Leetcode

Minimum Difference Between Largest and Smallest Value in Three Moves

  • Time:O(n\log n)
  • Space:O(1)

C++

class Solution {
 public:
  int minDifference(vector<int>& nums) {
    const int n = nums.size();
    if (n < 5)
      return 0;

    int ans = INT_MAX;

    sort(begin(nums), end(nums));

    // change nums[0..i) to nums[i] and
    // change nums[n - 3 + i..n) to nums[n - 4 + i]
    for (int i = 0; i <= 3; ++i)
      ans = min(ans, nums[n - 4 + i] - nums[i]);

    return ans;
  }
};

JAVA

class Solution {
  public int minDifference(int[] nums) {
    final int n = nums.length;
    if (n < 5)
      return 0;

    int ans = Integer.MAX_VALUE;

    Arrays.sort(nums);

    // change nums[0..i) to nums[i] and
    // change nums[n - 3 + i..n) to nums[n - 4 + i]
    for (int i = 0; i <= 3; ++i)
      ans = Math.min(ans, nums[n - 4 + i] - nums[i]);

    return ans;
  }
}

Python

class Solution:
  def minDifference(self, nums: List[int]) -> int:
    n = len(nums)
    if n < 5:
      return 0

    ans = math.inf

    nums.sort()

    # change nums[0..i) to nums[i] and
    # change nums[n - 3 + i..n) to nums[n - 4 + i]
    for i in range(4):
      ans = min(ans, nums[n - 4 + i] - nums[i])

    return ans