453.Minimum Moves to Equal Array Elements

Method 1

Given an integer array nums of size n, return the minimum number of moves required to make all array elements equal.
In one move, you can increment n - 1 elements of the array by 1.

Example 1

Input: nums = [1,2,3]
Output: 3
Explanation: Only three moves are needed (remember each move increments two elements):
[1,2,3]  =>  [2,3,3]  =>  [3,4,3]  =>  [4,4,4]

Example 2

Input: nums = [1,1,1]
Output: 0

Method 1

【O(n) time | O(1) space】

package Leetcode.Array;

/**
 * @author zhengstars
 * @date 2024/10/22
 */
public class MinimumMovesToEqualArrayElements {
    public static int minMoves(int[] nums) {
        // Handle edge cases: null array or array with length <= 1
        if (nums == null || nums.length <= 1) {
            return 0;
        }

        // Step 1: Find the minimum value in the array
        // This is our target value that all other elements will eventually equal to
        int min = nums[0];
        for (int num : nums) {
            min = Math.min(min, num);
        }

        // Step 2: Calculate the total moves needed
        // For each element, we need (current value - minimum value) moves to reach the minimum
        int moves = 0;
        for (int num : nums) {
            moves += num - min;  // Add the difference between current element and minimum
        }

        return moves;
    }

    public static void main(String[] args) {

        // Test Case 
        System.out.println("Test Case 1:  " + minMoves(new int[]{ 1,2,3 } ) );  // Expected output: 3
        System.out.println("Test Case 2:  " + minMoves(new int[]{ 1,1,1 } ) );  // Expected output: 0
        System.out.println("Test Case 3:  " + minMoves(new int[]{ 5,8,10 } ) );  // Expected output: 8
        System.out.println("Test Case 4:  " + minMoves(new int[]{ 1,2,2,2,3 } ) );  // Expected output: 6
    }
}

Method 1

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