923. 3Sum With Multiplicity

Problem

Given an integer array arr, and an integer target, return the number of tuples i, j, k such that i < j < k and arr[i] + arr[j] + arr[k] == target.
As the answer can be very large, return it modulo 109 + 7.

Example 1:
Input: arr = [1,1,2,2,3,3,4,4,5,5], target = 8
Output: 20
Explanation: 
Enumerating by the values (arr[i], arr[j], arr[k]):
(1, 2, 5) occurs 8 times;
(1, 3, 4) occurs 8 times;
(2, 2, 4) occurs 2 times;
(2, 3, 3) occurs 2 times.

Example 2:
Input: arr = [1,1,2,2,2,2], target = 5
Output: 12
Explanation: 
arr[i] = 1, arr[j] = arr[k] = 2 occurs 12 times:
We choose one 1 from [1,1] in 2 ways,
and two 2s from [2,2,2,2] in 6 ways.

Example 3:
Input: arr = [2,1,3], target = 6
Output: 1
Explanation: (1, 2, 3) occured one time in the array so we return 1.

Constraints:
3 <= arr.length <= 3000
0 <= arr[i] <= 100
0 <= target <= 300

Intuition

// TODO

Approach

// TODO

Implementation

class Solution {
    private static final int MOD = 1_000_000_007;
    public int threeSumMulti(int[] arr, int target) {
        int n = arr.length;

        Arrays.sort(arr);

        int count = 0;
        for (int i = 0; i < n; i++) {
            int remain = target - arr[i];
            if (remain < 0) continue;

            int l = i+1; int r = n-1;
            while (l < r) {
                int s = arr[l] + arr[r];
                if (s == remain) {
                    int countDupL = 1;
                    while (l < r && arr[l] == arr[l+1]) {
                        l++;
                        countDupL++;
                    }

                    int countDupR = 1;
                    while (l < r && arr[r] == arr[r-1]) {
                        r--;
                        countDupR++;
                    }

                    if (l == r) {
                        count = (count + toHop2(countDupL)) % MOD;
                    } else {
                        count = (count + countDupL * countDupR) % MOD;
                        l++;
                        r--;
                    }
                } else if (s > remain) {
                    r--;
                } else {
                    l++;
                }
            }
        }   

        return count;
    }

    // nC2
    private int toHop2(int n) {
        return n * (n-1) / 2;
    }
}