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2567. Minimum Score by Changing Two Elements

  • Time: $O(\texttt{sort})$
  • Space: $O(\texttt{sort})$
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class Solution {
 public:
  int minimizeSum(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    // Can always change the num to any other num in nums, so `low` becomes 0.
    // So, we can rephrase the problem as finding the min `high`.
    const int n = nums.size();
    const int highOfChangingTwoMins = nums.back() - nums[2];
    const int highOfChangingTwoMaxs = nums[n - 3] - nums[0];
    const int highOfChangingMinAndMax = nums[n - 2] - nums[1];
    return min({highOfChangingTwoMins, highOfChangingTwoMaxs,
                highOfChangingMinAndMax});
  }
};

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class Solution {
  public int minimizeSum(int[] nums) {
    Arrays.sort(nums);
    // Can always change the num to any other num in nums, so `low` becomes 0.
    // So, we can rephrase the problem as finding the min `high`.
    final int n = nums.length;
    final int highOfChangingTwoMins = nums[n - 1] - nums[2];
    final int highOfChangingTwoMaxs = nums[n - 3] - nums[0];
    final int highOfChangingMinAndMax = nums[n - 2] - nums[1];
    return Math.min(Math.min(highOfChangingTwoMins, highOfChangingTwoMaxs),
                    highOfChangingMinAndMax);
  }
}

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class Solution:
  def minimizeSum(self, nums: List[int]) -> int:
    nums.sort()
    # Can always change the num to any other num in nums, so `low` becomes 0.
    # So, we can rephrase the problem as finding the min `high`.
    highOfChangingTwoMins = nums[-1] - nums[2]
    highOfChangingTwoMaxs = nums[-3] - nums[0]
    highOfChangingMinAndMax = nums[-2] - nums[1]
    return min(highOfChangingTwoMins, highOfChangingTwoMaxs,
               highOfChangingMinAndMax)