feat: add solutions to lc problem: No.1726 (#869)

No.1726.Tuple with Same Product

Co-authored-by: Yang Libin <contact@yanglibin.info>

Author: poltao

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

@@ -46,22 +46,53 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：组合数+哈希表**
+
+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。
+
+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。
+
+时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def tupleSameProduct(self, nums: List[int]) -> int:
+        cnt = defaultdict(int)
+        for i in range(1, len(nums)):
+            for j in range(i):
+                x = nums[i] * nums[j]
+                cnt[x] += 1
+        return sum(v * (v - 1) // 2 for v in cnt.values()) << 3
 ```
 
-### **Java**
+### **Go**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
-```java
-
+```go
+func tupleSameProduct(nums []int) int {
+	product, n := make(map[int]int), len(nums)
+	for i := 1; i < n; i++ {
+		for j := 0; j < i; j++ {
+			multiplier := nums[i] * nums[j]
+			if _, ok := product[multiplier]; !ok {
+				product[multiplier] = 0
+			}
+			product[multiplier] += 1
+		}
+	}
+	ans := 0
+	for _, v := range product {
+		ans += v * (v - 1) / 2
+	}
+	return ans << 3
+}
 ```
 
 ### **...**

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

@@ -40,18 +40,45 @@
 
 ## Solutions
 
+**Approach 1: Number of Combinations + Hash Table**
+
+Time complexity $O(n^2)$, Space complexity $O(n^2)$.
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 ```python
-
+class Solution:
+    def tupleSameProduct(self, nums: List[int]) -> int:
+        cnt = defaultdict(int)
+        for i in range(1, len(nums)):
+            for j in range(i):
+                x = nums[i] * nums[j]
+                cnt[x] += 1
+        return sum(v * (v - 1) // 2 for v in cnt.values()) << 3
 ```
 
-### **Java**
-
-```java
-
+### **Go**
+
+```go
+func tupleSameProduct(nums []int) int {
+	product, n := make(map[int]int), len(nums)
+	for i := 1; i < n; i++ {
+		for j := 0; j < i; j++ {
+			multiplier := nums[i] * nums[j]
+			if _, ok := product[multiplier]; !ok {
+				product[multiplier] = 0
+			}
+			product[multiplier] += 1
+		}
+	}
+	ans := 0
+	for _, v := range product {
+		ans += v * (v - 1) / 2
+	}
+	return ans << 3
+}
 ```
 
 ### **...**

solution/1700-1799/1726.Tuple with Same Product/Solution.go

@@ -0,0 +1,17 @@
+func tupleSameProduct(nums []int) int {
+	product, n := make(map[int]int), len(nums)
+	for i := 1; i < n; i++ {
+		for j := 0; j < i; j++ {
+			multiplier := nums[i] * nums[j]
+			if _, ok := product[multiplier]; !ok {
+				product[multiplier] = 0
+			}
+			product[multiplier] += 1
+		}
+	}
+	ans := 0
+	for _, v := range product {
+		ans += v * (v - 1) / 2
+	}
+	return ans << 3
+}

solution/1700-1799/1726.Tuple with Same Product/Solution.py

@@ -0,0 +1,8 @@
+class Solution:
+    def tupleSameProduct(self, nums: List[int]) -> int:
+        cnt = defaultdict(int)
+        for i in range(1, len(nums)):
+            for j in range(i):
+                x = nums[i] * nums[j]
+                cnt[x] += 1
+        return sum(v * (v - 1) // 2 for v in cnt.values()) << 3