# C++ program to find maximum distance between two rival students after x swaps

Suppose we have four numbers n, x, a and b. There are n students in the row. There are two rivalling students among them. One of them is at position a and another one is at position b. Positions are numbered from 1 to n from left to right. We want to maximize the distance between these two students. We can perform the following operation x times: Select two adjacent students and then swap them. We have to find the maximum possible distance after x swaps.

So, if the input is like n = 5; x = 1; a = 3; b = 2, then the output will be 2, because we can swap students at position 3 and 4 so the distance between these two students is |4 - 2| = 2.

## Steps

To solve this, we will follow these steps −

return minimum of (|a - b| + x) and (n - 1)

## Example

Let us see the following implementation to get better understanding −

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

int solve(int n, int x, int a, int b) {
return min(abs(a - b) + x, n - 1);
}
int main() {
int n = 5;
int x = 1;
int a = 3;
int b = 2;
cout << solve(n, x, a, b) << endl;
}

## Input

5, 1, 3, 2

## Output

2

Updated on: 03-Mar-2022

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