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Commit e2c2633

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ifmagicyanglbme
andauthored
feat: add solutions to lc problem: No.1726 (doocs#869)
No.1726.Tuple with Same Product Co-authored-by: Yang Libin <contact@yanglibin.info>
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‎solution/1700-1799/1726.Tuple with Same Product/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:组合数+哈希表**
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假设存在 `n` 组数,对于其中任意两组数 `a、b` 和 `c、d`,均满足 $a * b = c * d$ 的条件,则这样的组合一共有 $\mathrm{C}_n^2 = \frac{n*(n-1)}{2}$ 个。
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根据题意每一组满足上述条件的组合可以构成 `8` 个满足题意的元组,故将各个相同乘积的组合数乘以 `8` 相加即可得到结果。
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时间复杂度 $O(n^2),ドル空间复杂度 $O(n^2)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def tupleSameProduct(self, nums: List[int]) -> int:
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cnt = defaultdict(int)
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for i in range(1, len(nums)):
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for j in range(i):
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x = nums[i] * nums[j]
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cnt[x] += 1
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return sum(v * (v - 1) // 2 for v in cnt.values()) << 3
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```
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### **Java**
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### **Go**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```go
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func tupleSameProduct(nums []int) int {
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product, n := make(map[int]int), len(nums)
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for i := 1; i < n; i++ {
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for j := 0; j < i; j++ {
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multiplier := nums[i] * nums[j]
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if _, ok := product[multiplier]; !ok {
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product[multiplier] = 0
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}
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product[multiplier] += 1
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}
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}
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ans := 0
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for _, v := range product {
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ans += v * (v - 1) / 2
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}
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return ans << 3
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}
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```
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### **...**

‎solution/1700-1799/1726.Tuple with Same Product/README_EN.md‎

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## Solutions
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**Approach 1: Number of Combinations + Hash Table**
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Time complexity $O(n^2),ドル Space complexity $O(n^2)$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def tupleSameProduct(self, nums: List[int]) -> int:
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cnt = defaultdict(int)
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for i in range(1, len(nums)):
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for j in range(i):
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x = nums[i] * nums[j]
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cnt[x] += 1
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return sum(v * (v - 1) // 2 for v in cnt.values()) << 3
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```
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### **Java**
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```java
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### **Go**
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```go
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func tupleSameProduct(nums []int) int {
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product, n := make(map[int]int), len(nums)
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for i := 1; i < n; i++ {
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for j := 0; j < i; j++ {
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multiplier := nums[i] * nums[j]
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if _, ok := product[multiplier]; !ok {
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product[multiplier] = 0
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}
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product[multiplier] += 1
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}
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}
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ans := 0
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for _, v := range product {
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ans += v * (v - 1) / 2
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}
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return ans << 3
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}
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```
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### **...**
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func tupleSameProduct(nums []int) int {
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product, n := make(map[int]int), len(nums)
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for i := 1; i < n; i++ {
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for j := 0; j < i; j++ {
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multiplier := nums[i] * nums[j]
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if _, ok := product[multiplier]; !ok {
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product[multiplier] = 0
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}
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product[multiplier] += 1
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}
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}
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ans := 0
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for _, v := range product {
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ans += v * (v - 1) / 2
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}
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return ans << 3
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}
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class Solution:
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def tupleSameProduct(self, nums: List[int]) -> int:
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cnt = defaultdict(int)
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for i in range(1, len(nums)):
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for j in range(i):
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x = nums[i] * nums[j]
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cnt[x] += 1
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return sum(v * (v - 1) // 2 for v in cnt.values()) << 3

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