Given two arrays of same size, we need to convert the first array into another with minimum operations. In an operation, we can either increment or decrement an element by one. Note that orders of appearance of elements do not need to be same.

Here to convert one number into another we can add or subtract 1 from it.

Examples:

Input :a = { 3, 1, 1 }, b = { 1, 2, 2 }

Output :2

Explanation :Here we can increase any 1 into 2 by 1 operation and 3 to 2 in one decrement operation. So a[] becomes {2, 2, 1} which is a permutation of b[].

Input :a = { 3, 1, 1 }, b = { 1, 1, 2 }

Output :1

**Algorithm :**

1. First sort both the arrays.

2. After sorting we will run a loop in which we compare the first and second array elements and calculate the required operation needed to make first array equal to second.

Below is C++ implementation of the above approach

// CPP program to find minimum increment/decrement // operations to make array elements same. #include <bits/stdc++.h> using namespace std; int MinOperation(int a[], int b[], int n) { // sorting both arrays in // ascending order sort(a, a + n); sort(b, b + n); // variable to store the // final result int result = 0; // After sorting both arrays // Now each array is in non- // decreasing order. Thus, // we will now compare each // element of the array and // do the increment or decrement // operation depending upon the // value of array b[]. for (int i = 0; i < n; ++i) { if (a[i] > b[i]) result = result + abs(a[i] - b[i]); else if (a[i] < b[i]) result = result + abs(a[i] - b[i]); } return result; } // Driver code int main() { int a[] = { 3, 1, 1 }; int b[] = { 1, 2, 2 }; int n = sizeof(a) / sizeof(a[0]); cout << MinOperation(a, b, n); return 0; }

**Output:**

2

Time Complexity : O(n Log n)

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