + n, when n is equal to the n-1 index number of the element you want to swap
As other already have stated it's all about indices. In a for-loop you are almost always in trouble if your stop condition is i <= size, because arrays in C++ are zero-indexed.
Where Black Moses alogrithm is far the easiest to understand (and probably the fastes), I read your code as if you try to swap the first value of the array through the array to the last position. Below I have tried to pin out this approach.
#include <stdio.h>
#include <tchar.h>
#include <iostream>
void ShiftLeft(int* pArr, size_t length)
{
for (size_t i = 1; i < length; i++)
{
int tmp = pArr[i - 1]; // Preserves the previous value
pArr[i - 1] = pArr[i]; // Overwrites the previous position with the current value
pArr[i] = tmp; // Stores the previous value in the current position
// All in all the first value is swapped down the array until it is at the length - 1 position
// and all the other values are swapped to the left.
/* For an array with 4 values the progression is as follows:
i = 0: 1 2 3 4
i = 1: 2 1 3 4
i = 2: 2 3 1 4
i = 3: 2 3 4 1
*/
}
}
void ShiftRight(int* pArr, size_t length)
{
for (size_t i = length - 1; i > 0; i--)
{
// This code does exactly the same as for ShiftLeft but the loop is running backwards
int tmp = pArr[i - 1];
pArr[i - 1] = pArr[i];
pArr[i] = tmp;
}
}
void Print(int* pArr, size_t length)
{
for (size_t i = 0; i < length; i++)
{
std::cout << pArr[i] << " ";
}
std::cout << std::endl;
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
size_t length = sizeof(arr) / sizeof(arr[0]);
Print(arr, length);
ShiftLeft(arr, length);
Print(arr, length);
ShiftRight(arr, length);
Print(arr, length);
return 0;
}
#include <iostream>
using namespace std;
int a[100], outR[100], outL[100], n, i;
int main() {
cin >> n;
for (i = 0; i < n; i++) cin >> a[i];
// Right
for (i = 0; i < n; i++) {
outR[i+1]= a[i];
}
outR[0] = a[n-1]; // add first number
// Left
for (i = 1; i < n; i++) {
outL[i-1]= a[i];
}
outL[n-1] = a[0]; // add last number
// Answer
cout << "Right:\n";
for(i=0; i<n; i++) {
cout << outR[i] << endl;
}
cout << "Left:\n";
for(i = 0; i < n; i++) {
cout << outL[i] << endl;
}
return 0;
}
Simple answer where you can easily see everything, good luck.
You may be interested in ,,vector coding", it seems be easier if you spend some time on this:
#include <iostream>
#include <vector>
using namespace std;
vector <int> a, outR, outL;
size_t i;
int main () {
int n, temp_int;
cin >> n;
while (n--) {
cin >> temp_int; // here you read number to your vector
a.push_back(temp_int); // here you add this to vector
// remember that vector start from element 0 as like arrays
}
// Left
// remember that last element will be first
// you may have acces to size of your vector easily
for (i = 0; i < (a.size()-1); i++) {
outL.push_back(a.at(i+1)); // here you create new vector
}
outL.push_back(a.at(0)); // add last elemet which rotated
// Right
// to rotate left first you have push last element so
outR.push_back(a.at(a.size()-1)); // add first elemet which rotated
for (i = 1; i < a.size(); i++) {
outR.push_back(a.at(i-1)); // here you push rest
}
cout << "Left" << "\n";
for (i = 0; i < a.size(); i++) {
cout << outL.at(i) << endl; // here you print value
}
cout << "Right" << "\n";
for (i = 0; i < a.size(); i++) {
cout << outR.at(i) << endl; // here you print value
}
return 0;
}
int* leftShiftOneByOneWIthoutTemp(int arr[], int sz)
{
for (int i=0 ;i < sz-1; i++)
{
arr[i] = arr[sz-1] + arr[i];
arr[sz-1] = arr[i] - arr[sz-1] ;
arr[i] = arr[i] - arr[sz-1] ;
std::cout << "iter "<< i << std::endl;
printArray(arr,5);
}
std::cout << "final "<< std::endl;
printArray(arr,5);
return arr;
}
Replace your code (to shift array left) with below code.
templ = a[0];
for(i=0;i<n-1;i++)
{
a[i] = a[i+1];
cout<<"Left: "<<a[i]<<endl;
}
a[n-1] = templ;
cout<<"Left: "<<a[n-1]<<endl;
So I have written a small program that calculates A[0]*B[0] - A[1]*B[1] - A[2]*B[2] - ... - A[n-1]*B[n-1] using vectors.
