diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README.md b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README.md index 0c99b95b8df00..188223e7e8c64 100644 --- a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README.md +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README.md @@ -64,7 +64,17 @@ tags: -### 方法一 +### 方法一：滑动窗口 + +我们可以维护一个滑动窗口，使得窗口内的字符不重复。初始时，我们用一个长度为 $26$ 的二进制整数 $\textit{mask}$ 表示窗口内的字符，其中第 $i$ 位为 $1$ 表示字符 $i$ 在窗口内出现过，否则表示字符 $i$ 在窗口内没有出现过。 + +然后，我们遍历字符串 $s$，对于每一个位置 $r$，如果 $\textit{s}[r]$ 在窗口内出现过，我们需要将窗口的左边界 $l$ 右移，直到窗口内不再有重复的字符。在这之后，我们将 $\textit{s}[r]$ 加入窗口内，此时如果窗口的长度大于等于 $3$，那么我们就找到了一个以 $\textit{s}[r]$ 结尾的长度为 $3$ 的好子字符串。 + +遍历结束后，我们就找到了所有的好子字符串的数量。 + +时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。 + +> 该解法可以拓展到长度为 $k$ 的好子字符串的数量。 @@ -73,10 +83,15 @@ tags: ```python class Solution: def countGoodSubstrings(self, s: str) -> int: - count, n = 0, len(s) - for i in range(n - 2): - count += s[i] != s[i + 1] and s[i] != s[i + 2] and s[i + 1] != s[i + 2] - return count + ans = mask = l = 0 + for r, x in enumerate(map(lambda c: ord(c) - 97, s)): + while mask >> x & 1: + y = ord(s[l]) - 97 + mask ^= 1 << y + l += 1 + mask |= 1 << x + ans += int(r - l + 1 >= 3) + return ans ``` #### Java @@ -84,33 +99,81 @@ class Solution: ```java class Solution { public int countGoodSubstrings(String s) { - int count = 0, n = s.length(); - for (int i = 0; i < n - 2; ++i) { - char a = s.charAt(i), b = s.charAt(i + 1), c = s.charAt(i + 2); - if (a != b && a != c && b != c) { - ++count; + int ans = 0; + int n = s.length(); + for (int l = 0, r = 0, mask = 0; r < n; ++r) { + int x = s.charAt(r) - 'a'; + while ((mask >> x & 1) == 1) { + int y = s.charAt(l++) - 'a'; + mask ^= 1 << y; } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; } - return count; + return ans; } } ``` +#### C++ + +```cpp +class Solution { +public: + int countGoodSubstrings(string s) { + int ans = 0; + int n = s.length(); + for (int l = 0, r = 0, mask = 0; r < n; ++r) { + int x = s[r] - 'a'; + while ((mask >> x & 1) == 1) { + int y = s[l++] - 'a'; + mask ^= 1 << y; + } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; + } + return ans; + } +}; +``` + +#### Go + +```go +func countGoodSubstrings(s string) (ans int) { + mask, l := 0, 0 + for r, c := range s { + x := int(c - 'a') + for (mask>>x)&1 == 1 { + y := int(s[l] - 'a') + l++ + mask ^= 1 << y + } + mask |= 1 << x + if r-l+1 >= 3 { + ans++ + } + } + return +} +``` + #### TypeScript ```ts function countGoodSubstrings(s: string): number { - const n: number = s.length; - let count: number = 0; - for (let i: number = 0; i < n - 2; ++i) { - let a: string = s.charAt(i), - b: string = s.charAt(i + 1), - c: string = s.charAt(i + 2); - if (a != b && a != c && b != c) { - ++count; + let ans = 0; + const n = s.length; + for (let l = 0, r = 0, mask = 0; r < n; ++r) { + const x = s.charCodeAt(r) - 'a'.charCodeAt(0); + while ((mask >> x) & 1) { + const y = s.charCodeAt(l++) - 'a'.charCodeAt(0); + mask ^= 1 << y; } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; } - return count; + return ans; } ``` @@ -123,13 +186,26 @@ class Solution { * @return Integer */ function countGoodSubstrings($s) { - $cnt = 0; - for ($i = 0; $i < strlen($s) - 2; $i++) { - if ($s[$i] != $s[$i + 1] && $s[$i] != $s[$i + 2] && $s[$i + 1] != $s[$i + 2]) { - $cnt++; + $ans = 0; + $n = strlen($s); + $l = 0; + $r = 0; + $mask = 0; + + while ($r < $n) { + $x = ord($s[$r]) - ord('a'); + while (($mask >> $x) & 1) { + $y = ord($s[$l++]) - ord('a'); + $mask ^= 1 << $y; } + $mask |= 1 << $x; + if ($r - $l + 1 >= 3) { + $ans++; + } + $r++; } - return $cnt++; + + return $ans; } } ``` diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README_EN.md b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README_EN.md index 969bd847d861a..eaecc71c46565 100644 --- a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README_EN.md +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/README_EN.md @@ -35,7 +35,7 @@ tags: 
 Input: s = "xyzzaz"
 Output: 1
-Explanation: There are 4 substrings of size 3: "xyz", "yzz", "zza", and "zaz". 
+Explanation: There are 4 substrings of size 3: "xyz", "yzz", "zza", and "zaz".
 The only good substring of length 3 is "xyz".
