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1726.tuple-with-same-product.cpp
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57 lines (53 loc) · 1.45 KB
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// Tag: Array, Hash Table, Counting
// Time: O(N^2)
// Space: O(N^2)
// Ref: -
// Note: -
// Video: https://youtu.be/6PvtDyA-IPw
// 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) , (4,3,2,6) , (3,4,6,2) , (4,3,6,2)
//
// Example 2:
//
// Input: nums = [1,2,4,5,10]
// Output: 16
// Explanation: There are 16 valid 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) , (10,2,5,4)
// (4,5,2,10) , (4,5,10,2) , (5,4,2,10) , (5,4,10,2)
//
//
// Constraints:
//
// 1 <= nums.length <= 1000
// 1 <= nums[i] <= 104
// All elements in nums are distinct.
//
//
class Solution {
public:
int tupleSameProduct(vector<int>& nums) {
int n = nums.size();
unordered_map<int, int> counter;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
counter[nums[i] * nums[j]] += 1;
}
}
int res = 0;
for (auto &[prod, count]: counter) {
if (count > 1) {
res += count * (count - 1) * 4;
}
}
return res;
}
};