Count number of distinct pairs whose sum exists in the given array in C++
We are given an array of, let's say, arr[] of integer values of any respective size and the task is to calculate the count of the number of distinct pairs available in a given array whose sum also exists in the same array.
Arrays a kind of data structure that can store a fixed-size sequential collection of elements of the same type. An array is used to store a collection of data, but it is often more useful to think of an array as a collection of variables of the same type.
Points to remember
A pair will be counted once with the same elements irrespective of their orders. For example, (3,2) and (2,3) will be counted as 1.
If there is a number occurring multiple times in an array then it will be considered exactly twice to form one pair. For example, if an array has elements as {2, 2, 2, 2} then the pair will be (2,2) and it will be counted as 1.
For Example
Input − int arr = {6, 4, 10, 14}
Output − count is 2Explanation − pairs with the sum in an array are (6,4) and (10,4) so count is 2
Input − int arr = {6, 6, 6 ,6, 6, 13}
Output − count is 0Explanation − there is no pair in an array with the sum in the same array. So, count is 0.
Approach used in the below program is as follows
Create an array let’s say, arr[]
Calculate the length of an array using the length() function that will return an integer value as per the elements in an array.
Take a temporary variable that will store the count of elements.
Create a map type variable let’s say mp
Start loop for i to 0 and i less than the size of an array
Create another map of pairs type variable let’s say par
Start loop for i to 0 and i less than the size of an array
Inside the loop, start another loop with j to i+1 and j less than the size of an array
Inside the loop, check if mp[arr[i]+arr[j]] > 0 AND pr[{arr[i], arr[j] }] =0 then increment the
count by 1
Increment par[{ arr[i], arr[j] }] by 1
Increment par[{ arr[j], arr[i] }] by 1
Return the count
Print the result.
Example
Live Demo
#include <iostream>
#include <map>
using namespace std;
// Returns number of pairs in ar[0..n-1] with
// sum equal to 'sum'
int countpairs(int ar[], int n){
// Store counts of all elements in map m
// to find pair (ar[i], sum-ar[i])
// because (ar[i]) + (sum - ar[i]) = sum
map<int, int> mymap;
for (int i = 0; i < n; i++){
mymap[ar[i]]++;
}
// To remove duplicate items we use result map
map<pair<int, int>, int> p;
int result = 0;
// Considering all pairs
for (int i = 0; i < n; i++){
for (int j = i + 1; j < n; j++){
// If sum of current pair exists
if (mymap[ar[i] + ar[j]] > 0 && p[{ ar[i], ar[j] }] ==0){
result++;
}
// Inserting the current pair both ways to avoid
// duplicates.
p[{ ar[i], ar[j] }]++;
p[{ ar[j], ar[i] }]++;
}
}
return result;
}
// main function
int main(){
int ar[] = { 6, 4, 10, 14 };
int n = sizeof(ar) / sizeof(ar[0]);
cout << "count is "<<countpairs(ar, n);
return 0;
}Output
If we run the above code we will get the following output −
count is 2
Published on 15-May-2020 15:23:09
