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 functionbool cmp(ll a, ll b) { return a > b; }// Function to find the minimum number// of operation required to satisfy the// given conditionsint 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 Codeint main(){ vector<ll> A = { 2, 3 }; vector<ll> B = { 3, 5 }; cout << FindMinMoves(A, B); return 0;} |
Java
// Java program for the above approachimport java.util.Arrays;class GFG{ // Comparator functionpublic static boolean cmp(int a, int b){ return a > b;}// Function to find the minimum number// of operation required to satisfy the// given conditionspublic 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 Codepublic 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 conditionsdef 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 CodeA = [ 2, 3 ]B = [ 3, 5 ]print(FindMinMoves(A, B))# This code is contributed by gfgking |
C#
// C# program for the above approachusing System;class GFG{// Comparator functionpublic static bool cmp(int a, int b){ return a > b;}// Function to find the minimum number// of operation required to satisfy the// given conditionspublic 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 Codepublic 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:
3
Time Complexity: O(K*log K), where the value of K is max(N, M).
Auxiliary Space: O(1)
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