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Count the number of pairs (i, j) such that nums1[i] is divisible by nums2[j] * k (i.e., nums1[i] % (nums2[j]*k) == 0). With n, m ≤ 50 you can brute-force all pairs or use a frequency map of nums2 values to tally matching divisibility checks.
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Leetcode 3162. Find the Number of Good Pairs I
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