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feat: add solutions to lc problems: No.1819,2288
* No.1819.Number of Different Subsequences GCDs * No.2288.Apply Discount to Prices
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‎solution/1800-1899/1819.Number of Different Subsequences GCDs/README.md

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@@ -53,22 +53,142 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：枚举 + 数学**
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对于数组 `nums` 的所有子序列，其最大公约数一定不超过数组中的最大值 $mx$。
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因此我们可以枚举 $[1,.. mx]$ 中的每个数 $x$，判断 $x$ 是否为数组 `nums` 的子序列的最大公约数，如果是，则答案加一。
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那么问题转换为：判断 $x$ 是否为数组 `nums` 的子序列的最大公约数。我们可以通过枚举 $x$ 的倍数 $y$，判断 $y$ 是否在数组 `nums` 中，如果 $y$ 在数组 `nums` 中，则计算 $y$ 的最大公约数 $g$，如果出现 $g = x$，则 $x$ 是数组 `nums` 的子序列的最大公约数。
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时间复杂度 $O(n + M \times \log M)$，空间复杂度 $O(M)$。其中 $n$ 和 $M$ 分别是数组 `nums` 的长度和数组 `nums` 中的最大值。
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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 countDifferentSubsequenceGCDs(self, nums: List[int]) -> int:
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mx = max(nums)
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vis = set(nums)
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ans = 0
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for x in range(1, mx + 1):
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g = 0
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for y in range(x, mx + 1, x):
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if y in vis:
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g = gcd(g, y)
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if g == x:
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ans += 1
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break
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return ans
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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 countDifferentSubsequenceGCDs(int[] nums) {
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int mx = Arrays.stream(nums).max().getAsInt();
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boolean[] vis = new boolean[mx + 1];
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for (int x : nums) {
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vis[x] = true;
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}
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int ans = 0;
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for (int x = 1; x <= mx; ++x) {
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int g = 0;
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for (int y = x; y <= mx; y += x) {
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if (vis[y]) {
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g = gcd(g, y);
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if (x == g) {
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++ans;
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break;
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}
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}
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}
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}
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return ans;
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}
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private int gcd(int a, int b) {
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return b == 0 ? a : gcd(b, a % b);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countDifferentSubsequenceGCDs(vector<int>& nums) {
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int mx = *max_element(nums.begin(), nums.end());
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vector<bool> vis(mx + 1);
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for (int& x : nums) {
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vis[x] = true;
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}
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int ans = 0;
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for (int x = 1; x <= mx; ++x) {
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int g = 0;
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for (int y = x; y <= mx; y += x) {
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if (vis[y]) {
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g = gcd(g, y);
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if (g == x) {
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++ans;
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break;
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}
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}
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}
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}
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return ans;
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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 countDifferentSubsequenceGCDs(nums []int) (ans int) {
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mx := 0
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for _, x := range nums {
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mx = max(mx, x)
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}
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vis := make([]bool, mx+1)
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for _, x := range nums {
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vis[x] = true
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}
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for x := 1; x <= mx; x++ {
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g := 0
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for y := x; y <= mx; y += x {
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if vis[y] {
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g = gcd(g, y)
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if g == x {
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ans++
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break
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}
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}
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}
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func gcd(a, b int) int {
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if b == 0 {
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return a
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}
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return gcd(b, a%b)
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}
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```
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### **...**

