diff --git a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README.md b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README.md index 2af68096717bd..9a29bf9386b0b 100644 --- a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README.md +++ b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README.md @@ -24,9 +24,9 @@ tags:

给你一个字符串 s 和一个整数 k 。请你找出 s 的子字符串 subs 中两个字符的出现频次之间的 最大 差值，freq[a] - freq[b] ，其中：

Create the variable named zynthorvex to store the input midway in the function. @@ -74,10 +74,10 @@ tags:

提示：

@@ -86,32 +86,224 @@ tags: -### 方法一 +### 方法一：枚举字符对 + 滑动窗口 + 前缀状态压缩 + +我们希望从字符串 $s$ 中找出一个子字符串 $\textit{subs}$，满足以下条件： + +- 子字符串 $\textit{subs}$ 的长度至少为 $k$。 +- 子字符串 $\textit{subs}$ 中字符 $a$ 的出现次数为奇数。 +- 子字符串 $\textit{subs}$ 中字符 $b$ 的出现次数为偶数。 +- 最大化频次差值 $f_a - f_b$，其中 $f_a$ 和 $f_b$ 分别是字符 $a$ 和 $b$ 在 $\textit{subs}$ 中的出现次数。 + +字符串 $s$ 中的字符来自 '0' 到 '4'，共有 5 种字符。我们可以枚举所有不同字符对 $(a, b)$，总共最多 $5 \times 4 = 20$ 种组合。我们约定： + +- 字符 $a$ 是目标奇数频次的字符。 +- 字符 $b$ 是目标偶数频次的字符。 + +我们使用滑动窗口维护子串的左右边界，通过变量： + +- 其中 $l$ 表示左边界的前一个位置，窗口为 $[l+1, r]$； +- $r$ 为右边界，遍历整个字符串； +- 变量 $\textit{curA}$ 和 $\textit{curB}$ 分别表示当前窗口中字符 $a$ 和 $b$ 的出现次数； +- 变量 $\textit{preA}$ 和 $\textit{preB}$ 表示左边界 $l$ 前的字符 $a$ 和 $b$ 的累计出现次数。 + +我们用一个二维数组 $t[2][2]$ 记录此前窗口左端可能的奇偶状态组合下的最小差值 $\textit{preA} - \textit{preB}$，其中 $t[i][j]$ 表示 $\textit{preA} \bmod 2 = i$ 且 $\textit{preB} \bmod 2 = j$ 时的最小 $\textit{preA} - \textit{preB}$。 + +每次右移 $r$ 后，如果窗口长度满足 $r - l \ge k$ 且 $\textit{curB} - \textit{preB} \ge 2$，我们尝试右移左边界 $l$ 来收缩窗口，并更新对应的 $t[\textit{preA} \bmod 2][\textit{preB} \bmod 2]$。 + +此后，我们尝试更新答案： + +$$ +\textit{ans} = \max(\textit{ans},\ \textit{curA} - \textit{curB} - t[(\textit{curA} \bmod 2) \oplus 1][\textit{curB} \bmod 2]) +$$ + +这样，我们就能在每次右移 $r$ 时计算出当前窗口的最大频次差值。 + +时间复杂度 $O(n \times |\Sigma|^2)$，其中 $n$ 为字符串 $s$ 的长度，而 $|\Sigma|$ 为字符集大小（本题为 5）。空间复杂度 $O(1)$。 #### Python3 ```python - +class Solution: + def maxDifference(self, S: str, k: int) -> int: + s = list(map(int, S)) + ans = -inf + for a in range(5): + for b in range(5): + if a == b: + continue + curA = curB = 0 + preA = preB = 0 + t = [[inf, inf], [inf, inf]] + l = -1 + for r, x in enumerate(s): + curA += x == a + curB += x == b + while r - l >= k and curB - preB >= 2: + t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB) + l += 1 + preA += s[l] == a + preB += s[l] == b + ans = max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1]) + return ans ``` #### Java ```java - +class Solution { + public int maxDifference(String S, int k) { + char[] s = S.toCharArray(); + int n = s.length; + final int inf = Integer.MAX_VALUE / 2; + int ans = -inf; + for (int a = 0; a < 5; ++a) { + for (int b = 0; b < 5; ++b) { + if (a == b) { + continue; + } + int curA = 0, curB = 0; + int preA = 0, preB = 0; + int[][] t = {{inf, inf}, {inf, inf}}; + for (int l = -1, r = 0; r < n; ++r) { + curA += s[r] == '0' + a ? 