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581. Shortest Unsorted Continuous Subarray

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Description

Given an integer array nums, you need to find one continuous subarray that if you only sort this subarray in ascending order, then the whole array will be sorted in ascending order.

Return the shortest such subarray and output its length.

 

Example 1:

Input: nums = [2,6,4,8,10,9,15]
Output: 5
Explanation: You need to sort [6, 4, 8, 10, 9] in ascending order to make the whole array sorted in ascending order.

Example 2:

Input: nums = [1,2,3,4]
Output: 0

Example 3:

Input: nums = [1]
Output: 0

 

Constraints:

• 1 <= nums.length <= 104
• -105 <= nums[i] <= 105

 

Follow up: Can you solve it in O(n) time complexity?

Solutions

Python3

class Solution:
    def findUnsortedSubarray(self, nums: List[int]) -> int:
        n = len(nums)
        numsSorted = sorted(nums)
        left, right = 0, n - 1
        while left < n and nums[left] == numsSorted[left]:
            left += 1
        while right >= 0 and nums[right] == numsSorted[right]:
            right -= 1
        return 0 if right == -1 else right - left + 1

Java

class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int n = nums.length;
        int[] numsSorted = new int[n];
        System.arraycopy(nums, 0, numsSorted, 0, n);
        Arrays.sort(numsSorted);
        int left = 0, right = n - 1;
        while (left < n && nums[left] == numsSorted[left]) {
            left++;
        }
        while (right >= 0 && nums[right] == numsSorted[right]) {
            right--;
        }
        return right == -1 ? 0 : right - left + 1;
    }
}

Go

func findUnsortedSubarray(nums []int) int {
	n := len(nums)
	maxn, minn := math.MinInt32, math.MaxInt32
	left, right := -1, -1
	for i := 0; i < n; i++ {
		if maxn > nums[i] {
			right = i
		} else {
			maxn = nums[i]
		}
		if minn < nums[n-i-1] {
			left = n - i - 1
		} else {
			minn = nums[n-i-1]
		}
	}
	if right == -1 {
		return 0
	}
	return right - left + 1
}

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