# Could someone confirm if this is O(n^2)?

• Although it has 3 loops, the innermost while loop does not reset its loop variable everytime. Here is my logic:

I keep 3 pointers n1p, n2p, n3p.

n2p is the pointer for the middle number, and increments by 1 in outermost loop.

For each n2p, I fix a n1p such that n1p < n2p. This increments in the inner loop. Already we have reached O(n^2).

Now comes the important part. For the very first n1p iteration (n1p==0), I start n3p from the rightmost position. I move it left until the desired condition w/ target is hit. However, n3p does not reset in the inner for/while loop for future iterations of n1p. So as n1p keeps moving right every inner iteration, n3p moves a few positions to the left as well. Since n1p and n2p are fixed, the number of combos for this pair of (n1p, n2p) is the difference between n2p and n3p.

``````class Solution {
public:
int threeSumSmaller(vector<int>& nums, int target) {
int combos = 0;
std::sort(nums.begin(), nums.end());
for (int n2p=0; n2p<nums.size(); n2p++) {       //this is the middle number
int n3p = nums.size()-1;                    //Don't reset this in inner loop! Important.
for (int n1p=0; n1p < n2p && n2p < n3p; n1p++) {//this is the first number
while(n3p>n2p && ((nums[n1p] + nums[n2p]) >= (target - nums[n3p]))) {
n3p--;
}
combos += (n3p - n2p);
}
}
return combos;
}
};``````

• Yeah, that's O(n^2). Interesting variation, to use the middle index for the main loop. I think most people use the first one, and I use the last one (because I then need nums.size() only once).

I'd just prefer index variables i,j,k, as they're really standard and are also used in the problem statement.

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