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923. 3Sum With Multiplicity

DifficultyMedium

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

Solution 1: Counting + Enumeration

Thinking

Count triples summing to \(\textit{target}\). Values lie in \([0,100]\) and \(n\le 3000\), so a cubic enumeration is tight. Count frequencies, then enumerate \(i<j\); the third value \(c=\textit{target}-a-b\) must lie after \(j\).

Decrement \(b\) before looking up \(c\), so the counter only represents the suffix, and each lookup is \(O(1)\).

We can use a hash table or an array \(\textit{cnt}\) of length \(101\) to count the occurrence of each element in the array \(\textit{arr}\).

Then, we enumerate each element \(\textit{arr}[j]\) in the array \(\textit{arr}\), first subtract one from \(\textit{cnt}[\textit{arr}[j]]\), and then enumerate the elements \(\textit{arr}[i]\) before \(\textit{arr}[j]\), calculate \(c = \textit{target} - \textit{arr}[i] - \textit{arr}[j]\). If \(c\) is in the range of \([0, 100]\), then the answer is increased by \(\textit{cnt}[c]\), and finally return the answer.

Note that the answer may exceed \({10}^9 + 7\), so take the modulus after each addition operation.

The time complexity is \(O(n^2)\), and the space complexity is \(O(C)\), where \(n\) is the length of the array \(\textit{arr}\), and \(C\) is the maximum value of the elements in the array \(\textit{arr}\), in this problem \(C = 100\).

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

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class Solution {
    public int threeSumMulti(int[] arr, int target) {
        final int mod = (int) 1e9 + 7;
        int[] cnt = new int[101];
        for (int x : arr) {
            ++cnt[x];
        }
        int n = arr.length;
        int ans = 0;
        for (int j = 0; j < n; ++j) {
            --cnt[arr[j]];
            for (int i = 0; i < j; ++i) {
                int c = target - arr[i] - arr[j];
                if (c >= 0 && c < cnt.length) {
                    ans = (ans + cnt[c]) % mod;
                }
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int threeSumMulti(vector<int>& arr, int target) {
        const int mod = 1e9 + 7;
        int cnt[101]{};
        for (int x : arr) {
            ++cnt[x];
        }
        int n = arr.size();
        int ans = 0;
        for (int j = 0; j < n; ++j) {
            --cnt[arr[j]];
            for (int i = 0; i < j; ++i) {
                int c = target - arr[i] - arr[j];
                if (c >= 0 && c <= 100) {
                    ans = (ans + cnt[c]) % mod;
                }
            }
        }
        return ans;
    }
};

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

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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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