Given an array arr[] of size N, the task is to print the minimum number of moves needed to make all array elements equal by selecting any two distinct indices and then increment the element at the first selected index and decrement the element at the other selected index by 1 in each move. If it is impossible to make all the array elements equal then print “-1“.
Examples:
Input: arr[] = {5, 4, 1, 10}
Output: 5
Explanation:
One of the possible way to perform operation is:
Operation 1: Select the indices 1 and 3 and then increment arr[1] by 1 and decrement arr[3] by 1. Thereafter, the array modifies to {5, 5, 1, 9}.
Operation 2: Select the indices 2 and 3 and then increment arr[2] by 1 and decrement arr[3] by 1. Thereafter, the array modifies to {5, 5, 2, 8}.
Operation 3: Select the indices 2 and 3 and then increment arr[2] by 1 and decrement arr[3] by 1. Thereafter, the array modifies to {5, 5, 3, 7}.
Operation 4: Select the indices 2 and 3 and then increment arr[2] by 1 and decrement arr[3] by 1. Thereafter, the array modifies to {5, 5, 4, 6}.
Operation 5: Select the indices 2 and 3 and then increment arr[2] by 1 and decrement arr[3] by 1. Thereafter, the array modifies to {5, 5, 5, 5}.
Therefore, the total number of move needed is 5. Also, it is the minimum possible moves needed.Input: arr[] = {1, 4}
Output: -1
Approach: The given problem can be solved based on the following observations:
- It can be observed that in one move the sum of the array remains the same therefore, if the sum of the array is not divisible N then it is impossible to make all the array elements equal.
- Otherwise, each array element will be equal to the sum of the array divided by N.
- Therefore, the idea is to use the two pointer technique to find the minimum count of moves needed to make all the array elements equal to the sum/N.
Follow the steps below to solve the problem:
- Initialize a variable, say ans as 0, to store the count of the moves needed.
- Find the sum of the array and store it in a variable say sum.
- Now if the sum is not divisible by N then print “-1“. Otherwise, update the sum as sum =sum/N.
- Sort the array in ascending order.
- Initialize variables, say i as 0 and j as N – 1 to iterate over the array.
- Iterate until i is less than j and perform the following steps:
- If increasing arr[i] to sum is less than decreasing arr[j] to sum then add sum – arr[i] to the ans, and then update arr[i], and arr[j] and then increment i by 1.
- Otherwise, add arr[j] – sum to the ans, and update arr[i] and arr[j] and then decrement j by 1.
- Finally, after completing the above steps, print the value of stored in ans.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach#include <algorithm>#include <iostream>using namespace std;// Function to find the minimum// operations to make the array// elements equalint find(int arr[], int N){ // Stores the sum of the // array int Sum = 0; for (int i = 0; i < N; i++) { Sum += arr[i]; } if (Sum % N) { return -1; } // update sum Sum /= N; // sort array sort(arr, arr + N); // Store the minimum // needed moves int ans = 0; int i = 0, j = N - 1; // Iterate until i // is less than j while (i < j) { if (Sum - arr[i] < arr[j] - Sum) { // Increment ans by // Sum-arr[i] ans += (Sum - arr[i]); // update arr[i] += (Sum - arr[i]); arr[j] -= (Sum - arr[i]); // Increment i by 1 i++; } else { // Increment ans by //arr[j]-Sum ans += (arr[j] - Sum); // Update arr[i] += (arr[j] - Sum); arr[j] -= (arr[j] - Sum); // Decrement j by 1 j--; } } // Return the value in ans return ans;}// Driver codeint main(){ // Given input int arr[] = { 5, 4, 1, 10 }; int N = sizeof(arr) / sizeof(int); // Function call cout << find(arr, N); return 0;}// This code is contributed by Parth Manchanda |
Java
import java.util.Arrays;// Java Program for the above approachclass GFG { // Function to find the minimum // operations to make the array // elements equal public static int find(int arr[], int N) { // Stores the sum of the // array int Sum = 0; for (int i = 0; i < N; i++) { Sum += arr[i]; } if (Sum % N > 0) { return -1; } // update sum Sum /= N; // sort array Arrays.sort(arr); // Store the minimum // needed moves int ans = 0; int i = 0, j = N - 1; // Iterate until i // is less than j while (i < j) { if (Sum - arr[i] < arr[j] - Sum) { // Increment ans by // Sum-arr[i] ans += (Sum - arr[i]); // update arr[i] += (Sum - arr[i]); arr[j] -= (Sum - arr[i]); // Increment i by 1 i++; } else { // Increment ans by // arr[j]-Sum ans += (arr[j] - Sum); // Update arr[i] += (arr[j] - Sum); arr[j] -= (arr[j] - Sum); // Decrement j by 1 j--; } } // Return the value in ans return ans; } // Driver code public static void main(String args[]) { // Given input int arr[] = { 5, 4, 1, 10 }; int N = arr.length; // Function call System.out.println(find(arr, N)); }}// This code is contributed by gfgking |
