Number of K-Divisible Pairs - Amazon Top Interview Questions
Problem Statement :
You are given a list of integers nums and an integer k. Return the number of pairs i < j such that (nums[i] + nums[j]) % k = 0 Constraints n ≤ 100,000 where n is the length of nums 1 ≤ k ≤ 100 Example 1 Input nums = [2, 4, 5, 1, 2] k = 6 Output 3 Explanation We have the following pairs [2, 4] [4, 2] [5, 1] Example 2 Input nums = [3, 7, 2] k = 7 Output 0
Solution :
Solution in C++ :
int solve(vector<int>& nums, int k) {
unordered_map<int, int> m;
int ans = 0;
for (auto& n : nums) {
ans += m[(k - n % k) % k];
m[n % k] += 1;
}
return ans;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int k) {
int res = 0;
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
int mod = nums[i] % k;
if (map.containsKey((k - mod) % k)) {
res += map.get((k - mod) % k);
}
map.put(mod, map.getOrDefault(mod, 0) + 1);
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, nums, k):
c = 0
a = {}
for x in nums:
y = x % k
c += a.get((k - y) % k, 0)
a[y] = a.get(y, 0) + 1
return c
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