세 가지 동작의 최대값과 최소값 간의 최소 차이

In the first move, change 2 to 3. nums becomes [5,3,3,4]. In the second move, change 4 to 3. nums becomes [5,3,3,3]. In the third move, change 5 to 3. nums becomes [3,3,3,3]. After performing 3 moves, the difference between the minimum and maximum is 3 - 3 = 0.

In the first move, change 5 to 0. nums becomes [1,0,0,10,14]. In the second move, change 10 to 0. nums becomes [1,0,0,0,14]. In the third move, change 14 to 1. nums becomes [1,0,0,0,1]. After performing 3 moves, the difference between the minimum and maximum is 1 - 0 = 1. It can be shown that there is no way to make the difference 0 in 3 moves.

In the first move, change 100 to 7. nums becomes [3,7,20]. In the second move, change 20 to 7. nums becomes [3,7,7]. In the third move, change 3 to 7. nums becomes [7,7,7]. After performing 3 moves, the difference between the minimum and maximum is 7 - 7 = 0.

class Solution { /** * @param Integer[] $nums * @return Integer */ function minDifference($nums) { $n = count($nums); // If the array has 4 or fewer elements, the difference is zero because we can remove all but one element. if ($n <= 4) { return 0; } // Sort the array to facilitate the calculation of differences after removals. sort($nums); // We consider removing 0, 1, 2, or 3 elements from the start or the end. // Calculate the differences: // 1. Remove 3 from start: nums[n-1] - nums[3] // 2. Remove 2 from start, 1 from end: nums[n-2] - nums[2] // 3. Remove 1 from start, 2 from end: nums[n-3] - nums[1] // 4. Remove 3 from end: nums[n-4] - nums[0] $differences = [ $nums[$n - 1] - $nums[3], $nums[$n - 2] - $nums[2], $nums[$n - 3] - $nums[1], $nums[$n - 4] - $nums[0] ]; // Return the minimum difference. return min($differences); } }