# [C++] Clean Code 2 Solution - Sort O(NlgN) & max min vectors O(N)

• Sort

``````class Solution {
public:
int findUnsortedSubarray(vector<int>& nums) {
vector<int> sorted(nums);
sort(sorted.begin(), sorted.end());
int n = nums.size(), i = 0, j = n - 1;
while (i < n && nums[i] == sorted[i]) {
i++;
}
while (j > i && nums[j] == sorted[j]) {
j--;
}
return j + 1 - i;
}
};
``````

max on left, min on right - O(N)
The idea is that:

1. From the left, for a number to stay untouched, there need to be no number less than it on the right hand side;
2. From the right, for a number to stay untouched, there need to be no number greater than it on the left hand side;

Those 2 can be easily checked if we build 2 vectors:

1. max so far from the left,
2. and min so far from the right;
``````/**
*            /------------\
* nums:  [2, 6, 4, 8, 10, 9, 15]
* minr:   2  4  4  8   9  9  15
*         <--------------------
* maxl:   2  6  6  8  10 10  15
*         -------------------->
*/
class Solution {
public:
int findUnsortedSubarray(vector<int>& nums) {
int n = nums.size();
vector<int> maxlhs(n);   // max number from left to cur
vector<int> minrhs(n);   // min number from right to cur
for (int i = n - 1, minr = INT_MAX; i >= 0; i--) minrhs[i] = minr = min(minr, nums[i]);
for (int i = 0,     maxl = INT_MIN; i < n;  i++) maxlhs[i] = maxl = max(maxl, nums[i]);

int i = 0, j = n - 1;
while (i < n && nums[i] <= minrhs[i]) i++;
while (j > i && nums[j] >= maxlhs[j]) j--;

return j + 1 - i;
}
};
``````

• Given input - 2 6 4 8 10 9 15
Please explain. I'm dry running in this code and getting output 3.
Thanks

• @aryan_gupta I get 5, can you retry?

• @alexander i tried. Thank alot. I understood.

• This post is deleted!

• @alexander The first solution is easy to think of. And the second solution is awesome to solve problem in O(n) space and O(n) time. Thank you for your post!

• nice solution!

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