Problem statement

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 10^9 + 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 1: Bruteforce

Find all tuples that satisfy the target sum.

Code

#include <vector>
#include <iostream>
using namespace std;
int threeSumMulti(vector<int>& arr, int target) {
    const int MOD = 1000000007;
    int count = 0;
    for (int i = 0; i < arr.size() - 2; i++) {
        for (int j = i + 1; j < arr.size() - 1; j++) {
            for (int k = j + 1; k < arr.size(); k++) {
                if (arr[i] + arr[j] + arr[k] == target) {
                    count += 1 % MOD;
                }
            }
        }
    }
    return count;
}
int main() {
    vector<int> arr{1,1,2,2,3,3,4,4,5,5};
    cout << threeSumMulti(arr, 8) << endl;
    arr = {1,1,2,2,2,2};
    cout << threeSumMulti(arr, 5) << endl;
}

Output:
20
12

Complexity

• Runtime: O(N^3), where N = arr.length.
• Extra space: O(1).

Solution 2: Store one of the values in the tuples

Rewrite the condition arr[i] + arr[j] + arr[k] == target as arr[i] == target - arr[j] - arr[k], then you can reduce one of the for loops by using a map to store the count of the values arr[i] that have been visited.

Code

#include <vector>
#include <iostream>
#include <unordered_map>
using namespace std;
int threeSumMulti(vector<int>& arr, int target) {
    unordered_map<int, int> m;
    const int MOD = 1000000007;
    int count = 0;
    for (int i = 0; i < arr.size() - 1; i++) {
        for (int j = i + 1; j < arr.size(); j++) {
            auto it = m.find(target - arr[i] - arr[j]);
            if (it != m.end()) {
                count = (count + it->second) % MOD;
            }
        }
        m[arr[i]]++;
    }
    return count;
}
int main() {
    vector<int> arr{1,1,2,2,3,3,4,4,5,5};
    cout << threeSumMulti(arr, 8) << endl;
    arr = {1,1,2,2,2,2};
    cout << threeSumMulti(arr, 5) << endl;
}

Output:
20
12

Complexity

• Runtime: O(N^2), where N = arr.length.
• Extra space: O(N).