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581. Shortest Unsorted Continuous Subarray 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  int findUnsortedSubarray(vector<int>& nums) {
    const int n = nums.size();
    int mn = INT_MAX;
    int mx = INT_MIN;
    bool meetDecrease = false;
    bool meetIncrease = false;

    for (int i = 1; i < n; ++i) {
      if (nums[i] < nums[i - 1])
        meetDecrease = true;
      if (meetDecrease)
        mn = min(mn, nums[i]);
    }

    for (int i = n - 2; i >= 0; --i) {
      if (nums[i] > nums[i + 1])
        meetIncrease = true;
      if (meetIncrease)
        mx = max(mx, nums[i]);
    }

    int l;
    for (l = 0; l < n; ++l)
      if (nums[l] > mn)
        break;

    int r;
    for (r = n - 1; r >= 0; --r)
      if (nums[r] < mx)
        break;

    return l < r ? r - l + 1 : 0;
  }
};

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class Solution {
  public int findUnsortedSubarray(int[] nums) {
    final int n = nums.length;
    int mn = Integer.MAX_VALUE;
    int mx = Integer.MIN_VALUE;
    boolean meetDecrease = false;
    boolean meetIncrease = false;

    for (int i = 1; i < n; ++i) {
      if (nums[i] < nums[i - 1])
        meetDecrease = true;
      if (meetDecrease)
        mn = Math.min(mn, nums[i]);
    }

    for (int i = n - 2; i >= 0; --i) {
      if (nums[i] > nums[i + 1])
        meetIncrease = true;
      if (meetIncrease)
        mx = Math.max(mx, nums[i]);
    }

    int l = 0;
    for (l = 0; l < n; ++l)
      if (nums[l] > mn)
        break;

    int r = 0;
    for (r = n - 1; r >= 0; --r)
      if (nums[r] < mx)
        break;

    return l > r ? 0 : r - l + 1;
  }
}

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class Solution:
  def findUnsortedSubarray(self, nums: list[int]) -> int:
    mn = math.inf
    mx = -math.inf
    flag = False

    for i in range(1, len(nums)):
      if nums[i] < nums[i - 1]:
        flag = True
      if flag:
        mn = min(mn, nums[i])

    flag = False

    for i in reversed(range(len(nums) - 1)):
      if nums[i] > nums[i + 1]:
        flag = True
      if flag:
        mx = max(mx, nums[i])

    for l in range(len(nums)):
      if nums[l] > mn:
        break

    for r, num in reversed(list(enumerate(nums))):
      if num < mx:
        break

    return 0 if l >= r else r - l + 1