# O(n2) time complexity java 2 pointers w explaination

• a, b, a and are triangle sides when a + b > c, a+c > b, and b+c > a. If a, b, c in ascending order then only thing needs to check is a+b < c.

Sorting array helps reduce number of checks and can use 2 pointers. Same as 3sum problem.
lo and hi pointer only visits nums elements once. For each lo increasing, hi doesn't need to be reset because if nums[i] + nums[lo] < nums[hi] then of course nums[i] + nums[lo+1] < nums[hi]. So when increase lo, hi should try to go further to the end of array.

``````    public int triangleNumber(int[] nums) {
if (nums.length < 3)
return 0;

int count = 0;
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i ++){

int lo = i + 1;
int hi = lo;

while(lo < nums.length - 1){

while(hi + 1 < nums.length && nums[hi + 1] < nums[i] + nums[lo])
hi++;
if (hi - lo > 0)
count += hi - lo;

lo++;

}
}

return count;

}

``````

• @bellevue of course you meant a + b > c

• @zinking yes that's what i meant :)

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