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923. 3Sum With Multiplicity

Description

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

Solutions

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

    public int threeSumMulti(int[] arr, int target) {
        int[] cnt = new int[101];
        for (int v : arr) {
            ++cnt[v];
        }
        long ans = 0;
        for (int j = 0; j < arr.length; ++j) {
            int b = arr[j];
            --cnt[b];
            for (int i = 0; i < j; ++i) {
                int a = arr[i];
                int c = target - a - b;
                if (c >= 0 && c <= 100) {
                    ans = (ans + cnt[c]) % MOD;
                }
            }
        }
        return (int) ans;
    }
}

  • class Solution {
public:
    const int mod = 1e9 + 7;

    int threeSumMulti(vector<int>& arr, int target) {
        int cnt[101] = {0};
        for (int& v : arr) {
            ++cnt[v];
        }
        long ans = 0;
        for (int j = 0; j < arr.size(); ++j) {
            int b = arr[j];
            --cnt[b];
            for (int i = 0; i < j; ++i) {
                int a = arr[i];
                int c = target - a - b;
                if (c >= 0 && c <= 100) {
                    ans += cnt[c];
                    ans %= mod;
                }
            }
        }
        return ans;
    }
};

  • class Solution:
    def threeSumMulti(self, arr: List[int], target: int) -> int:
        cnt = Counter(arr)
        ans = 0
        mod = 10**9 + 7
        for j, b in enumerate(arr):
            cnt[b] -= 1
            for i in range(j):
                a = arr[i]
                c = target - a - b
                ans = (ans + cnt[c]) % mod
        return ans


  • func threeSumMulti(arr []int, target int) int {
	const mod int = 1e9 + 7
	cnt := [101]int{}
	for _, v := range arr {
		cnt[v]++
	}
	ans := 0
	for j, b := range arr {
		cnt[b]--
		for i := 0; i < j; i++ {
			a := arr[i]
			c := target - a - b
			if c >= 0 && c <= 100 {
				ans += cnt[c]
				ans %= mod
			}
		}
	}
	return ans
}

  • function threeSumMulti(arr: number[], target: number): number {
    const mod = 10 ** 9 + 7;
    const cnt: number[] = Array(101).fill(0);
    for (const x of arr) {
        ++cnt[x];
    }
    let ans = 0;
    const n = arr.length;
    for (let j = 0; j < n; ++j) {
        --cnt[arr[j]];
        for (let i = 0; i < j; ++i) {
            const c = target - arr[i] - arr[j];
            if (c >= 0 && c < cnt.length) {
                ans = (ans + cnt[c]) % mod;
            }
        }
    }
    return ans;
}

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