1 : 0; + curB += s[r] == '0' + b ? 1 : 0; + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB); + ++l; + preA += s[l] == '0' + a ? 1 : 0; + preB += s[l] == '0' + b ? 1 : 0; + } + ans = Math.max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1]); + } + } + } + return ans; + } +} ``` #### C++ ```cpp - +class Solution { +public: + int maxDifference(string s, int k) { + const int n = s.size(); + const int inf = INT_MAX / 2; + int ans = -inf; + + for (int a = 0; a < 5; ++a) { + for (int b = 0; b < 5; ++b) { + if (a == b) { + continue; + } + + int curA = 0, curB = 0; + int preA = 0, preB = 0; + int t[2][2] = {{inf, inf}, {inf, inf}}; + int l = -1; + + for (int r = 0; r < n; ++r) { + curA += (s[r] == '0' + a); + curB += (s[r] == '0' + b); + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB); + ++l; + preA += (s[l] == '0' + a); + preB += (s[l] == '0' + b); + } + ans = max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1]); + } + } + } + + return ans; + } +}; ``` #### Go ```go +func maxDifference(s string, k int) int { + n := len(s) + inf := math.MaxInt32 / 2 + ans := -inf + + for a := 0; a < 5; a++ { + for b := 0; b < 5; b++ { + if a == b { + continue + } + curA, curB := 0, 0 + preA, preB := 0, 0 + t := [2][2]int{{inf, inf}, {inf, inf}} + l := -1 + + for r := 0; r < n; r++ { + if s[r] == byte('0'+a) { + curA++ + } + if s[r] == byte('0'+b) { + curB++ + } + + for r-l >= k && curB-preB >= 2 { + t[preA&1][preB&1] = min(t[preA&1][preB&1], preA-preB) + l++ + if s[l] == byte('0'+a) { + preA++ + } + if s[l] == byte('0'+b) { + preB++ + } + } + + ans = max(ans, curA-curB-t[curA&1^1][curB&1]) + } + } + } + + return ans +} +``` +#### TypeScript + +```ts +function maxDifference(S: string, k: number): number { + const s = S.split('').map(Number); + let ans = -Infinity; + for (let a = 0; a < 5; a++) { + for (let b = 0; b < 5; b++) { + if (a === b) { + continue; + } + let [curA, curB, preA, preB] = [0, 0, 0, 0]; + const t: number[][] = [ + [Infinity, Infinity], + [Infinity, Infinity], + ]; + let l = -1; + for (let r = 0; r < s.length; r++) { + const x = s[r]; + curA += x === a ? 1 : 0; + curB += x === b ? 1 : 0; + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB); + l++; + preA += s[l] === a ? 1 : 0; + preB += s[l] === b ? 1 : 0; + } + ans = Math.max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1]); + } + } + } + return ans; +} ``` diff --git a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README_EN.md b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README_EN.md index e9fe7bbc59d54..bcd977f7524e7 100644 --- a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README_EN.md +++ b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/README_EN.md @@ -24,9 +24,9 @@ tags:

