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feat: add solutions to lc problem: No.1577
No.1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers
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‎solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/README.md

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@@ -59,22 +59,126 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表**
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我们用哈希表 `cnt1` 统计 `nums1` 中每个数出现的次数，用哈希表 `cnt2` 统计 `nums2` 中每个数出现的次数。
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然后我们双重循环遍历两个哈希表，记当前 `cnt1` 遍历到的键值对为 $(a, x)$，当前 `cnt2` 遍历到的键值对为 $(b, y)$。接下来分情况讨论：
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- 如果 $a^2$ 能被 $b$ 整除，设 $c=\frac{a^2}{b}$，若 $b=c$，那么答案加上 $x \times y \times (y - 1)$，否则答案加上 $x \times y \times cnt2[c]$。
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- 如果 $b^2$ 能被 $a$ 整除，设 $c=\frac{b^2}{a}$，若 $a=c$，那么答案加上 $x \times (x - 1) \times y$，否则答案加上 $x \times cnt1[c] \times y$。
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最后将答案除以 $2$ 返回即可。
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时间复杂度 $O(n \times m)$，空间复杂度 $O(n + m)$。其中 $n$ 和 $m$ 分别为数组 `nums1` 和 `nums2` 的长度。
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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 numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
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cnt1 = Counter(nums1)
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cnt2 = Counter(nums2)
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ans = 0
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for a, x in cnt1.items():
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for b, y in cnt2.items():
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if a * a % b == 0:
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c = a * a // b
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if b == c:
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ans += x * y * (y - 1)
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else:
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ans += x * y * cnt2[c]
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if b * b % a == 0:
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c = b * b // a
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if a == c:
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ans += x * (x - 1) * y
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else:
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ans += x * y * cnt1[c]
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return ans >> 1
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```
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### **Java**
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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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class Solution {
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public int numTriplets(int[] nums1, int[] nums2) {
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Map<Integer, Integer> cnt1 = new HashMap<>();
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Map<Integer, Integer> cnt2 = new HashMap<>();
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for (int v : nums1) {
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cnt1.put(v, cnt1.getOrDefault(v, 0) + 1);
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}
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for (int v : nums2) {
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cnt2.put(v, cnt2.getOrDefault(v, 0) + 1);
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}
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long ans = 0;
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for (var e1 : cnt1.entrySet()) {
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long a = e1.getKey(), x = e1.getValue();
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for (var e2 : cnt2.entrySet()) {
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long b = e2.getKey(), y = e2.getValue();
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if ((a * a) % b == 0) {
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long c = a * a / b;
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if (b == c) {
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ans += x * y * (y - 1);
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} else {
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ans += x * y * cnt2.getOrDefault((int) c, 0);
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}
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}
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if ((b * b) % a == 0) {
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long c = b * b / a;
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if (a == c) {
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ans += x * (x - 1) * y;
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} else {
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ans += x * y * cnt1.getOrDefault((int) c, 0);
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}
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}
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}
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}
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return (int) (ans >> 1);
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}
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}
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```
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### **Go**
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```go
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func numTriplets(nums1 []int, nums2 []int) (ans int) {
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cnt1 := map[int]int{}
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cnt2 := map[int]int{}
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for _, v := range nums1 {
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cnt1[v]++
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}
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for _, v := range nums2 {
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cnt2[v]++
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}
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for a, x := range cnt1 {
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for b, y := range cnt2 {
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if a*a%b == 0 {
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c := a * a / b
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if b == c {
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ans += x * y * (y - 1)
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} else {
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ans += x * y * cnt2[c]
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}
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}
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if b*b%a == 0 {
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c := b * b / a
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if a == c {
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ans += x * (x - 1) * y
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} else {
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ans += x * y * cnt1[c]
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}
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}
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}
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}
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ans /= 2
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return
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}
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```
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### **...**

