README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0900-0999/0923.3Sum%20With%20Multiplicity/README_EN.md	Array Hash Table Two Pointers Counting Sorting

923. 3Sum With Multiplicity

中文文档

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

We can use a hash table or an array $cnt$ of length $101$ to count the occurrence of each element in the array $arr$.

Then, we enumerate each element $arr[j]$ in the array $arr$, first subtract one from $cnt[arr[j]]$, and then enumerate the elements $arr[i]$ before $arr[j]$, calculate $c = target - arr[i] - arr[j]$. If $c$ is in the range of $[0, 100]$, then the answer is increased by $cnt[c]$, 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 $O(n^2)$, where $n$ is the length of the array $arr$. The space complexity is $O(C)$, where $C$ is the maximum value of the elements in the array $arr$, in this problem $C = 100$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}