Python3
# Python program for the above approach# Function to find the minimum# operations to make the array# elements equaldef find(arr, N): # Stores the sum of the # array Sum = sum(arr) # If sum is not divisible # by N if Sum % N: return -1 else: # Update sum Sum //= N # Sort the array arr.sort() # Store the minimum # needed moves ans = 0 i = 0 j = N-1 # Iterate until i # is less than j while i < j: # If the Sum-arr[i] # is less than the # arr[j]-sum if Sum-arr[i] < arr[j]-Sum: # Increment ans by # Sum-arr[i] ans += Sum-arr[i] # Update arr[i] += Sum-arr[i] arr[j] -= Sum-arr[i] # Increment i by 1 i += 1 # Otherwise, else: # Increment ans by # arr[j]-Sum ans += arr[j]-Sum # Update arr[i] += arr[j]-Sum arr[j] -= arr[j]-Sum # Decrement j by 1 j -= 1 # Return the value in ans return ans# Driver Codeif __name__ == '__main__': # Given Input arr = [5, 4, 1, 10] N = len(arr) # Function Call print(find(arr, N)) |
C#
// C# program for the above approachusing System;class GFG{ // Function to find the minimum// operations to make the array// elements equalpublic static int find(int[] arr, int N){ // Stores the sum of the // array int Sum = 0; int i = 0; for(i = 0; i < N; i++) { Sum += arr[i]; } if (Sum % N > 0) { return -1; } // update sum Sum /= N; // sort array Array.Sort(arr); // Store the minimum // needed moves int ans = 0; i = 0; int j = N - 1; // Iterate until i // is less than j while (i < j) { if (Sum - arr[i] < arr[j] - Sum) { // Increment ans by // Sum-arr[i] ans += (Sum - arr[i]); // update arr[i] += (Sum - arr[i]); arr[j] -= (Sum - arr[i]); // Increment i by 1 i++; } else { // Increment ans by // arr[j]-Sum ans += (arr[j] - Sum); // Update arr[i] += (arr[j] - Sum); arr[j] -= (arr[j] - Sum); // Decrement j by 1 j--; } } // Return the value in ans return ans;}// Driver codestatic public void Main(){ // Given input int[] arr = { 5, 4, 1, 10 }; int N = arr.Length; // Function call Console.WriteLine(find(arr, N));}}// This code is contributed by target_2 |
Javascript
<script> // JavaScript Program for the above approach // Function to find the minimum // operations to make the array // elements equal function find(arr, N) { // Stores the sum of the // array let Sum = 0; for (i = 0; i < arr.length; i++) { Sum = Sum + arr[i]; } // If sum is not divisible // by N if (Sum % N) return -1; else { // Update sum Sum = Math.floor(Sum / N); // Sort the array arr.sort(function (a, b) { return a - b }); // Store the minimum // needed moves ans = 0 i = 0 j = N - 1 // Iterate until i // is less than j while (i < j) { // If the Sum-arr[i] // is less than the // arr[j]-sum if (Sum - arr[i] < arr[j] - Sum) { // Increment ans by // Sum-arr[i] ans += Sum - arr[i] // Update arr[i] += Sum - arr[i] arr[j] -= Sum - arr[i] // Increment i by 1 i += 1 } // Otherwise, else { // Increment ans by // arr[j]-Sum ans += arr[j] - Sum // Update arr[i] += arr[j] - Sum arr[j] -= arr[j] - Sum // Decrement j by 1 j -= 1 } } } // Return the value in ans return ans; } // Driver Code // Given Input let arr = [5, 4, 1, 10]; let N = arr.length; // Function Call document.write(find(arr, N)); // This code is contributed by Potta Lokesh </script> |
Output
5
Time Complexity: O(N*log(N))
Auxiliary Space: O(1)
Another Approach:
- Compute the sum of all elements in the array and divide it by N to find the target value (the value that all elements in the array should be equal to).
- Iterate over the array, and for each element:
- Compute the difference between the current element and the target value.
- If the difference is odd, return -1 (because we can’t make all elements equal by incrementing and decrementing in pairs).
- Otherwise, add the absolute value of the difference to a variable count (this represents the number of moves needed to make the current element equal to the target value).
- Return the count.
