2183. Count Array Pairs Divisible by K
Description
Given a 0-indexed integer array nums of length n and an integer k, return the number of pairs (i, j) such that:
0 <= i < j <= n - 1andnums[i] * nums[j]is divisible byk.
Example 1:
Input: nums = [1,2,3,4,5], k = 2 Output: 7 Explanation: The 7 pairs of indices whose corresponding products are divisible by 2 are (0, 1), (0, 3), (1, 2), (1, 3), (1, 4), (2, 3), and (3, 4). Their products are 2, 4, 6, 8, 10, 12, and 20 respectively. Other pairs such as (0, 2) and (2, 4) have products 3 and 15 respectively, which are not divisible by 2.
Example 2:
Input: nums = [1,2,3,4], k = 5 Output: 0 Explanation: There does not exist any pair of indices whose corresponding product is divisible by 5.
Constraints:
1 <= nums.length <= 1051 <= nums[i], k <= 105
Solutions
Solution: Hash Map + Math
- Time complexity: O(n√k)
- Space complexity: O(n)
JavaScript
js
/**
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
const countPairs = function (nums, k) {
const gcdMap = new Map();
let result = 0;
const gcd = (a, b) => (b ? gcd(b, a % b) : a);
for (const num of nums) {
const gcdA = gcd(num, k);
for (const [gcdB, count] of gcdMap) {
if ((gcdA * gcdB) % k === 0) {
result += count;
}
}
const count = gcdMap.get(gcdA) ?? 0;
gcdMap.set(gcdA, count + 1);
}
return result;
};