0581. Shortest Unsorted Continuous Subarray

https://leetcode.com/problems/shortest-unsorted-continuous-subarray

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?

ac1: stack with 2 passes

Learnt: for array, think about two pass: from left and from right.

class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int left = nums.length, right = 0;
        Deque<Integer> stack = new ArrayDeque<>();
        
        for (int i = 0; i < nums.length; i++) {
            while (!stack.isEmpty() && nums[stack.peek()] > nums[i]) {
                int idx = stack.pop();
                left = Math.min(left, idx);
            }
            stack.push(i);
        }
        
        stack = new ArrayDeque<>();
        for (int i = nums.length - 1; i >= 0; i--) {
            while (!stack.isEmpty() && nums[i] > nums[stack.peek()]) {
                int idx = stack.pop();
                right = Math.max(right, idx);
            }
            stack.push(i);
        }
        
        return Math.max(0, right - left + 1);
    }
}

// O(N) time, O(N) space
// To find left and right bound.

ac2: similar idea without stack

We don't really need a stack to find the left/right boundary.

class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int left = nums.length, right = 0;
        
        // 1st pass from left
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < nums.length; i++) {
            max = Math.max(max, nums[i]);
            if (nums[i] < max) {
                // nums[i] needs to swap with a number on the left, so i is the right boundary.
                right = i;
            }
        }
        
        // 2nd pass from right
        int min = Integer.MAX_VALUE;
        for (int i = nums.length - 1; i >= 0; i--) {
            min = Math.min(min, nums[i]);
            if (nums[i] > min) {
                // nums[i] needs to swap with a number on the right, so i is the left boundary.
                left = i;
            }
        }
        
        return Math.max(0, right - left + 1);
    }
}

// O(N) time, O(1) space