Formatted question description: https://leetcode.ca/all/1726.html
1726. Tuple with Same Product
Level
Medium
Description
Given an array nums
of distinct positive integers, return the number of tuples (a, b, c, d)
such that a * b = c * d
where a
, b
, c
, and d
are elements of nums
, and a != b != c != d
.
Example 1:
Input: nums = [2,3,4,6]
Output: 8
Explanation: There are 8 valid tuples:
(2,6,3,4) , (2,6,4,3) , (6,2,3,4) , (6,2,4,3)
(3,4,2,6) , (3,4,2,6) , (3,4,6,2) , (4,3,6,2)
Example 2:
Input: nums = [1,2,4,5,10]
Output: 16
Explanation: There are 16 valids tuples:
(1,10,2,5) , (1,10,5,2) , (10,1,2,5) , (10,1,5,2)
(2,5,1,10) , (2,5,10,1) , (5,2,1,10) , (5,2,10,1)
(2,10,4,5) , (2,10,5,4) , (10,2,4,5) , (10,2,4,5)
(4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)
Example 3:
Input: nums = [2,3,4,6,8,12]
Output: 40
Example 4:
Input: nums = [2,3,5,7]
Output: 0
Constraints:
1 <= nums.length <= 1000
1 <= nums[i] <= 10^4
- All elements in
nums
are distinct.
Solution
Loop over all pairs of elements in nums
. For each pair of elements, calculate the two elements’ product. Use a hash map to store each product and the number of pairs (a, b)
such that a * b
equals the product.
Then loop over all entries of the map. For each product, if the number of pairs is count
, then the number of tuples of the same product as the current product is count * (count - 1) * 4
.
Finally, return the number of tuples.
class Solution {
public int tupleSameProduct(int[] nums) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
int length = nums.length;
for (int i = 0; i < length; i++) {
int num1 = nums[i];
for (int j = i + 1; j < length; j++) {
int num2 = nums[j];
int product = num1 * num2;
int count = map.getOrDefault(product, 0) + 1;
map.put(product, count);
}
}
int tuples = 0;
Set<Integer> keySet = map.keySet();
for (int product : keySet) {
int count = map.get(product);
tuples += count * (count - 1) * 4;
}
return tuples;
}
}