Shortest Sublist to Sort - Google Top Interview Questions


Problem Statement :


Given a list of integers nums, return the length of the shortest sublist in nums which if sorted would make nums sorted in ascending order.

Constraints

n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [0, 1, 4, 3, 8, 9]

Output

2

Explanation

Sorting the sublist [4, 3] would get us [0, 1, 3, 4, 8, 9]



Example 2

Input

nums = [5, 4, 3, 2, 8, 9]

Output

4

Explanation

Sorting the sublist [5, 4, 3, 2] would get us [2, 3, 4, 5, 8, 9]



Example 3

Input

nums = [1, 2, 3, 5, 9, 8, 5]

Output

3

Explanation

Sorting the sublist [9, 8, 5] would get us [1, 2, 3, 5, 5, 8, 9]


Solution :



title-img 




                        Solution in C++ :

int solve(vector<int>& nums) {
    int left, right, lmax, rmin, N;

    N = nums.size();
    left = right = -1;
    lmax = INT_MIN;
    rmin = INT_MAX;

    for (int i = 0; i < N; ++i) {
        if (nums[i] >= lmax)
            lmax = nums[i];
        else
            right = i;

        if (nums[N - i - 1] <= rmin)
            rmin = nums[N - i - 1];
        else
            left = N - i - 1;
    }

    if (left == -1) return 0;

    return right - left + 1;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums) {
        int[] minRight = new int[nums.length];
        int[] maxLeft = new int[nums.length];
        for (int i = 0; i < nums.length; i++) {
            maxLeft[i] = (i == 0) ? nums[i] : Math.max(maxLeft[i - 1], nums[i]);
            minRight[nums.length - 1 - i] = (nums.length - 1 - i == nums.length - 1)
                ? nums[nums.length - 1]
                : Math.min(minRight[nums.length - i], nums[nums.length - 1 - i]);
        }
        int start = -1, end = -1;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != minRight[i] || nums[i] != maxLeft[i]) {
                start = i;
                break;
            }
        }
        if (start == -1)
            return 0;
        for (int i = nums.length - 1; i >= 0; i--) {
            if (nums[i] != minRight[i] || nums[i] != maxLeft[i]) {
                end = i;
                break;
            }
        }
        return end - start + 1;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums):
        if not nums:
            return 0
        left_max = nums[0]
        n = len(nums)
        end = -1
        for i in range(1, n):
            if nums[i] < left_max:
                end = i
            left_max = max(left_max, nums[i])
        if end == -1:
            return 0

        right_min = nums[-1]
        start = -1
        for i in range(n - 2, -1, -1):
            if nums[i] > right_min:
                start = i
            right_min = min(right_min, nums[i])

        return end - start + 1


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