#include <iostream>
#include <string>
#include <vector>
using namespace std;
double CalcProduct(vector<double> v1,vector<double> v2){
size_t size; int i = 0;
vector<double> product;
double input, input2;
cout << "Enter vector size: ";
cin >> size;
cout << "Enter values for first vector: ";
while (i++ != size) {
cin >> input;
v1.push_back(input);
}
i = 0;
cout << "Enter values for second vector: ";
while (i++ != size) {
cin >> input2;
v2.push_back(input2);
}
double result = 0.0;
if (size >= 1) {
result += v1[0] * v2[0];
for (int i = 1; i < size; ++i)
result -= v1[i] * v2[i];
}
cout << "\nThe 'happy' product of both vectors is: " << result << endl;
}
int main()
{
vector<double> v1;
vector<double> v2;
CalcProduct(v1,v2);
return 0;
}
However, what would happen if I want to compute A[1]*B[1] - A[3]*B[3] - A[5]*B[5] - ....? I tried adding +1 to v1 and v2 before they sum up in result but my logic seems to be wrong. Sample input:
5
1 2 3 4 5
4 5 6 7 5
Expected output: -18
You're doing the calculation in this loop
for(int i = 1; i < size; ++i)
result -= v1[i] * v2[i];
so instead of incrementing i, you can make jumps of 2:
for(int i = 1; i < size; i+=2) {
result -= v1[i] * v2[i];
}
Note that it's a good practice to check that you're not running out of bounds, also I like to surround body of for loop with { and } even if it contains one line.
Edit: I didn't note that you already added the first element before the loop, you can easily solve this by starting from i=3 instead of 1.
First of all, calcProduct is doing three things at once: Reading the values, calculating the product and writing the result to the output. Hence, you should split that into two methods and maybe do the printout from main().
Secondly, once you have generated the vectors, you can use vector.size() to avoid out of bounds situations by checking loop counters against it.
// give vector as reference
void ReadInput(vector<double>& v1,vector<double>& v2)
{
size_t size = 0; // use size_t here as this will be well defined
size_t i = 0; // on 32 and 64 bit architectures
// maybe you should have more error handling here as
// one might desire size to be in some application
// specific range
cout << "Enter vector size: ";
cin >> size;
cout << "Enter values for first vector: ";
while (i++ != size)
{
cin >> input;
v1.push_back(input);
}
i = 0;
cout << "Enter values for second vector: ";
while (i++ != size)
{
cin >> input2;
v2.push_back(input2);
}
}
// giving the parameters as (const) references avoids unneccessary copying
// of data
double CalcProduct(const vector<double>& v1, const vector<double>& v2,
size_t start, size_t step)
{
double result = 0.0;
// do parameter checking before the calculation in order to keep
// the code more structured (e.g. this can be move to a seperate
// function if it gets too complex
if ( v1.size() != v2.size()
|| start > v1.size()
|| start+step > v1.size())
|| start < 1 )
{
throw out_of_range("CalcProduc(): parameter mismatch");
}
result = v1[start] * v2[start];
for (size_t i = start+step; i < v1.size(); i+=step)
{
result -= v1[i] * v2[i];
}
return result;
}
int main()
{
vector<double> v1;
vector<double> v2;
ReadInput(v1, v2);
cout << "\nThe 'happy' product of both vectors is: " << CalcProduct(v1, v2, 1, 2) << endl;
return 0;
}
The problem is that, I have an array of 10 integers, having some duplicates. The task is to copy this array to another array of same size, but without duplicate values. That is, read one element from array1, compare it with all the elements in array2, if it's already in array2, just skip it or print that it's already in array2, go to second element of array1, and repeat the process.
Now, I've tried this but don't know where's the problem:
#include <iostream>
using namespace std;
int main()
{
int temp;
int array1[] = {10,2,5,4,10,5,6,9,8,10};
int array2[11] = {0};
for(int i = 1; i <= 10; i++)
{
temp = array1[i-1];
for(int j = 1; j <= 10; j++)
{
if(temp == array2[j])
{
cout << "Duplicate " << temp << endl;
i++;
break;
}
}
array2[i] = array1[i-1];
}
for(int k = 1; k <= 10; k++)
cout << array2[k] << " " << endl;
system("pause");
}
array1 has 10 elements and array2 has 11, so right away the requirements haven't been met. Presumably, having 11 elements was a workaround for using incorrect index values in the for loops; the index should run from 0 to 9, not from 1 to 10.