Below is the implementation of above approach:
C++
#include <algorithm>#include <iostream>using namespace std; int find(int arr[], int N) { int Sum = 0; for (int i = 0; i < N; i++) { Sum += arr[i]; } if (Sum % N) { return -1; } Sum /= N; sort(arr, arr + N); int ans = 0, i = 0; while (i < N) { int j = i + 1; while (j < N && arr[j] <= Sum) { j++; } if (j == N) { break; } ans += j - i; arr[i] += j - i; arr[j] -= j - i; i++; } return ans;} int main() { int arr[] = { 5, 4, 1, 10 }; int N = sizeof(arr) / sizeof(int); cout << find(arr, N); // Output: 5 return 0;} |
Java
// Java implementation of a function to// find the minimum number of moves required// to make all elements of an array equalimport java.util.Arrays;public class Main {// Function to find the minimum number of moves required// to make all elements of an array equalstatic int find(int arr[], int N) { // Calculate the sum of all the elements of the array int Sum = 0; for (int i = 0; i < N; i++) { Sum += arr[i]; } // If the sum is not divisible by N, we cannot make all // elements equal by adding or subtracting from each other. // Hence, return -1. if (Sum % N != 0) { return -1; } // Calculate the target sum by dividing the sum of // all elements by N. int targetSum = Sum / N; // Sort the array in non-decreasing order Arrays.sort(arr); int ans = 0, i = 0; while (i < N) { // Find the index j of the first element in the array // that is greater than the target sum. int j = i + 1; while (j < N && arr[j] <= targetSum) { j++; } // If j equals N, it means that all the remaining // elements in the array are less than or equal to // the target sum, and hence, we cannot make them // equal to the target sum by adding or subtracting // from each other. Therefore, break out of the loop. if (j == N) { break; } // Calculate the number of moves required to make // elements arr[i] to arr[j-1] equal to the target sum. ans += j - i; arr[i] += j - i; arr[j] -= j - i; i++; } return ans;}public static void main(String[] args) { int arr[] = { 5, 4, 1, 10 }; int N = arr.length; System.out.println(find(arr, N)); // Output: 5}} |
Python3
def find(arr, N): Sum = sum(arr) if Sum % N: return -1 Sum //= N arr.sort() ans, i = 0, 0 while i < N: j = i + 1 while j < N and arr[j] <= Sum: j += 1 if j == N: break ans += j - i arr[i:i+j-i] = [arr[i] + j - i] * (j - i) arr[j] -= j - i i += 1 return ansarr = [5, 4, 1, 10]N = len(arr)print(find(arr, N)) # Output: 5 |
Javascript
// Javascript implementation of a function to// find the minimum number of moves required// to make all elements of an array equal// Function to find the minimum number of moves required// to make all elements of an array equalfunction find(arr, N) { // Calculate the sum of all the elements of the array let Sum = 0; for (let i = 0; i < N; i++) { Sum += arr[i]; } // If the sum is not divisible by N, we cannot make all // elements equal by adding or subtracting from each other. // Hence, return -1. if (Sum % N) { return -1; } // Calculate the target sum by dividing the sum of // all elements by N. Sum /= N; // Sort the array in non-decreasing order arr.sort((a, b) => a - b); let ans = 0, i = 0; while (i < N) { // Find the index j of the first element in the array // that is greater than the target sum. let j = i + 1; while (j < N && arr[j] <= Sum) { j++; } // If j equals N, it means that all the remaining // elements in the array are less than or equal to // the target sum, and hence, we cannot make them // equal to the target sum by adding or subtracting // from each other. Therefore, break out of the loop. if (j == N) { break; } // Calculate the number of moves required to make // elements arr[i] to arr[j-1] equal to the target sum. ans += j - i; arr[i] += j - i; arr[j] -= j - i; i++; } return ans;}let arr = [5, 4, 1, 10];let N = arr.length;console.log(find(arr, N)); // Output: 5 |
C#
using System;using System.Collections.Generic;class Gfg{ static int find(int[] arr, int N) { int Sum = 0, i=0; for ( i = 0; i < N; i++) { Sum += arr[i]; } if (Sum % N !=0) { return -1; } Sum /= N; Array.Sort(arr); int ans = 0; i = 0; while (i < N) { int j = i + 1; while (j < N && arr[j] <= Sum) { j++; } if (j == N) { break; } ans += j - i; arr[i] += j - i; arr[j] -= j - i; i++; } return ans; } public static void Main(String[] args) { int[] arr = { 5, 4, 1, 10 }; int N = arr.Length; Console.Write(find(arr, N)); // Output: 5 }} |
Output
5
Time Complexity: O(N), where N is the size of the input array
Auxiliary Space: O(1)
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