Reference: LeetCode
Difficulty: Medium

My Post: [Java] It’s Easy Because It Allows Duplicate Results!

Problem

Given an array of n integers nums and a target, find the number of index triplets i, j, k with 0 <= i < j < k < n that satisfy the condition nums[i] + nums[j] + nums[k] < target.

Note: We also count the duplicate pairs!

Example: 

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Input: nums = [-2,0,1,3], and target = 2
Output: 2 
Explanation: Because there are two triplets which sums are less than 2:
             [-2,0,1]
             [-2,0,3]

Follow up: Could you solve it in $O(N^2)$ runtime?

Analysis

Two Pointers

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public int threeSumSmaller(int[] nums, int target) {
  int n = nums.length;
  Arrays.sort(nums); // sort
  int count = 0;
  for (int i = 0; i < n; ++i) {
    // two pointers
    int j = i + 1, k = n - 1;  
    int t = target - nums[i];
    while (j < k) { // each element is only used once
      if (nums[j] + nums[k] < t) {
        //  0  1  2  3  4 <- index
        //  1  2  3  4  5
        //  i  j        k <- say we have a valid pair [1, 2, 5]
        // we have [1 2 5], [1 2 4], [1 2 3]. The count equals k - j
        count += (k - j);
        ++j;
      } else { // nums[j] + nums[k] >= t
        --k;
      }
    }
  }
  return count;
}

Time: $O(N^2)$
Space: $O(1)$