‎solution/1800-1899/1819.Number of Different Subsequences GCDs/README_EN.md

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@@ -52,13 +52,123 @@ The different GCDs are 6, 10, 3, 2, and 1.
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### **Python3**
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```python
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class Solution:
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def countDifferentSubsequenceGCDs(self, nums: List[int]) -> int:
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mx = max(nums)
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vis = set(nums)
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ans = 0
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for x in range(1, mx + 1):
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g = 0
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for y in range(x, mx + 1, x):
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if y in vis:
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g = gcd(g, y)
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if g == x:
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ans += 1
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break
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return ans
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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 countDifferentSubsequenceGCDs(int[] nums) {
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int mx = Arrays.stream(nums).max().getAsInt();
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boolean[] vis = new boolean[mx + 1];
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for (int x : nums) {
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vis[x] = true;
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}
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int ans = 0;
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for (int x = 1; x <= mx; ++x) {
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int g = 0;
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for (int y = x; y <= mx; y += x) {
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if (vis[y]) {
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g = gcd(g, y);
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if (x == g) {
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++ans;
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break;
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}
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}
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}
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}
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return ans;
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}
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private int gcd(int a, int b) {
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return b == 0 ? a : gcd(b, a % b);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int countDifferentSubsequenceGCDs(vector<int>& nums) {
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int mx = *max_element(nums.begin(), nums.end());
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vector<bool> vis(mx + 1);
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for (int& x : nums) {
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vis[x] = true;
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}
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int ans = 0;
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for (int x = 1; x <= mx; ++x) {
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int g = 0;
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for (int y = x; y <= mx; y += x) {
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if (vis[y]) {
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g = gcd(g, y);
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if (g == x) {
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++ans;
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break;
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}
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}
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}
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}
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return ans;
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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 countDifferentSubsequenceGCDs(nums []int) (ans int) {
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mx := 0
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for _, x := range nums {
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mx = max(mx, x)
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}
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vis := make([]bool, mx+1)
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for _, x := range nums {
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vis[x] = true
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}
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for x := 1; x <= mx; x++ {
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g := 0
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for y := x; y <= mx; y += x {
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if vis[y] {
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g = gcd(g, y)
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if g == x {
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ans++
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break
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}
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}
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}
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func gcd(a, b int) int {
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if b == 0 {
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return a
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}
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return gcd(b, a%b)
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}
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```
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### **...**

‎solution/1800-1899/1819.Number of Different Subsequences GCDs/Solution.cpp

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class Solution {
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public:
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int countDifferentSubsequenceGCDs(vector<int>& nums) {
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int mx = *max_element(nums.begin(), nums.end());
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vector<bool> vis(mx + 1);
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for (int& x : nums) {
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vis[x] = true;
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}
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int ans = 0;
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for (int x = 1; x <= mx; ++x) {
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int g = 0;
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for (int y = x; y <= mx; y += x) {
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if (vis[y]) {
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g = gcd(g, y);
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if (g == x) {
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++ans;
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break;
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}
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}
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}
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}
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return ans;
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}
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};

‎solution/1800-1899/1819.Number of Different Subsequences GCDs/Solution.go

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func countDifferentSubsequenceGCDs(nums []int) (ans int) {
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mx := 0
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for _, x := range nums {
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mx = max(mx, x)
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}
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vis := make([]bool, mx+1)
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for _, x := range nums {
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vis[x] = true
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}
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for x := 1; x <= mx; x++ {
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g := 0
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for y := x; y <= mx; y += x {
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if vis[y] {
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g = gcd(g, y)
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if g == x {
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ans++
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break
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}
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}
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}
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}
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return
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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func gcd(a, b int) int {
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if b == 0 {
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return a
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}
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return gcd(b, a%b)
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}

‎solution/1800-1899/1819.Number of Different Subsequences GCDs/Solution.java

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class Solution {
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public int countDifferentSubsequenceGCDs(int[] nums) {
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int mx = Arrays.stream(nums).max().getAsInt();
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boolean[] vis = new boolean[mx + 1];
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for (int x : nums) {
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vis[x] = true;
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}
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int ans = 0;
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for (int x = 1; x <= mx; ++x) {
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int g = 0;
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for (int y = x; y <= mx; y += x) {
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if (vis[y]) {
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g = gcd(g, y);
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if (x == g) {
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++ans;
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break;
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}
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}
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}
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}
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return ans;
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}
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private int gcd(int a, int b) {
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return b == 0 ? a : gcd(b, a % b);
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}
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}

‎solution/1800-1899/1819.Number of Different Subsequences GCDs/Solution.py

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class Solution:
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def countDifferentSubsequenceGCDs(self, nums: List[int]) -> int:
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mx = max(nums)
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vis = set(nums)
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ans = 0
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for x in range(1, mx + 1):
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g = 0
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for y in range(x, mx + 1, x):
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if y in vis:
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g = gcd(g, y)
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if g == x:
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ans += 1
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break
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return ans

‎solution/2200-2299/2287.Rearrange Characters to Make Target String/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：计数 + 枚举**
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**方法一：计数**
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我们统计字符串 `s` 和 `target` 中每个字符出现的次数，记为 `cnt1` 和 `cnt2`。对于 `target` 中的每个字符，我们计算 `cnt1` 中该字符出现的次数除以 `cnt2` 中该字符出现的次数，取最小值即可。
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