581

581. Shortest Unsorted Continuous Subarray (Medium)

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?

Related Topics:
Array, Two Pointers, Stack, Greedy, Sorting, Monotonic Stack

Solution 1. Greedy

The goal is to:

• find the greatest index L such that A[L] <= min(A[L .. N-1]) and A[0 .. L] is non-decreasing
• find the smallest index R such that A[R] >= max(A[0 .. R-1]) and A[R .. N-1] is non-decreasing.

Then, we need to sort A[L+1 .. R-1], R - L - 1 elements in total.

// OJ: https://leetcode.com/problems/shortest-unsorted-continuous-subarray/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int findUnsortedSubarray(vector<int>& A) {
        int N = A.size(), L = N - 1, R = 0;
        for (int i = N - 2, mn = A.back(); i >= 0; --i) {
            mn = min(mn, A[i]);
            if (mn != A[i]) L = i - 1;
        }
        for (int i = 1, mx = A[0]; i < N; ++i) {
            mx = max(mx, A[i]);
            if (mx != A[i]) R = i + 1;
        }
        return max(0, R - L - 1);
    }
};