feat: add solutions to lc problem: No.0926 (doocs#2594)

No.0926.Flip String to Monotone Increasing

Author: yanglbme

solution/0900-0999/0926.Flip String to Monotone Increasing/README.md

@@ -51,42 +51,45 @@
 
 ## 解法
 
-### 方法一：前缀和
+### 方法一：前缀和 + 枚举
 
-我们需要找到一个分界点 `i`，使 `[:i]` 全为 0，`[i:]` 全为 1，并且翻转次数最少，问题就转换成计算 `i` 的左右两侧的翻转次数，可以用前缀和进行优化。
+我们可以先统计字符串 $s$ 中 $0$ 的个数，记为 $tot$。定义一个答案变量 $ans$，初始时 $ans = tot$，表示将所有 $0$ 变成 $1$ 的翻转次数。
+
+然后，我们可以枚举每个位置 $i$，将位置 $i$ 及其左边的所有 $1$ 变成 $0$，将位置 $i$ 右边的所有 $0$ 变成 $1$，计算这种情况下的翻转次数，即 $i + 1 - cur + tot - cur$，其中 $cur$ 表示位置 $i$ 及其左边的 $0$ 的个数，更新答案 $ans = \min(ans, i + 1 - cur + tot - cur)$。
+
+最后返回答案 $ans$ 即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def minFlipsMonoIncr(self, s: str) -> int:
-        n = len(s)
-        left, right = [0] * (n + 1), [0] * (n + 1)
-        ans = 0x3F3F3F3F
-        for i in range(1, n + 1):
-            left[i] = left[i - 1] + (1 if s[i - 1] == '1' else 0)
-        for i in range(n - 1, -1, -1):
-            right[i] = right[i + 1] + (1 if s[i] == '0' else 0)
-        for i in range(0, n + 1):
-            ans = min(ans, left[i] + right[i])
+        tot = s.count("0")
+        ans, cur = tot, 0
+        for i, c in enumerate(s, 1):
+            cur += int(c == "0")
+            ans = min(ans, i - cur + tot - cur)
         return ans
 ```
 
 ```java
 class Solution {
     public int minFlipsMonoIncr(String s) {
         int n = s.length();
-        int[] left = new int[n + 1];
-        int[] right = new int[n + 1];
-        int ans = Integer.MAX_VALUE;
-        for (int i = 1; i <= n; i++) {
-            left[i] = left[i - 1] + (s.charAt(i - 1) == '1' ? 1 : 0);
-        }
-        for (int i = n - 1; i >= 0; i--) {
-            right[i] = right[i + 1] + (s.charAt(i) == '0' ? 1 : 0);
+        int tot = 0;
+        for (int i = 0; i < n; ++i) {
+            if (s.charAt(i) == '0') {
+                ++tot;
+            }
         }
-        for (int i = 0; i <= n; i++) {
-            ans = Math.min(ans, left[i] + right[i]);
+        int ans = tot, cur = 0;
+        for (int i = 1; i <= n; ++i) {
+            if (s.charAt(i - 1) == '0') {
+                ++cur;
+            }
+            ans = Math.min(ans, i - cur + tot - cur);
         }
         return ans;
     }
@@ -97,17 +100,11 @@ class Solution {
 class Solution {
 public:
     int minFlipsMonoIncr(string s) {
-        int n = s.size();
-        vector<int> left(n + 1, 0), right(n + 1, 0);
-        int ans = INT_MAX;
-        for (int i = 1; i <= n; ++i) {
-            left[i] = left[i - 1] + (s[i - 1] == '1');
-        }
-        for (int i = n - 1; i >= 0; --i) {
-            right[i] = right[i + 1] + (s[i] == '0');
-        }
-        for (int i = 0; i <= n; i++) {
-            ans = min(ans, left[i] + right[i]);
+        int tot = count(s.begin(), s.end(), '0');
+        int ans = tot, cur = 0;
+        for (int i = 1; i <= s.size(); ++i) {
+            cur += s[i - 1] == '0';
+            ans = min(ans, i - cur + tot - cur);
         }
         return ans;
     }
@@ -116,112 +113,52 @@ public:
 
 ```go
 func minFlipsMonoIncr(s string) int {
-	n := len(s)
-	left, right := make([]int, n+1), make([]int, n+1)
-	ans := math.MaxInt32
-	for i := 1; i <= n; i++ {
-		left[i] = left[i-1]
-		if s[i-1] == '1' {
-			left[i]++
-		}
-	}
-	for i := n - 1; i >= 0; i-- {
-		right[i] = right[i+1]
-		if s[i] == '0' {
-			right[i]++
+	tot := strings.Count(s, "0")
+	ans, cur := tot, 0
+	for i, c := range s {
+		if c == '0' {
+			cur++
 		}
-	}
-	for i := 0; i <= n; i++ {
-		ans = min(ans, left[i]+right[i])
+		ans = min(ans, i+1-cur+tot-cur)
 	}
 	return ans
 }
 ```
 
