453. Minimum Moves to Equal Array Elements

Difficulty: Medium

Topics: Junior, Array, Math

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

Example 3:

Input: nums = [5]
Output: 0

Example 4:

Input: nums = [-2,-1,0]
Output: 3

Example 5:

Input: nums = [0,10]
Output: 10

Example 6:

Input: nums = [5,5,10]
Output: 5

Example 7:

Input: nums = [1,2,100]
Output: 100

Example 8:

Input: nums = [-1000000000, 1000000000]
Output: 2000000000

Constraints:

n == nums.length
1 <= nums.length <= 10⁵
-10⁹ <= nums[i] <= 10⁹
The answer is guaranteed to fit in a 32-bit integer.

Similar Questions:

Solution:

We can solve this by recognizing that incrementing n - 1 elements is equivalent to decrementing exactly one element. Therefore, the minimum moves needed to make all elements equal is the total amount required to reduce every element to the array minimum.

Approach

Find the minimum value in nums.
For each element, calculate how far it is above the minimum.
Sum all those differences.
Return the sum as the minimum number of moves.

Let's implement this solution in PHP: 453. Minimum Moves to Equal Array Elements

<?php /** * @param Integer[] $nums * @return Integer */ function minMoves(array $nums): int { ... ... ... /** * go to ./solution.php */ } // Test cases echo minMoves([1, 2, 3]) . "\n"; // Output: 3 echo minMoves([1, 1, 1]) . "\n"; // Output: 0 echo minMoves([5]) . "\n"; // Output: 0 echo minMoves([-2, -1, 0]) . "\n"; // Output: 3 echo minMoves([0, 10]) . "\n"; // Output: 10 echo minMoves([5, 5, 10]) . "\n"; // Output: 5 echo minMoves([1, 2, 100]) . "\n"; // Output: 100 echo minMoves([-1000000000, 1000000000]) . "\n"; // Output: 2000000000 ?>

Explanation:

In one move, incrementing n - 1 elements by 1 is mathematically equivalent to decrementing the remaining one element by 1.
To make all elements equal using only this operation, the final target value must be the current minimum element.
Any element larger than the minimum must be reduced by nums[i] - min(nums) moves.
Summing these differences gives the total minimum moves.
The provided PHP solution correctly implements this logic using min($nums) and a single loop.

Complexity Analysis

Time Complexity: O(n), where n is the length of nums. We scan the array once to find the minimum and once to sum differences.
Space Complexity: O(1), since only a few scalar variables are used.

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