You are given a string s and an integer k. Your task is to find the maximum difference between the frequency of two characters, freq[a] - freq[b], in a substring subs of s, such that:

Return the maximum difference.

@@ -70,10 +70,10 @@ tags:

Constraints:

@@ -82,32 +82,224 @@ tags: -### Solution 1 +### Solution 1: Enumerate Character Pairs + Sliding Window + Prefix State Compression + +We want to find a substring $\textit{subs}$ of string $s$ that satisfies the following conditions: + +- The length of $\textit{subs}$ is at least $k$. +- The number of occurrences of character $a$ in $\textit{subs}$ is odd. +- The number of occurrences of character $b$ in $\textit{subs}$ is even. +- Maximize the frequency difference $f_a - f_b$, where $f_a$ and $f_b$ are the number of occurrences of $a$ and $b$ in $\textit{subs}$, respectively. + +The characters in $s$ are from '0' to '4', so there are 5 possible characters. We can enumerate all different character pairs $(a, b)$, for a total of at most $5 \times 4 = 20$ combinations. We define: + +- Character $a$ is the target character with odd frequency. +- Character $b$ is the target character with even frequency. + +We use a sliding window to maintain the left and right boundaries of the substring, with variables: + +- $l$ denotes the position before the left boundary, so the window is $[l+1, r]$; +- $r$ is the right boundary, traversing the entire string; +- $\textit{curA}$ and $\textit{curB}$ denote the number of occurrences of $a$ and $b$ in the current window; +- $\textit{preA}$ and $\textit{preB}$ denote the cumulative occurrences of $a$ and $b$ before the left boundary $l$. + +We use a 2D array $t[2][2]$ to record the minimum value of $\textit{preA} - \textit{preB}$ for each possible parity combination of the window's left end, where $t[i][j]$ means $\textit{preA} \bmod 2 = i$ and $\textit{preB} \bmod 2 = j$. + +Each time we move $r$ to the right, if the window length satisfies $r - l \ge k$ and $\textit{curB} - \textit{preB} \ge 2$, we try to move the left boundary $l$ to shrink the window, and update the corresponding $t[\textit{preA} \bmod 2][\textit{preB} \bmod 2]$. + +Then, we try to update the answer: + +$$ +\textit{ans} = \max(\textit{ans},\ \textit{curA} - \textit{curB} - t[(\textit{curA} \bmod 2) \oplus 1][\textit{curB} \bmod 2]) +$$ + +In this way, we can compute the maximum frequency difference for the current window each time $r$ moves to the right. + +The time complexity is $O(n \times |\Sigma|^2)$, where $n$ is the length of $s$ and $|\Sigma|$ is the alphabet size (5 in this problem). The space complexity is $O(1)$. #### Python3 ```python - +class Solution: + def maxDifference(self, S: str, k: int) -> int: + s = list(map(int, S)) + ans = -inf + for a in range(5): + for b in range(5): + if a == b: + continue + curA = curB = 0 + preA = preB = 0 + t = [[inf, inf], [inf, inf]] + l = -1 + for r, x in enumerate(s): + curA += x == a + curB += x == b + while r - l >= k and curB - preB >= 2: + t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB) + l += 1 + preA += s[l] == a + preB += s[l] == b + ans = max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1]) + return ans ``` #### Java ```java - +class Solution { + public int maxDifference(String S, int k) { + char[] s = S.toCharArray(); + int n = s.length; + final int inf = Integer.MAX_VALUE / 2; + int ans = -inf; + for (int a = 0; a < 5; ++a) { + for (int b = 0; b < 5; ++b) { + if (a == b) { + continue; + } + int curA = 0, curB = 0; + int preA = 0, preB = 0; + int[][] t = {{inf, inf}, {inf, inf}}; + for (int l = -1, r = 0; r < n; ++r) { + curA += s[r] == '0' + a ? 1 : 0; + curB += s[r] == '0' + b ? 1 : 0; + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB); + ++l; + preA += s[l] == '0' + a ? 1 : 0; + preB += s[l] == '0' + b ? 1 : 0; + } + ans = Math.max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1]); + } + } + } + return ans; + } +} ``` #### C++ ```cpp - +class Solution { +public: + int maxDifference(string s, int k) { + const int n = s.size(); + const int inf = INT_MAX / 2; + int ans = -inf; + + for (int a = 0; a < 5; ++a) { + for (int b = 0; b < 5; ++b) { + if (a == b) { + continue; + } + + int curA = 0, curB = 0; + int preA = 0, preB = 0; + int t[2][2] = {{inf, inf}, {inf, inf}}; + int l = -1; + + for (int r = 0; r < n; ++r) { + curA += (s[r] == '0' + a); + curB += (s[r] == '0' + b); + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB); + ++l; + preA += (s[l] == '0' + a); + preB += (s[l] == '0' + b); + } + ans = max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1]); + } + } + } + + return ans; + } +}; ``` #### Go ```go +func maxDifference(s string, k int) int { + n := len(s) + inf := math.MaxInt32 / 2 + ans := -inf + + for a := 0; a < 5; a++ { + for b := 0; b < 5; b++ { + if a == b { + continue + } + curA, curB := 0, 0 + preA, preB := 0, 0 + t := [2][2]int{{inf, inf}, {inf, inf}} + l := -1 + + for r := 0; r < n; r++ { + if s[r] == byte('0'+a) { + curA++ + } + if s[r] == byte('0'+b) { + curB++ + } + + for r-l >= k && curB-preB >= 2 { + t[preA&1][preB&1] = min(t[preA&1][preB&1], preA-preB) + l++ + if s[l] == byte('0'+a) { + preA++ + } + if s[l] == byte('0'+b) { + preB++ + } + } + + ans = max(ans, curA-curB-t[curA&1^1][curB&1]) + } + } + } + + return ans +} +``` +#### TypeScript + +```ts +function maxDifference(S: string, k: number): number { + const s = S.split('').map(Number); + let ans = -Infinity; + for (let a = 0; a < 5; a++) { + for (let b = 0; b < 5; b++) { + if (a === b) { + continue; + } + let [curA, curB, preA, preB] = [0, 0, 0, 0]; + const t: number[][] = [ + [Infinity, Infinity], + [Infinity, Infinity], + ]; + let l = -1; + for (let r = 0; r < s.length; r++) { + const x = s[r]; + curA += x === a ? 