# Shortest Unsorted Continuous Subarray

Given an integer array nums, you need to find one **continuous subarray** such that if you only sort this subarray in non-decreasing order, then the whole array will be sorted in non-decreasing order.

Return *the shortest such subarray and output its length*.

```
class Solution {
public int findUnsortedSubarray(int[] nums) {
int end = -1;
int max = nums[0];
for(int i = 1; i < nums.length; i++){
if(max > nums[i]){ // the left value is greater then current value
end = i; // mark that index with end
}
else{
max = nums[i];
}
}
int start = 0;
int min = nums[nums.length - 1];
for(int i = nums.length - 2; i >= 0; i--){
if(min < nums[i]){ // the right value is smaller then current value
start = i; // mark that index with start
}
else{
min = nums[i];
}
}
return end - start + 1;
}
}
```

## Related Problems

Given a string s, find the length of the **longest** **substring **without repeating characters.

Given a string **S** and an integer **K**, return *the length of the longest substring of* **S** *that contains at most* **K** **distinct*** characters*.

Given an array of strings strs, group **the anagrams** together. You can return the answer in **any order**.

An **Anagram** is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals* after the insertion*.

**Note** that you don't need to modify intervals in-place. You can make a new array and return it.