When you add an element to the second array, you should only check it value against the elements that have already been added, not against the values in the entire array.
Finally, there's an underspecification. Once you've eliminated duplicates, you have fewer than 10 elements; array2 has 10 elements; what values should the extra elements have?
std::unique_copy is your friend:
http://en.cppreference.com/w/cpp/algorithm/unique_copy
remember to sort the source array first
In C++, break immediately ends one loop structure, and starts execution immediately after it. Thus, the line array2[i] = array1[i-1]; executes redardless of whether the inner for loop finds a duplicate. One solution is to set a variable indicating that the value is a duplicate:
int main() {
int temp;
bool isDuplicate; //added this line
int array1[] = {10,2,5,4,10,5,6,9,8,10};
int array2[11] = {0};
for(int i = 1; i <= 10; i++)
{
temp = array1[i-1];
isDuplicate=false;//added this line
for(int j = 1; j <= 10; j++)
{
if(temp == array2[j])
{
cout << "Duplicate " << temp << endl;
i++;
isDuplicate=true; //added this line
break;
}
}
if(!isDuplicate) //added this line
array2[i] = array1[i-1];
}
for(int k = 1; k <= 10; k++)
cout << array2[k] << " " << endl; system("pause"); }
Alternatively (though many programmers would disagree with this practice) you could use a goto statement instead of a break statement:
int main()
{
int temp;
int array1[] = {10,2,5,4,10,5,6,9,8,10};
int array2[11] = {0};
for(int i = 1; i <= 10; i++)
{
temp = array1[i-1];
for(int j = 1; j <= 10; j++)
{
if(temp == array2[j])
{
cout << "Duplicate " << temp << endl;
i++;
goto duplicate; //added this line
}
}
array2[i] = array1[i-1];
//added next line
duplicate:
}
for(int k = 1; k <= 10; k++)
cout << array2[k] << " " << endl;
system("pause");
}
You could use a std::set to ensure uniqueness for you.
http://en.cppreference.com/w/cpp/container/set
You have three approaches:
compare each element one by one (O(N^2) performance)
sort your reference array and use a binary search to determine if the element exists (O(N*lnN) performance)
create a lookup hash (O(1) performance)
I can see two main sources of problems in your code: 1) the break statement, as it is, does not solve the problem of differentiating between the case when duplicate is found, and when the element in array1 should be added to array2. 2) There is no counter which would store the number of elements inserted so far into array2, this way they could not be copied to array2 next to each other. The code which fixes both is:
#include <iostream>
using namespace std;
int main()
{
int array1[] = {10,2,5,4,10,5,6,9,8,10};
int array2[10];
int array2_elements_inserted = 0;
for(int i = 0; i < 10; i++)
{
int temp = array1[i];
bool isDuplicate = false;
for(int j = 0; j < array2_elements_inserted; j++)
{
if(temp == array2[j])
{
cout << "Duplicate " << temp << endl;
isDuplicate = true;
break;
}
}
if (!isDuplicate)
{
array2[array2_elements_inserted] = temp;
++array2_elements_inserted;
}
}
for(int k = 0; k < array2_elements_inserted; k++)
cout << array2[k] << " " << endl;
// system("pause");
}
Output:
10
2
5
4
6
9
8
First of all, use dynamic containers. Especially have a look at those provide by
the standard library, e.g. std::vector. Second, you should use a set data structure
to keep track of the elements you have seen before, e.g., std::set.
Then it's just an iteration on the input array and appending new elements to the
output array.
Here's an example:
#include <vector>
#include <set>
#include <iostream>
int main() {
// define and print input data
std::vector<int> v1 = {10,2,5,4,10,5,6,9,8,10};
for (int i : v1)
std::cout << i << " ";
std::cout << "\n";
// this will soon contain the output data
std::vector<int> v2;
// a set to keep track of the already seen elements
std::set<int> set;
// iterate the input array using range-based for loop
for (int i : v1) {
// check for duplicates
if (set.find(i) == set.end()) {
// first occurrence, insert to set, append to output data
set.insert(i);
v2.push_back(i);
}
else {
// seen before, do nothing
}
}
// print output data
for (int i : v2)
std::cout << i << " ";
std::cout << "\n";
}
The output:
$ g++ test.cc -std=c++11 && ./a.out
10 2 5 4 10 5 6 9 8 10
10 2 5 4 6 9 8
For reference:
http://en.cppreference.com/w/cpp/container/vector
http://en.cppreference.com/w/cpp/language/range-for
http://en.cppreference.com/w/cpp/container/set
http://en.cppreference.com/w/cpp/container/set/find