LeetCode 0923. 3Sum With Multiplicity Solution in Java, C++, Python & More | Explanation + Code

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(n²), 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.

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)