K-diff Pairs | C++ | 2 pointer O(nlog(n)) O(1) | Beats 100% in time, 99.90 in mem.
        while(i<j && j<nums.size())
        {
            if(i>0 && nums[i]==nums[i-1])
            {
                i++;
                if(i==j)
                    j++;
            }
            else if(nums[j]-nums[i]>k)
            {
                i++;
                if(i==j)
                    j++;
            }
            else if(nums[j]-nums[i]<k)
            {
                j++;
            }
            else
            {
                count++;
                i++;
                j++;
            }
        }
	```
Comments (0)