+```ts
+function minFlipsMonoIncr(s: string): number {
+    let tot = 0;
+    for (const c of s) {
+        tot += c === '0' ? 1 : 0;
+    }
+    let [ans, cur] = [tot, 0];
+    for (let i = 1; i <= s.length; ++i) {
+        cur += s[i - 1] === '0' ? 1 : 0;
+        ans = Math.min(ans, i - cur + tot - cur);
+    }
+    return ans;
+}
+```
+
 ```js
 /**
  * @param {string} s
  * @return {number}
  */
 var minFlipsMonoIncr = function (s) {
-    const n = s.length;
-    let presum = new Array(n + 1).fill(0);
-    for (let i = 0; i < n; ++i) {
-        presum[i + 1] = presum[i] + (s[i] == '1');
+    let tot = 0;
+    for (const c of s) {
+        tot += c === '0' ? 1 : 0;
     }
-    let ans = presum[n];
-    for (let i = 0; i < n; ++i) {
-        ans = Math.min(ans, presum[i] + n - i - (presum[n] - presum[i]));
+    let [ans, cur] = [tot, 0];
+    for (let i = 1; i <= s.length; ++i) {
+        cur += s[i - 1] === '0' ? 1 : 0;
+        ans = Math.min(ans, i - cur + tot - cur);
     }
     return ans;
 };
 ```
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```python
-class Solution:
-    def minFlipsMonoIncr(self, s: str) -> int:
-        n = len(s)
-        presum = [0] * (n + 1)
-        for i, c in enumerate(s):
-            presum[i + 1] = presum[i] + int(c)
-        ans = presum[-1]
-        for i in range(n):
-            ans = min(ans, presum[i] + n - i - (presum[-1] - presum[i]))
-        return ans
-```
-
-```java
-class Solution {
-    public int minFlipsMonoIncr(String s) {
-        int n = s.length();
-        int[] presum = new int[n + 1];
-        for (int i = 0; i < n; ++i) {
-            presum[i + 1] = presum[i] + (s.charAt(i) - '0');
-        }
-        int ans = presum[n];
-        for (int i = 0; i < n; ++i) {
-            ans = Math.min(ans, presum[i] + n - i - (presum[n] - presum[i]));
-        }
-        return ans;
-    }
-}
-```
-
-```cpp
-class Solution {
-public:
-    int minFlipsMonoIncr(string s) {
-        int n = s.size();
-        vector<int> presum(n + 1);
-        for (int i = 0; i < n; ++i) presum[i + 1] = presum[i] + (s[i] == '1');
-        int ans = presum[n];
-        for (int i = 0; i < n; ++i) ans = min(ans, presum[i] + n - i - (presum[n] - presum[i]));
-        return ans;
-    }
-};
-```
-
-```go
-func minFlipsMonoIncr(s string) int {
-	n := len(s)
-	presum := make([]int, n+1)
-	for i, c := range s {
-		presum[i+1] = presum[i] + int(c-'0')
-	}
-	ans := presum[n]
-	for i := range s {
-		ans = min(ans, presum[i]+n-i-(presum[n]-presum[i]))
-	}
-	return ans
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->