581. 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.

Constraints

• 1 <= nums.length <= 104

• -105 <= nums[i] <= 105

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Solutions

Solution 1

/**
 * Time complexity : O(n^2). Two nested loops are there.
 * Space complexity : O(1). Constant space is used.
 */

public class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int l = nums.length, r = 0;
        for (int i = 0; i < nums.length - 1; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[j] < nums[i]) {
                    r = Math.max(r, j);
                    l = Math.min(l, i);
                }
            }
        }
        return r - l < 0 ? 0 : r - l + 1;
    }
}

Solution 2

/**
 * Time complexity : O(nlogn). Sorting takes nlogn time.
 * Space complexity : O(n). We are making copy of original array.
 */
 
 public class Solution {
    public int findUnsortedSubarray(int[] nums) {
        int[] snums = nums.clone();
        Arrays.sort(snums);
        int start = snums.length, end = 0;
        for (int i = 0; i < snums.length; i++) {
            if (snums[i] != nums[i]) {
                start = Math.min(start, i);
                end = Math.max(end, i);
            }
        }
        return (end - start >= 0 ? end - start + 1 : 0);
    }
}

Solution 3

/**
 * Time complexity : O(n). Stack of size n is filled.
 * Space complexity : O(n). Stack size grows upto n.
 */
 
 public class Solution {
    public int findUnsortedSubarray(int[] nums) {
        Stack < Integer > stack = new Stack < Integer > ();
        int l = nums.length, r = 0;
        for (int i = 0; i < nums.length; i++) {
            while (!stack.isEmpty() && nums[stack.peek()] > nums[i])
                l = Math.min(l, stack.pop());
            stack.push(i);
        }
        stack.clear();
        for (int i = nums.length - 1; i >= 0; i--) {
            while (!stack.isEmpty() && nums[stack.peek()] < nums[i])
                r = Math.max(r, stack.pop());
            stack.push(i);
        }
        return r - l > 0 ? r - l + 1 : 0;
    }
}

Solution 4

/**
 * Time complexity : O(n). Two O(n) loops are used.
 * Space complexity : O(1). Constant space is used.
 */
 
 class Solution {
    public int findUnsortedSubarray(int[] nums) {
        if(nums == null || nums.length <= 1) {
            return 0;
        }
        int n = nums.length;
        
        int end = 0;
        int max = nums[0];
        for(int i = 0; i < n; i++) {
            if(nums[i] >= max) {
                max = nums[i];
            } else {
                end = i;
            }
        }
        
        int start = n;
        int min = nums[n-1];
        for(int i = n-1; i >= 0; i--) {
            if(nums[i] <= min) {
                min = nums[i];
            } else {
                start = i;
            }
        }
        
        return (start == n)? 0:end-start+1;
    }
}

Follow up

• Can you solve it in O(n) time complexity?