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-rw-r--r-- | count_equal_div_pair_arr.c | 23 | ||||
-rw-r--r-- | count_equal_div_pair_arr.py | 37 |
2 files changed, 60 insertions, 0 deletions
diff --git a/count_equal_div_pair_arr.c b/count_equal_div_pair_arr.c new file mode 100644 index 0000000..d6aeb12 --- /dev/null +++ b/count_equal_div_pair_arr.c @@ -0,0 +1,23 @@ +// 2176. Count Equal and Divisible Pairs in an Array +#include "leetcode.h" + +/* + * given a 0-indexed integer array '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 divisble by 'k'. + */ + +int countPairs(int *nums, int numsSize, int k) { + int cnt = 0; + for (int i = 0; i < numsSize; i++) + for (int j = i + 1; j < numsSize; j++) + if (nums[i] == nums[j] && !((i * j) % k)) + cnt++; + return cnt; +} + +int main() { + int n1[] = {3, 1, 2, 2, 2, 1, 3}, n2[] = {1, 2, 3, 4}; + printf("%d\n", countPairs(n1, ARRAY_SIZE(n1), 2)); // expect: 4 + printf("%d\n", countPairs(n2, ARRAY_SIZE(n2), 1)); // expect: 0 +} diff --git a/count_equal_div_pair_arr.py b/count_equal_div_pair_arr.py new file mode 100644 index 0000000..84d4a55 --- /dev/null +++ b/count_equal_div_pair_arr.py @@ -0,0 +1,37 @@ +# 2176. Count Equal and Divisible Pairs in an Array +import collections +from math import gcd + +""" +given a 0-indexed integer array '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 divisble by 'k'. +""" + + +class Solution(object): + def countPairs(self, nums, k): + """ + :type nums: List[int] + :type k: int + :rtype: int + """ + ids = collections.defaultdict(list) + for i, j in enumerate(nums): + ids[j].append(i) + ans = 0 + for i in ids.values(): + gcds = collections.Counter() + for j in i: + curr_i = gcd(j, k) + ans += sum( + cnt for curr_j, cnt in gcds.items() if curr_i * curr_j % k == 0 + ) + gcds[curr_i] += 1 + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.countPairs(nums=[3, 1, 2, 2, 2, 1, 3], k=2)) + print(obj.countPairs(nums=[1, 2, 3, 4], k=1)) |