C++ code to check reengagements can be done so elements sum is at most x

Suppose we have two arrays A and B of size n, and another number x. We have to check whether we can rearrange the elements in B, so that A[i] + B[1] <= x, for all i in range 0 to n-1.

So, if the input is like A = [1, 2, 3]; B = [1, 1, 2]; x = 4, then the output will be True, because if we rearrange B like [1, 2, 1], the sum values will be 1 + 1 <= 4, 2 + 2 <= 4, and 3 + 1 <= 4.

Steps

To solve this, we will follow these steps −

n := size of A
ans := 1
sum := 0
for initialize i := 0, when i < n, update (increase i by 1), do:
   sum := A[i] + B[n - i - 1]
   if sum > x, then:
      ans := 0
if ans is non-zero, then:
   return true
Otherwise
   return false

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
bool solve(vector<int> A, vector<int> B, int x){
   int n = A.size();
   int ans = 1;
   int sum = 0;
   for (int i = 0; i < n; ++i){
      sum = A[i] + B[n - i - 1];
      if (sum > x)
      ans = 0;
   }
   if (ans)
      return true;
   else
      return false;
}
int main(){
   vector<int> A = { 1, 2, 3 };
   vector<int> B = { 1, 1, 2 };
   int x = 4;
   cout << solve(A, B, x) << endl;
}

Input

{ 1, 2, 3 }, { 1, 1, 2 }, 4

Output

1