LeetCode-in-Java

923. 3Sum With Multiplicity

Medium

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.

Constraints:

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

Solution

public class Solution {
    private static final int MOD = (int) 1e9 + 7;
    private static final int MAX = 100;

    public int threeSumMulti(int[] arr, int target) {
        int answer = 0;
        int[] countRight = new int[MAX + 1];
        for (int num : arr) {
            ++countRight[num];
        }
        int[] countLeft = new int[MAX + 1];
        for (int j = 0; j < arr.length - 1; j++) {
            --countRight[arr[j]];
            int remains = target - arr[j];
            if (remains <= 2 * MAX) {
                for (int v = 0; v <= Math.min(remains, MAX); v++) {
                    if (remains - v <= MAX) {
                        int count = countRight[v] * countLeft[remains - v];
                        if (count > 0) {
                            answer = (answer + count) % MOD;
                        }
                    }
                }
            }
            ++countLeft[arr[j]];
        }
        return answer;
    }
}

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