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feat: add solutions to lc problem: No.1653
No.1653.Minimum Deletions to Make String Balanced
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‎solution/1600-1699/1653.Minimum Deletions to Make String Balanced/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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定义 `dp[i]` 表示字符串 `s[0..i)` 达到平衡时，需要删除的最少字符数，答案为 `dp[n]`。用变量 $b$ 统计字符 `b` 的个数。
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遍历字符串 $s$：
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如果当前字符为 `b`，此时不影响平衡串的最小删除次数，那么 $dp[i]=dp[i-1]$，并且累加 $b$。
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如果当前字符为 `a`，此时，要想达到平衡，要么把前面的所有 `b` 都删掉，操作次数为 $b$；要么把当前的 `a` 删除，操作次数为 $dp[i-1]+1$。取两者的最小值即可。即 $dp[i]=\min(b,dp[i-1]+1)$。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串 $s$ 的长度。
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我们发现 $dp[i]$ 只与 $dp[i-1]$ 有关，因此可以使用变量 `ans` 来代替数组 `dp`。空间复杂度优化为 $O(1)$。
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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 minimumDeletions(self, s: str) -> int:
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n = len(s)
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dp = [0] * (n + 1)
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b = 0
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for i, c in enumerate(s, 1):
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if c == 'b':
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dp[i] = dp[i - 1]
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b += 1
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else:
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dp[i] = min(dp[i - 1] + 1, b)
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return dp[n]
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```
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```python
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class Solution:
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def minimumDeletions(self, s: str) -> int:
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ans = b = 0
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for c in s:
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if c == 'b':
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b += 1
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else:
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ans = min(b, ans + 1)
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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 minimumDeletions(String s) {
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int n = s.length();
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int[] dp = new int[n + 1];
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int b = 0;
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for (int i = 0; i < n; ++i) {
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if (s.charAt(i) == 'b') {
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dp[i + 1] = dp[i];
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++b;
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} else {
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dp[i + 1] = Math.min(dp[i] + 1, b);
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}
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}
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return dp[n];
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}
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}
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```
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```java
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class Solution {
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public int minimumDeletions(String s) {
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int ans = 0, b = 0;
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for (char c : s.toCharArray()) {
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if (c == 'b') {
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++b;
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} else {
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ans = Math.min(b, ans + 1);
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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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### **C++**
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```cpp
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class Solution {
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public:
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int minimumDeletions(string s) {
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int n = s.size();
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vector<int> dp(n + 1);
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int b = 0;
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for (int i = 0; i < n; ++i) {
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if (s[i] == 'b') {
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dp[i + 1] = dp[i];
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++b;
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} else {
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dp[i + 1] = min(dp[i] + 1, b);
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}
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}
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return dp[n];
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}
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};
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```
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```cpp
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class Solution {
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public:
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int minimumDeletions(string s) {
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int ans = 0, b = 0;
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for (char c : s) {
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if (c == 'b') {
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++b;
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} else {
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ans = min(b, ans + 1);
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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 minimumDeletions(s string) int {
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n := len(s)
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dp := make([]int, n+1)
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b := 0
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for i, c := range s {
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if c == 'b' {
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b++
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dp[i+1] = dp[i]
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} else {
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dp[i+1] = min(dp[i]+1, b)
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}
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}
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return dp[n]
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}
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func min(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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```
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```go
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func minimumDeletions(s string) int {
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ans, b := 0, 0
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for _, c := range s {
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if c == 'b' {
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b++
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} else {
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ans = min(b, ans+1)
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}
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}
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return ans
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}
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func min(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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```
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### **...**

‎solution/1600-1699/1653.Minimum Deletions to Make String Balanced/README_EN.md

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## Solutions
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Dynamic programming.
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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 minimumDeletions(self, s: str) -> int:
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n = len(s)
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dp = [0] * (n + 1)
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b = 0
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for i, c in enumerate(s, 1):
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if c == 'b':
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dp[i] = dp[i - 1]
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b += 1
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else:
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dp[i] = min(dp[i - 1] + 1, b)
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return dp[n]
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```
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```python
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class Solution:
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def minimumDeletions(self, s: str) -> int:
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ans = b = 0
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for c in s:
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if c == 'b':
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b += 1
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else:
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ans = min(b, ans + 1)
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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 minimumDeletions(String s) {
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int n = s.length();
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int[] dp = new int[n + 1];
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int b = 0;
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for (int i = 0; i < n; ++i) {
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if (s.charAt(i) == 'b') {
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dp[i + 1] = dp[i];
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++b;
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} else {
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dp[i + 1] = Math.min(dp[i] + 1, b);
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}
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}
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return dp[n];
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}
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}
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```
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```java
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class Solution {
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public int minimumDeletions(String s) {
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int ans = 0, b = 0;
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for (char c : s.toCharArray()) {
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if (c == 'b') {
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++b;
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} else {
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ans = Math.min(b, ans + 1);
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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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### **C++**
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```cpp
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class Solution {
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public:
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int minimumDeletions(string s) {
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int n = s.size();
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vector<int> dp(n + 1);
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int b = 0;
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for (int i = 0; i < n; ++i) {
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if (s[i] == 'b') {
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dp[i + 1] = dp[i];
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++b;
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} else {
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dp[i + 1] = min(dp[i] + 1, b);
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}
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}
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return dp[n];
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}
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};
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```
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```cpp
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class Solution {
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public:
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int minimumDeletions(string s) {
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int ans = 0, b = 0;
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for (char c : s) {
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if (c == 'b') {
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++b;
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} else {
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ans = min(b, ans + 1);
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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 minimumDeletions(s string) int {
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n := len(s)
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dp := make([]int, n+1)
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b := 0
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for i, c := range s {
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if c == 'b' {
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b++
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dp[i+1] = dp[i]
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} else {
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dp[i+1] = min(dp[i]+1, b)
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}
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}
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return dp[n]
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}
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func min(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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```
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```go
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func minimumDeletions(s string) int {
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ans, b := 0, 0
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for _, c := range s {
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if c == 'b' {
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b++
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} else {
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ans = min(b, ans+1)
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}
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}
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return ans
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}
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func min(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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```
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### **...**

‎solution/1600-1699/1653.Minimum Deletions to Make String Balanced/Solution.cpp

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class Solution {
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public:
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int minimumDeletions(string s) {
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int ans = 0, b = 0;
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for (char c : s) {
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if (c == 'b') {
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++b;
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} else {
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ans = min(b, ans + 1);
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}
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}
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return ans;
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}
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};

‎solution/1600-1699/1653.Minimum Deletions to Make String Balanced/Solution.go

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func minimumDeletions(s string) int {
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ans, b := 0, 0
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for _, c := range s {
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if c == 'b' {
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b++
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} else {
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ans = min(b, ans+1)
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}
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}
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return ans
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}
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func min(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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}

‎solution/1600-1699/1653.Minimum Deletions to Make String Balanced/Solution.java

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class Solution {
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public int minimumDeletions(String s) {
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int ans = 0, b = 0;
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for (char c : s.toCharArray()) {
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if (c == 'b') {
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++b;
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} else {
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ans = Math.min(b, ans + 1);
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}
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}
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return ans;
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}
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}

‎solution/1600-1699/1653.Minimum Deletions to Make String Balanced/Solution.py

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class Solution:
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def minimumDeletions(self, s: str) -> int:
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ans = b = 0
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for c in s:
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if c == 'b':
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b += 1
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else:
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ans = min(b, ans + 1)
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return ans

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