Find Number of Pairs in an Array that Satisfy a Given Condition in C++

Suppose, we are given n numbers in array nums. We have to choose a pair of two numbers from the array, and there is a condition that the difference of their positions in the array is equal to the sum of the two numbers. There can be a total of n(n - 1)/2 number of total pairs from the given array of numbers. We have to find out the total number of such pairs from the array.

So, if the input is like n = 8, nums = {4, 2, 1, 0, 1, 2, 3, 3}, then the output will be 13.

There can be 13 such pairs in the array.

To solve this, we will follow these steps −

Define an array vals(n)
for initialize i := 0, when i < n, update (increase i by 1), do:
   vals[i] := i + 1 - nums[i]
sort the array vals
res := 0
for initialize i := 0, when i < n, update (increase i by 1), do:
   k := nums[i] + i + 1
res := res + (position of first occurrence of a value not less than k in array vals - position of first occurrence of a value not greater than k in array vals)
return res

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;

int solve(int n, vector<int> nums){
   vector<int> vals(n);
   for( int i = 0; i < n; i++)
      vals[i] = i + 1 - nums[i]; 
   sort(vals.begin(), vals.end());
   int res = 0;
   for( int i = 0; i < n; i++ ) {
      int k = nums[i] + i + 1;
      res += upper_bound(vals.begin(), vals.end(), k) - lower_bound(vals.begin(), vals.end(), k);
   }
   return res;
}
int main() {
   int n = 8;
   vector<int> nums = {4, 2, 1, 0, 1, 2, 3, 3};
   cout<< solve(n, nums);
   return 0;
}

Input

8, {4, 2, 1, 0, 1, 2, 3, 3}

Output

13