{ "data": { "question": { "questionId": "1616", "questionFrontendId": "1509", "categoryTitle": "Algorithms", "boundTopicId": 322324, "title": "Minimum Difference Between Largest and Smallest Value in Three Moves", "titleSlug": "minimum-difference-between-largest-and-smallest-value-in-three-moves", "content": "
You are given an integer array nums
.
In one move, you can choose one element of nums
and change it to any value.
Return the minimum difference between the largest and smallest value of nums
after performing at most three moves.
\n
Example 1:
\n\n\nInput: nums = [5,3,2,4]\nOutput: 0\nExplanation: We can make at most 3 moves.\nIn the first move, change 2 to 3. nums becomes [5,3,3,4].\nIn the second move, change 4 to 3. nums becomes [5,3,3,3].\nIn the third move, change 5 to 3. nums becomes [3,3,3,3].\nAfter performing 3 moves, the difference between the minimum and maximum is 3 - 3 = 0.\n\n\n
Example 2:
\n\n\nInput: nums = [1,5,0,10,14]\nOutput: 1\nExplanation: We can make at most 3 moves.\nIn the first move, change 5 to 0. nums becomes [1,0,0,10,14].\nIn the second move, change 10 to 0. nums becomes [1,0,0,0,14].\nIn the third move, change 14 to 1. nums becomes [1,0,0,0,1].\nAfter performing 3 moves, the difference between the minimum and maximum is 1 - 0 = 1.\nIt can be shown that there is no way to make the difference 0 in 3 moves.\n\n
Example 3:
\n\n\nInput: nums = [3,100,20]\nOutput: 0\nExplanation: We can make at most 3 moves.\nIn the first move, change 100 to 7. nums becomes [3,7,20].\nIn the second move, change 20 to 7. nums becomes [3,7,7].\nIn the third move, change 3 to 7. nums becomes [7,7,7].\nAfter performing 3 moves, the difference between the minimum and maximum is 7 - 7 = 0.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 105
-109 <= nums[i] <= 109
给你一个数组 nums
。
每次操作你可以选择 nums
中的任意一个元素并将它改成 任意值 。
在 执行最多三次移动后 ,返回 nums
中最大值与最小值的最小差值。
\n\n
示例 1:
\n\n\n输入:nums = [5,3,2,4]\n输出:0\n解释:我们最多可以走 3 步。\n第一步,将 2 变为 3 。 nums 变成 [5,3,3,4] 。\n第二步,将 4 改为 3 。 nums 变成 [5,3,3,3] 。\n第三步,将 5 改为 3 。 nums 变成 [3,3,3,3] 。\n执行 3 次移动后,最小值和最大值之间的差值为 3 - 3 = 0 。\n\n
示例 2:
\n\n\n输入:nums = [1,5,0,10,14]\n输出:1\n解释:我们最多可以走 3 步。\n第一步,将 5 改为 0 。 nums变成 [1,0,0,10,14] 。\n第二步,将 10 改为 0 。 nums变成 [1,0,0,0,14] 。\n第三步,将 14 改为 1 。 nums变成 [1,0,0,0,1] 。\n执行 3 步后,最小值和最大值之间的差值为 1 - 0 = 1 。\n可以看出,没有办法可以在 3 步内使差值变为0。\n\n\n
示例 3:
\n\n\n输入:nums = [3,100,20]\n输出:0\n解释:我们最多可以走 3 步。\n第一步,将 100 改为 7 。 nums 变成 [3,7,20] 。\n第二步,将 20 改为 7 。 nums 变成 [3,7,7] 。\n第三步,将 3 改为 7 。 nums 变成 [7,7,7] 。\n执行 3 步后,最小值和最大值之间的差值是 7 - 7 = 0。\n\n
\n\n
提示:
\n\n1 <= nums.length <= 105
-109 <= nums[i] <= 109
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