Minimize operations to make minimum value of one array greater than maximum value of the other

Namachila November 18, 2025

0 0 5 minutes read

Given two arrays A[] and B[] consisting of N and M integers, the task is to find the minimum number of operations required to make the minimum element of the array A[] at least the maximum element of the array B[] such that in each operation any array element A[] can be incremented by 1 or any array element B[] can be decremented by 1.

Examples:

Input: A[] = {2, 3}, B[] = {3, 5}
Output: 3
Explanation:
Following are the operations performed:

Increase the value of A[1] by 1 modifies the array A[] = {3, 3}.

Decrease the value of B[2] by 1 modifies the array B[] = {3, 4}.

Decrease the value of B[2] by 1 modifies the array B[] = {3, 3}.

After the above operations, the minimum elements of the array A[] is 3 which is greater than or equal to the maximum element of the array B[] is 3. Therefore, the total number of operations is 3.

Input: A[] = {1, 2, 3}, B[] = {4}
Output: 3

Approach: The problem can be solved by using the Greedy Approach. Follow the steps below to solve the given problem:

Sort the array A[] in increasing order.
Sort the array B[] in decreasing order.
Traverse both of the arrays while A[i] < B[i], in order to make all elements of the arrays A[] and B[], till index i, equal, say x, then the total number of operations is given by:

=> (B[0] + B[1] + … + B[i]) – i*x + (A[0] + A[1] + … + A[i]) + i*x
=> (B[0] – A[0]) + (B[1] – A[1]) + … + (B[i] – A[i]).

Traverse both the arrays until the value of A[i] is smaller than B[i], and the value of (B[i] – A[i]) to the variable, say ans.
After completing the above steps, print the value of ans as the minimum number of operations required.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long
 
// Comparator function
bool cmp(ll a, ll b) { return a > b; }
 
// Function to find the minimum number
// of operation required to satisfy the
// given conditions
int FindMinMoves(vector<ll> A, vector<ll> B)
{
    int n, m;
    n = A.size();
    m = B.size();
 
    // sort the array A and B in the
    // ascending and descending order
    sort(A.begin(), A.end());
    sort(B.begin(), B.end(), cmp);
 
    ll ans = 0;
 
    // Iterate over both the arrays
    for (int i = 0; i < min(m, n)
                    && A[i] < B[i];
         ++i) {
 
        // Add the difference to the
        // variable answer
        ans += (B[i] - A[i]);
    }
 
    // Return the resultant operations
    return ans;
}
 
// Driver Code
int main()
{
    vector<ll> A = { 2, 3 };
    vector<ll> B = { 3, 5 };
    cout << FindMinMoves(A, B);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;
 
class GFG{
     
// Comparator function
public static boolean cmp(int a, int b)
{
    return a > b;
}
 
// Function to find the minimum number
// of operation required to satisfy the
// given conditions
public static int FindMinMoves(int[] A, int[] B)
{
    int n, m;
    n = A.length;
    m = B.length;
 
    // Sort the array A and B in the
    // ascending and descending order
    Arrays.sort(A);
    Arrays.sort(B);
 
    int ans = 0;
 
    // Iterate over both the arrays
    for(int i = 0; 
            i < Math.min(m, n) && A[i] < B[i]; ++i) 
    {
         
        // Add the difference to the
        // variable answer
        ans += (B[i] - A[i]);
    }
 
    // Return the resultant operations
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int[] A = { 2, 3 };
    int[] B = { 3, 5 };
     
    System.out.println(FindMinMoves(A, B));
}
}
 
// This code is contributed by _saurabh_jaiswal

Python3

# Python3 program for the above approach
 
# Function to find the minimum number
# of operation required to satisfy the
# given conditions
def FindMinMoves(A, B):
     
    n = len(A)
    m = len(B)
 
    # sort the array A and B in the
    # ascending and descending order
    A.sort()
    B.sort(reverse = True)
    ans = 0
 
    # Iterate over both the arrays
    i = 0
     
    for i in range(min(m, n)):
 
        # Add the difference to the
        # variable answer
        if A[i] < B[i]:
            ans += (B[i] - A[i])
 
    # Return the resultant operations
    return ans
 
# Driver Code
A = [ 2, 3 ]
B = [ 3, 5 ]
 
print(FindMinMoves(A, B))
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Comparator function
public static bool cmp(int a, int b)
{
    return a > b;
}
 
// Function to find the minimum number
// of operation required to satisfy the
// given conditions
public static int FindMinMoves(int[] A, int[] B)
{
    int n, m;
    n = A.Length;
    m = B.Length;
 
    // Sort the array A and B in the
    // ascending and descending order
    Array.Sort(A);
    Array.Sort(B);
 
    int ans = 0;
 
    // Iterate over both the arrays
    for(int i = 0; 
            i < Math.Min(m, n) && A[i] < B[i]; ++i) 
    {
         
        // Add the difference to the
        // variable answer
        ans += (B[i] - A[i]);
    }
 
    // Return the resultant operations
    return ans;
}
 
// Driver Code
public static void Main()
{
    int[] A = { 2, 3 };
    int[] B = { 3, 5 };
     
    Console.Write(FindMinMoves(A, B));
}
}
 
// This code is contributed by target_2.

Javascript

<script>
 
       // JavaScript program for the above approach
 
       // Function to find the minimum number
       // of operation required to satisfy the
       // given conditions
       function FindMinMoves(A, B) 
       {
           let n, m;
           n = A.length;
           m = B.length;
 
           // sort the array A and B in the
           // ascending and descending order
           A.sort(function (a, b) { return a - b; });
           B.sort(function (a, b) { return b - a; });
 
           let ans = 0;
 
           // Iterate over both the arrays
           for (let i = 0; i < Math.min(m, n)
               && A[i] < B[i];
               ++i) {
 
               // Add the difference to the
               // variable answer
               ans += (B[i] - A[i]);
           }
 
           // Return the resultant operations
           return ans;
       }
 
       // Driver Code
       let A = [2, 3];
       let B = [3, 5];
       document.write(FindMinMoves(A, B));
 
   // This code is contributed by Potta Lokesh
   </script>

Output:

Time Complexity: O(K*log K), where the value of K is max(N, M).
Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!