16.Shortest Unsorted Continuous Subarray

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

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

Solution I : (Using sorting)

class Solution
{
public:
    int findUnsortedSubarray(vector<int> &nums)
    {

        vector<int> v = nums;

        sort(v.begin(), v.end());

        int start = -1, end = -1;
        for (int i = 0; i < nums.size(); i++)
        {
            if (nums[i] != v[i])
            {
                start = i;
                break;
            }
        }

        for (int i = nums.size() - 1; i >= 0; i--)
        {
            if (nums[i] != v[i])
            {
                end = i;
                break;
            }
        }
        
        if(start == end){
            return 0;
        }
        return (end - start) + 1;
    }
};

Time Complexity : O(N log N), Space Complexity: O(N)

Solution II: (Using Stack)

Approach: Finding the correct position of the minimum element in the unsorted subarray helps to determine the required left boundary. Similarly, the correct position of the maximum element in the unsorted subarray helps to determine the required right boundary.

Unsorted_subarray

class Solution
{
public:
    int findUnsortedSubarray(vector<int> &nums)
    {

        int n = nums.size();
        stack<int> s;
        s.push(0);
        int minStart = n;
        int maxEnd = -1;

        for (int i = 1; i < nums.size(); i++)
        {
            if (nums[i] < nums[s.top()])
            {

                while (!s.empty() && nums[s.top()] > nums[i])
                {
                    s.pop();
                }

                if (!s.empty() && s.top() < minStart)
                {
                    minStart = s.top() + 1;
                }
                else if (s.empty())
                {
                    minStart = 0;
                }
            }

            s.push(i);
        }

        stack<int> rs;
        rs.push(n - 1);

        for (int i = n - 2; i >= 0; i--)
        {
            if (nums[i] > nums[rs.top()])
            {
                while (!rs.empty() && nums[rs.top()] < nums[i])
                {
                    rs.pop();
                }

                if (!rs.empty() && rs.top() > maxEnd)
                {
                    maxEnd = rs.top() - 1;
                }
                else if (rs.empty())
                {
                    maxEnd = n - 1;
                }
            }

            rs.push(i);
        }

        return ((maxEnd - minStart) > 0) ? (maxEnd - minStart + 1) : 0;
    }
};

Time Complexity : O(N), Space Complexity: O(N)