C++ Solution to Coding Challenge 581. Shortest Unsorted Continuous Subarray

Another way of sorting special arrays

Nhut Nguyen

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Problem statement

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 <= 10^4.
  • -10^5 <= nums[i] <= 10^5.

Follow up: Can you solve it in O(n) time complexity?

Solution 1: Sort and compare the difference

Example 1

Comparing nums = [2,6,4,8,10,9,15] with its sorted one sortedNums = [2,4,6,8,9,10,15]:

  • The first position that makes the difference is left = 1, where 6 != 4.
  • The last position that makes the difference is right = 5, where 9 != 10.
  • The length of that shortest subarray is right - left + 1 = 5.

Code

#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int findUnsortedSubarray(vector<int>& nums) {
vector<int> sortedNums = nums;
sort(sortedNums.begin(), sortedNums.end());
int left = 0;
while (left < nums.size() && nums[left] == sortedNums[left]) {
left++;
}
int right = nums.size() - 1;
while (right >= 0 && nums[right] == sortedNums[right]) {
right--;
}
return left >= right ? 0 : right - left + 1;
}
int main() {
vector<int> nums{2,6,4,8,10,9,15};
cout << findUnsortedSubarray(nums) << endl;
nums = {1,2,3,4};
cout << findUnsortedSubarray(nums) << endl;
nums = {1};
cout << findUnsortedSubarray(nums)…

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