LeetCode 1577: Number of Ways Where Square of Number Is Equal to Product of Two Numbers
LeetCode 1577 Solution Explanation
Explanation:
To solve this problem, we can iterate through all possible combinations of (i, j, k) for both Type 1 and Type 2 triplets, and check if the conditions specified in the problem statement are satisfied. We will maintain two hash maps to store the frequencies of squares in nums1 and products in nums2 respectively, to make the checking process more efficient.
- For Type 1 triplets, we iterate through all pairs (j, k) in nums2 and calculate the square of nums2[j] * nums2[k]. We then check if this square exists in the map for squares of nums1.
- For Type 2 triplets, we iterate through all pairs (j, k) in nums1 and calculate the square of nums1[j] * nums1[k]. We then check if this square exists in the map for products in nums2.
Finally, we return the total count of valid triplets found.
Time Complexity: O(N^2), where N is the size of the larger array between nums1 and nums2.
Space Complexity: O(N), to store the frequencies of squares and products in hash maps.
:
LeetCode 1577 Solutions in Java, C++, Python
class Solution {
public int numTriplets(int[] nums1, int[] nums2) {
return countTriplets(nums1, nums2) + countTriplets(nums2, nums1);
}
private int countTriplets(int[] nums1, int[] nums2) {
int count = 0;
Map<Long, Integer> squares = new HashMap<>();
for (int i = 0; i < nums1.length; i++) {
long square = (long) nums1[i] * nums1[i];
squares.put(square, squares.getOrDefault(square, 0) + 1);
}
for (int j = 0; j < nums2.length; j++) {
for (int k = j + 1; k < nums2.length; k++) {
long product = (long) nums2[j] * nums2[k];
if (squares.containsKey(product)) {
count += squares.get(product);
}
}
}
return count;
}
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