feat: add solutions to lc problem: No.1726 (#1845)

No.1726.Tuple with Same Product

Author: yanglbme

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

@@ -46,13 +46,13 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-**方法一：组合数+哈希表**
+**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。
+假设存在 $n$ 组数，对于其中任意两组数 $a, b$ 和 $c, d$，均满足 $a \times b = c \times d$ 的条件，则这样的组合一共有 $\mathrm{C}_n^2 = \frac{n \times (n-1)}{2}$ 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。
+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（等价于：左移 $3$ 位）即可得到结果。
 
-时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。
+时间复杂度 $O(n^2)$，空间复杂度 $O(n^2)$。其中 $n$ 为数组长度。
 
 <!-- tabs:start -->
 
@@ -82,7 +82,7 @@ class Solution {
         for (int i = 1; i < nums.length; ++i) {
             for (int j = 0; j < i; ++j) {
                 int x = nums[i] * nums[j];
-                cnt.put(x, cnt.getOrDefault(x, 0) + 1);
+                cnt.merge(x, 1, Integer::sum);
             }
         }
         int ans = 0;
@@ -135,6 +135,51 @@ func tupleSameProduct(nums []int) int {
 }
 ```
 
+### **Rust**
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn tuple_same_product(nums: Vec<i32>) -> i32 {
+        let mut cnt: HashMap<i32, i32> = HashMap::new();
+        let mut ans = 0;
+
+        for i in 1..nums.len() {
+            for j in 0..i {
+                let x = nums[i] * nums[j];
+                *cnt.entry(x).or_insert(0) += 1;
+            }
+        }
+
+        for v in cnt.values() {
+            ans += v * (v - 1) / 2;
+        }
+
+        ans << 3
+    }
+}
+```
+
+### **TypeScript**
+
+```ts
+function tupleSameProduct(nums: number[]): number {
+    const cnt: Map<number, number> = new Map();
+    for (let i = 1; i < nums.length; ++i) {
+        for (let j = 0; j < i; ++j) {
+            const x = nums[i] * nums[j];
+            cnt.set(x, (cnt.get(x) ?? 0) + 1);
+        }
+    }
+    let ans = 0;
+    for (const [_, v] of cnt) {
+        ans += (v * (v - 1)) / 2;
+    }
+    return ans << 3;
+}
+```
+
 ### **...**
 
 ```

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

@@ -40,9 +40,13 @@
 
 ## Solutions
 
-**Solution 1: Number of Combinations + Hash Table**
+**Solution 1: Combination + Hash Table**
 
-Time complexity $O(n^2)$, Space complexity $O(n^2)$.
+Assuming there are $n$ pairs of numbers, for any two pairs of numbers $a, b$ and $c, d$ that satisfy the condition $a \times b = c \times d$, there are a total of $\mathrm{C}_n^2 = \frac{n \times (n-1)}{2}$ such combinations.
+
+According to the problem description, each combination that satisfies the above condition can form $8$ tuples that satisfy the problem requirements. Therefore, we can multiply the number of combinations with the same product by $8$ (equivalent to left shifting by $3$ bits) and add them up to get the result.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the array.
 
 <!-- tabs:start -->
 
@@ -68,7 +72,7 @@ class Solution {
         for (int i = 1; i < nums.length; ++i) {
             for (int j = 0; j < i; ++j) {
                 int x = nums[i] * nums[j];
-                cnt.put(x, cnt.getOrDefault(x, 0) + 1);
+                cnt.merge(x, 1, Integer::sum);
             }
         }
         int ans = 0;
@@ -121,6 +125,51 @@ func tupleSameProduct(nums []int) int {
 }
 ```
 
+### **Rust**
+
+```rust
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn tuple_same_product(nums: Vec<i32>) -> i32 {
+        let mut cnt: HashMap<i32, i32> = HashMap::new();
+        let mut ans = 0;
+
+        for i in 1..nums.len() {
+            for j in 0..i {
+                let x = nums[i] * nums[j];
+                *cnt.entry(x).or_insert(0) += 1;
+            }
+        }
+
+        for v in cnt.values() {
+            ans += v * (v - 1) / 2;
+        }
+
+        ans << 3
+    }
+}
+```
+
+### **TypeScript**
+
+```ts
+function tupleSameProduct(nums: number[]): number {
+    const cnt: Map<number, number> = new Map();
+    for (let i = 1; i < nums.length; ++i) {
+        for (let j = 0; j < i; ++j) {
+            const x = nums[i] * nums[j];
+            cnt.set(x, (cnt.get(x) ?? 0) + 1);
+        }
+    }
+    let ans = 0;
+    for (const [_, v] of cnt) {
+        ans += (v * (v - 1)) / 2;
+    }
+    return ans << 3;
+}
+```
+
 ### **...**
 
 ```

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

@@ -1,12 +1,11 @@
-func tupleSameProduct(nums []int) int {
+func tupleSameProduct(nums []int) (ans int) {
 	cnt := map[int]int{}
 	for i := 1; i < len(nums); i++ {
 		for j := 0; j < i; j++ {
 			x := nums[i] * nums[j]
 			cnt[x]++
 		}
 	}
-	ans := 0
 	for _, v := range cnt {
 		ans += v * (v - 1) / 2
 	}

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

@@ -1,16 +1,16 @@
-class Solution {
-    public int tupleSameProduct(int[] nums) {
-        Map<Integer, Integer> cnt = new HashMap<>();
-        for (int i = 1; i < nums.length; ++i) {
-            for (int j = 0; j < i; ++j) {
-                int x = nums[i] * nums[j];
-                cnt.put(x, cnt.getOrDefault(x, 0) + 1);
-            }
-        }
-        int ans = 0;
-        for (int v : cnt.values()) {
-            ans += v * (v - 1) / 2;
-        }
-        return ans << 3;
-    }
+class Solution {
+    public int tupleSameProduct(int[] nums) {
+        Map<Integer, Integer> cnt = new HashMap<>();
+        for (int i = 1; i < nums.length; ++i) {
+            for (int j = 0; j < i; ++j) {
+                int x = nums[i] * nums[j];
+                cnt.merge(x, 1, Integer::sum);
+            }
+        }
+        int ans = 0;
+        for (int v : cnt.values()) {
+            ans += v * (v - 1) / 2;
+        }
+        return ans << 3;
+    }
 }

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

@@ -0,0 +1,21 @@
+use std::collections::HashMap;
+
+impl Solution {
+    pub fn tuple_same_product(nums: Vec<i32>) -> i32 {
+        let mut cnt: HashMap<i32, i32> = HashMap::new();
+        let mut ans = 0;
+
+        for i in 1..nums.len() {
+            for j in 0..i {
+                let x = nums[i] * nums[j];
+                *cnt.entry(x).or_insert(0) += 1;
+            }
+        }
+
+        for v in cnt.values() {
+            ans += v * (v - 1) / 2;
+        }
+
+        ans << 3
+    }
+}

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

@@ -0,0 +1,14 @@
+function tupleSameProduct(nums: number[]): number {
+    const cnt: Map<number, number> = new Map();
+    for (let i = 1; i < nums.length; ++i) {
+        for (let j = 0; j < i; ++j) {
+            const x = nums[i] * nums[j];
+            cnt.set(x, (cnt.get(x) ?? 0) + 1);
+        }
+    }
+    let ans = 0;
+    for (const [_, v] of cnt) {
+        ans += (v * (v - 1)) / 2;
+    }
+    return ans << 3;
+}