# 2964. Number of Divisible Triplet Sums

## Description

Given a 0-indexed integer array nums and an integer d, return the number of triplets (i, j, k) such that i < j < k and (nums[i] + nums[j] + nums[k]) % d == 0.

Example 1:

Input: nums = [3,3,4,7,8], d = 5
Output: 3
Explanation: The triplets which are divisible by 5 are: (0, 1, 2), (0, 2, 4), (1, 2, 4).
It can be shown that no other triplet is divisible by 5. Hence, the answer is 3.

Example 2:

Input: nums = [3,3,3,3], d = 3
Output: 4
Explanation: Any triplet chosen here has a sum of 9, which is divisible by 3. Hence, the answer is the total number of triplets which is 4.

Example 3:

Input: nums = [3,3,3,3], d = 6
Output: 0
Explanation: Any triplet chosen here has a sum of 9, which is not divisible by 6. Hence, the answer is 0.

Constraints:

• 1 <= nums.length <= 1000
• 1 <= nums[i] <= 109
• 1 <= d <= 109

## Solutions

Solution 1: Hash Table + Enumeration

We can use a hash table $cnt$ to record the occurrence times of $nums[i] \bmod d$, then enumerate $j$ and $k$, calculate the value of $nums[i] \bmod d$ that makes the equation $(nums[i] + nums[j] + nums[k]) \bmod d = 0$ hold, which is $(d - (nums[j] + nums[k]) \bmod d) \bmod d$, and accumulate its occurrence times to the answer. Then we increase the occurrence times of $nums[j] \bmod d$ by one. Continue to enumerate $j$ and $k$ until $j$ reaches the end of the array.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public int divisibleTripletCount(int[] nums, int d) {
Map<Integer, Integer> cnt = new HashMap<>();
int ans = 0, n = nums.length;
for (int j = 0; j < n; ++j) {
for (int k = j + 1; k < n; ++k) {
int x = (d - (nums[j] + nums[k]) % d) % d;
ans += cnt.getOrDefault(x, 0);
}
cnt.merge(nums[j] % d, 1, Integer::sum);
}
return ans;
}
}

• class Solution {
public:
int divisibleTripletCount(vector<int>& nums, int d) {
unordered_map<int, int> cnt;
int ans = 0, n = nums.size();
for (int j = 0; j < n; ++j) {
for (int k = j + 1; k < n; ++k) {
int x = (d - (nums[j] + nums[k]) % d) % d;
ans += cnt[x];
}
cnt[nums[j] % d]++;
}
return ans;
}
};

• class Solution:
def divisibleTripletCount(self, nums: List[int], d: int) -> int:
cnt = defaultdict(int)
ans, n = 0, len(nums)
for j in range(n):
for k in range(j + 1, n):
x = (d - (nums[j] + nums[k]) % d) % d
ans += cnt[x]
cnt[nums[j] % d] += 1
return ans

• func divisibleTripletCount(nums []int, d int) (ans int) {
n := len(nums)
cnt := map[int]int{}
for j := 0; j < n; j++ {
for k := j + 1; k < n; k++ {
x := (d - (nums[j]+nums[k])%d) % d
ans += cnt[x]
}
cnt[nums[j]%d]++
}
return
}

• function divisibleTripletCount(nums: number[], d: number): number {
const n = nums.length;
const cnt: Map<number, number> = new Map();
let ans = 0;
for (let j = 0; j < n; ++j) {
for (let k = j + 1; k < n; ++k) {
const x = (d - ((nums[j] + nums[k]) % d)) % d;
ans += cnt.get(x) || 0;
}
cnt.set(nums[j] % d, (cnt.get(nums[j] % d) || 0) + 1);
}
return ans;
}