‎solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/README_EN.md

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@@ -55,13 +55,104 @@ Type 2: (3,0,1). nums2[3]<sup>2</sup> = nums1[0] * nums1[1].
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### **Python3**
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```python
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class Solution:
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def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
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cnt1 = Counter(nums1)
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cnt2 = Counter(nums2)
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ans = 0
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for a, x in cnt1.items():
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for b, y in cnt2.items():
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if a * a % b == 0:
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c = a * a // b
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if b == c:
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ans += x * y * (y - 1)
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else:
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ans += x * y * cnt2[c]
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if b * b % a == 0:
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c = b * b // a
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if a == c:
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ans += x * (x - 1) * y
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else:
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ans += x * y * cnt1[c]
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return ans >> 1
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```
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### **Java**
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```java
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class Solution {
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public int numTriplets(int[] nums1, int[] nums2) {
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Map<Integer, Integer> cnt1 = new HashMap<>();
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Map<Integer, Integer> cnt2 = new HashMap<>();
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for (int v : nums1) {
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cnt1.put(v, cnt1.getOrDefault(v, 0) + 1);
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}
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for (int v : nums2) {
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cnt2.put(v, cnt2.getOrDefault(v, 0) + 1);
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}
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long ans = 0;
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for (var e1 : cnt1.entrySet()) {
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long a = e1.getKey(), x = e1.getValue();
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for (var e2 : cnt2.entrySet()) {
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long b = e2.getKey(), y = e2.getValue();
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if ((a * a) % b == 0) {
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long c = a * a / b;
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if (b == c) {
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ans += x * y * (y - 1);
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} else {
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ans += x * y * cnt2.getOrDefault((int) c, 0);
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}
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}
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if ((b * b) % a == 0) {
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long c = b * b / a;
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if (a == c) {
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ans += x * (x - 1) * y;
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} else {
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ans += x * y * cnt1.getOrDefault((int) c, 0);
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}
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}
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}
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}
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return (int) (ans >> 1);
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}
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}
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```
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### **Go**
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```go
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func numTriplets(nums1 []int, nums2 []int) (ans int) {
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cnt1 := map[int]int{}
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cnt2 := map[int]int{}
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for _, v := range nums1 {
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cnt1[v]++
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}
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for _, v := range nums2 {
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cnt2[v]++
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}
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for a, x := range cnt1 {
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for b, y := range cnt2 {
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if a*a%b == 0 {
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c := a * a / b
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if b == c {
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ans += x * y * (y - 1)
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} else {
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ans += x * y * cnt2[c]
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}
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}
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if b*b%a == 0 {
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c := b * b / a
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if a == c {
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ans += x * (x - 1) * y
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} else {
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ans += x * y * cnt1[c]
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}
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}
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}
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}
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ans /= 2
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return
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}
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```
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### **...**

‎solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/Solution.go

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func numTriplets(nums1 []int, nums2 []int) (ans int) {
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cnt1 := map[int]int{}
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cnt2 := map[int]int{}
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for _, v := range nums1 {
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cnt1[v]++
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}
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for _, v := range nums2 {
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cnt2[v]++
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}
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for a, x := range cnt1 {
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for b, y := range cnt2 {
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if a*a%b == 0 {
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c := a * a / b
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if b == c {
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ans += x * y * (y - 1)
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} else {
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ans += x * y * cnt2[c]
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}
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}
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if b*b%a == 0 {
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c := b * b / a
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if a == c {
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ans += x * (x - 1) * y
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} else {
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ans += x * y * cnt1[c]
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}
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}
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}
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}
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ans /= 2
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return
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}

‎solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/Solution.java

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class Solution {
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public int numTriplets(int[] nums1, int[] nums2) {
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Map<Integer, Integer> cnt1 = new HashMap<>();
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Map<Integer, Integer> cnt2 = new HashMap<>();
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for (int v : nums1) {
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cnt1.put(v, cnt1.getOrDefault(v, 0) + 1);
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}
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for (int v : nums2) {
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cnt2.put(v, cnt2.getOrDefault(v, 0) + 1);
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}
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long ans = 0;
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for (var e1 : cnt1.entrySet()) {
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long a = e1.getKey(), x = e1.getValue();
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for (var e2 : cnt2.entrySet()) {
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long b = e2.getKey(), y = e2.getValue();
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if ((a * a) % b == 0) {
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long c = a * a / b;
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if (b == c) {
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ans += x * y * (y - 1);
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} else {
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ans += x * y * cnt2.getOrDefault((int) c, 0);
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}
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}
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if ((b * b) % a == 0) {
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long c = b * b / a;
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if (a == c) {
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ans += x * (x - 1) * y;
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} else {
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ans += x * y * cnt1.getOrDefault((int) c, 0);
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}
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}
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}
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}
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return (int) (ans >> 1);
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}
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}

‎solution/1500-1599/1577.Number of Ways Where Square of Number Is Equal to Product of Two Numbers/Solution.py

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class Solution:
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def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:
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cnt1 = Counter(nums1)
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cnt2 = Counter(nums2)
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ans = 0
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for a, x in cnt1.items():
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for b, y in cnt2.items():
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if a * a % b == 0:
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c = a * a // b
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if b == c:
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ans += x * y * (y - 1)
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else:
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ans += x * y * cnt2[c]
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if b * b % a == 0:
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c = b * b // a
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if a == c:
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ans += x * (x - 1) * y
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else:
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ans += x * y * cnt1[c]
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return ans >> 1

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