nums of length n and an integer k, return the number of pairs (i, j) where 0 <= i < j < n, such that nums[i] == nums[j] and (i * j) is divisible by k.
Example 1:
Input: nums = [3,1,2,2,2,1,3], k = 2 Output: 4 Explanation: There are 4 pairs that meet all the requirements: - nums[0] == nums[6], and 0 * 6 == 0, which is divisible by 2. - nums[2] == nums[3], and 2 * 3 == 6, which is divisible by 2. - nums[2] == nums[4], and 2 * 4 == 8, which is divisible by 2. - nums[3] == nums[4], and 3 * 4 == 12, which is divisible by 2.
Example 2:
Input: nums = [1,2,3,4], k = 1 Output: 0 Explanation: Since no value in nums is repeated, there are no pairs (i,j) that meet all the requirements.
Constraints:
1 <= nums.length <= 1001 <= nums[i], k <= 100class Solution {
public int countPairs(int[] nums, int k) {
int n = nums.length, count = 0;
for (int i = 1; i < n; i++)
for (int j = 0; j < i; j++)
if (nums[i] == nums[j] && (i * j) % k == 0)
count++;
return count;
}
}class Solution {
public int countPairs(int[] nums, int k) {
List<Integer>[] seen = new List[101];
int n = nums.length, count = 0;
for (int i = 0; i < 101; i++)
seen[i] = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j : seen[nums[i]])
if ((i * j) % k == 0)
count++;
seen[nums[i]].add(i);
}
return count;
}
}class Solution {
public int countPairs(int[] nums, int k) {
List<Integer>[] seen = new List[101];
int n = nums.length, count = 0;
for (int i = 0; i < 101; i++)
seen[i] = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j : seen[nums[i]])
if ((i * j) % k == 0)
count++;
seen[nums[i]].add(i);
}
return count;
}
}