# C++ Program to find out the minimum difference value in n integer pairs

C++Server Side ProgrammingProgramming

Suppose, we are given two arrays a and b that have a total of n and m values in them respectively. We have to make n or m number of pairs (whichever is minimum) using values from the two arrays. A pair must contain a value from array a and another one from array b. We have to make the pairs in such a way so that the difference of the value in the pairs is minimum and the same. We print the value as output.

So, if the input is like n = 4, m = 4, a = {2, 3, 4, 7}, b = {3, 4, 6, 5}, then the output will be 1.

The pairs that can be made are −

(3, 4), (4, 5), (7, 6), (2, 3).

The difference in the value in all pairs is 1.

## Steps

To solve this, we will follow these steps −

sort the array a
Define an array s1 initialized with 0
Define an array s2 initialized with 0
for initialize i := 1, when i < n, update i := i + 2, do:
insert last element of s1 + a[i] - a[i - 1] at the end of s1
for initialize i := 2, when i < n, update i := i + 2, do:
insert last element of s2 + a[i] - a[i - 1] at the end of s2
ans := infinity
for each value w in b, do:
diff := first element in the array a not less than w - first value of a
sub := last element of s1[diff / 2] + s2
ans := minimum of ans and sub
print(ans)

## Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;

void solve(int n, int m, vector<int> a, vector<int> b){
sort(a.begin(), a.end());
vector<int> s1 = {0};
vector<int> s2 = {0};
for (int i = 1; i < n; i += 2)
s1.push_back(a[i] - a[i - 1] + s1.back());
for (int i = 2; i < n; i += 2)
s2.push_back(a[i] - a[i - 1] + s2.back());
int ans = INF;
for (const auto & w : b) {
int diff = lower_bound(a.begin(), a.end(), w) - a.begin();
int sub = s1[diff / 2] + s2.back() - s2[diff / 2] + abs(a[diff / 2 * 2] - w);
ans = min(ans, sub);
}
cout << ans << endl;
}
int main() {
int n = 4, m = 4;
vector<int> a = {2, 3, 4, 7}, b = {3, 4, 6, 5};
solve(n, m, a, b);
return 0;
}

## Input

4, 4, {2, 3, 4, 7}, {3, 4, 6, 5}

## Output

1
Updated on 02-Mar-2022 11:09:42