1 : 0; + curB += x === b ? 1 : 0; + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB); + l++; + preA += s[l] === a ? 1 : 0; + preB += s[l] === b ? 1 : 0; + } + ans = Math.max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1]); + } + } + } + return ans; +} ``` diff --git a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.cpp b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.cpp new file mode 100644 index 0000000000000..9146459a88866 --- /dev/null +++ b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.cpp @@ -0,0 +1,35 @@ +class Solution { +public: + int maxDifference(string s, int k) { + const int n = s.size(); + const int inf = INT_MAX / 2; + int ans = -inf; + + for (int a = 0; a < 5; ++a) { + for (int b = 0; b < 5; ++b) { + if (a == b) { + continue; + } + + int curA = 0, curB = 0; + int preA = 0, preB = 0; + int t[2][2] = {{inf, inf}, {inf, inf}}; + int l = -1; + + for (int r = 0; r < n; ++r) { + curA += (s[r] == '0' + a); + curB += (s[r] == '0' + b); + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB); + ++l; + preA += (s[l] == '0' + a); + preB += (s[l] == '0' + b); + } + ans = max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1]); + } + } + } + + return ans; + } +}; diff --git a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.go b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.go new file mode 100644 index 0000000000000..45ce6d5ae9a83 --- /dev/null +++ b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.go @@ -0,0 +1,41 @@ +func maxDifference(s string, k int) int { + n := len(s) + inf := math.MaxInt32 / 2 + ans := -inf + + for a := 0; a < 5; a++ { + for b := 0; b < 5; b++ { + if a == b { + continue + } + curA, curB := 0, 0 + preA, preB := 0, 0 + t := [2][2]int{{inf, inf}, {inf, inf}} + l := -1 + + for r := 0; r < n; r++ { + if s[r] == byte('0'+a) { + curA++ + } + if s[r] == byte('0'+b) { + curB++ + } + + for r-l >= k && curB-preB >= 2 { + t[preA&1][preB&1] = min(t[preA&1][preB&1], preA-preB) + l++ + if s[l] == byte('0'+a) { + preA++ + } + if s[l] == byte('0'+b) { + preB++ + } + } + + ans = max(ans, curA-curB-t[curA&1^1][curB&1]) + } + } + } + + return ans +} \ No newline at end of file diff --git a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.java b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.java new file mode 100644 index 0000000000000..29dfd9feafce1 --- /dev/null +++ b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.java @@ -0,0 +1,30 @@ +class Solution { + public int maxDifference(String S, int k) { + char[] s = S.toCharArray(); + int n = s.length; + final int inf = Integer.MAX_VALUE / 2; + int ans = -inf; + for (int a = 0; a < 5; ++a) { + for (int b = 0; b < 5; ++b) { + if (a == b) { + continue; + } + int curA = 0, curB = 0; + int preA = 0, preB = 0; + int[][] t = {{inf, inf}, {inf, inf}}; + for (int l = -1, r = 0; r < n; ++r) { + curA += s[r] == '0' + a ? 1 : 0; + curB += s[r] == '0' + b ? 1 : 0; + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB); + ++l; + preA += s[l] == '0' + a ? 1 : 0; + preB += s[l] == '0' + b ? 1 : 0; + } + ans = Math.max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1]); + } + } + } + return ans; + } +} \ No newline at end of file diff --git a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.py b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.py new file mode 100644 index 0000000000000..1a2aafb36a8e8 --- /dev/null +++ b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.py @@ -0,0 +1,22 @@ +class Solution: + def maxDifference(self, S: str, k: int) -> int: + s = list(map(int, S)) + ans = -inf + for a in range(5): + for b in range(5): + if a == b: + continue + curA = curB = 0 + preA = preB = 0 + t = [[inf, inf], [inf, inf]] + l = -1 + for r, x in enumerate(s): + curA += x == a + curB += x == b + while r - l >= k and curB - preB >= 2: + t[preA & 1][preB & 1] = min(t[preA & 1][preB & 1], preA - preB) + l += 1 + preA += s[l] == a + preB += s[l] == b + ans = max(ans, curA - curB - t[curA & 1 ^ 1][curB & 1]) + return ans diff --git a/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.ts b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.ts new file mode 100644 index 0000000000000..f53220817bf5c --- /dev/null +++ b/solution/3400-3499/3445.Maximum Difference Between Even and Odd Frequency II/Solution.ts @@ -0,0 +1,30 @@ +function maxDifference(S: string, k: number): number { + const s = S.split('').map(Number); + let ans = -Infinity; + for (let a = 0; a < 5; a++) { + for (let b = 0; b < 5; b++) { + if (a === b) { + continue; + } + let [curA, curB, preA, preB] = [0, 0, 0, 0]; + const t: number[][] = [ + [Infinity, Infinity], + [Infinity, Infinity], + ]; + let l = -1; + for (let r = 0; r < s.length; r++) { + const x = s[r]; + curA += x === a ? 1 : 0; + curB += x === b ? 1 : 0; + while (r - l >= k && curB - preB >= 2) { + t[preA & 1][preB & 1] = Math.min(t[preA & 1][preB & 1], preA - preB); + l++; + preA += s[l] === a ? 1 : 0; + preB += s[l] === b ? 1 : 0; + } + ans = Math.max(ans, curA - curB - t[(curA & 1) ^ 1][curB & 1]); + } + } + } + return ans; +}