@@ -62,7 +62,17 @@ The good substrings are "abc", "bca", "cab", and & -### Solution 1 +### Solution 1: Sliding Window + +We can maintain a sliding window such that the characters within the window are not repeated. Initially, we use a binary integer $\textit{mask}$ of length $26$ to represent the characters within the window, where the $i$-th bit being $1$ indicates that character $i$ has appeared in the window, otherwise it indicates that character $i$ has not appeared in the window. + +Then, we traverse the string $s$. For each position $r$, if $\textit{s}[r]$ has appeared in the window, we need to move the left boundary $l$ of the window to the right until there are no repeated characters in the window. After this, we add $\textit{s}[r]$ to the window. At this point, if the length of the window is greater than or equal to $3$, then we have found a good substring of length $3$ ending at $\textit{s}[r]$. + +After the traversal, we have found the number of all good substrings. + +The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$. + +> This solution can be extended to find the number of good substrings of length $k$. @@ -71,10 +81,15 @@ The good substrings are "abc", "bca", "cab", and & ```python class Solution: def countGoodSubstrings(self, s: str) -> int: - count, n = 0, len(s) - for i in range(n - 2): - count += s[i] != s[i + 1] and s[i] != s[i + 2] and s[i + 1] != s[i + 2] - return count + ans = mask = l = 0 + for r, x in enumerate(map(lambda c: ord(c) - 97, s)): + while mask >> x & 1: + y = ord(s[l]) - 97 + mask ^= 1 << y + l += 1 + mask |= 1 << x + ans += int(r - l + 1 >= 3) + return ans ``` #### Java @@ -82,33 +97,81 @@ class Solution: ```java class Solution { public int countGoodSubstrings(String s) { - int count = 0, n = s.length(); - for (int i = 0; i < n - 2; ++i) { - char a = s.charAt(i), b = s.charAt(i + 1), c = s.charAt(i + 2); - if (a != b && a != c && b != c) { - ++count; + int ans = 0; + int n = s.length(); + for (int l = 0, r = 0, mask = 0; r < n; ++r) { + int x = s.charAt(r) - 'a'; + while ((mask >> x & 1) == 1) { + int y = s.charAt(l++) - 'a'; + mask ^= 1 << y; } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; } - return count; + return ans; } } ``` +#### C++ + +```cpp +class Solution { +public: + int countGoodSubstrings(string s) { + int ans = 0; + int n = s.length(); + for (int l = 0, r = 0, mask = 0; r < n; ++r) { + int x = s[r] - 'a'; + while ((mask >> x & 1) == 1) { + int y = s[l++] - 'a'; + mask ^= 1 << y; + } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; + } + return ans; + } +}; +``` + +#### Go + +```go +func countGoodSubstrings(s string) (ans int) { + mask, l := 0, 0 + for r, c := range s { + x := int(c - 'a') + for (mask>>x)&1 == 1 { + y := int(s[l] - 'a') + l++ + mask ^= 1 << y + } + mask |= 1 << x + if r-l+1 >= 3 { + ans++ + } + } + return +} +``` + #### TypeScript ```ts function countGoodSubstrings(s: string): number { - const n: number = s.length; - let count: number = 0; - for (let i: number = 0; i < n - 2; ++i) { - let a: string = s.charAt(i), - b: string = s.charAt(i + 1), - c: string = s.charAt(i + 2); - if (a != b && a != c && b != c) { - ++count; + let ans = 0; + const n = s.length; + for (let l = 0, r = 0, mask = 0; r < n; ++r) { + const x = s.charCodeAt(r) - 'a'.charCodeAt(0); + while ((mask >> x) & 1) { + const y = s.charCodeAt(l++) - 'a'.charCodeAt(0); + mask ^= 1 << y; } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; } - return count; + return ans; } ``` @@ -121,13 +184,26 @@ class Solution { * @return Integer */ function countGoodSubstrings($s) { - $cnt = 0; - for ($i = 0; $i < strlen($s) - 2; $i++) { - if ($s[$i] != $s[$i + 1] && $s[$i] != $s[$i + 2] && $s[$i + 1] != $s[$i + 2]) { - $cnt++; + $ans = 0; + $n = strlen($s); + $l = 0; + $r = 0; + $mask = 0; + + while ($r < $n) { + $x = ord($s[$r]) - ord('a'); + while (($mask >> $x) & 1) { + $y = ord($s[$l++]) - ord('a'); + $mask ^= 1 << $y; } + $mask |= 1 << $x; + if ($r - $l + 1 >= 3) { + $ans++; + } + $r++; } - return $cnt++; + + return $ans; } } ``` diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.cpp b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.cpp new file mode 100644 index 0000000000000..39627616b9aef --- /dev/null +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.cpp @@ -0,0 +1,17 @@ +class Solution { +public: + int countGoodSubstrings(string s) { + int ans = 0; + int n = s.length(); + for (int l = 0, r = 0, mask = 0; r < n; ++r) { + int x = s[r] - 'a'; + while ((mask >> x & 1) == 1) { + int y = s[l++] - 'a'; + mask ^= 1 << y; + } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; + } + return ans; + } +}; diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.go b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.go new file mode 100644 index 0000000000000..e5cbee12d2e9d --- /dev/null +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.go @@ -0,0 +1,16 @@ +func countGoodSubstrings(s string) (ans int) { + mask, l := 0, 0 + for r, c := range s { + x := int(c - 'a') + for (mask>>x)&1 == 1 { + y := int(s[l] - 'a') + l++ + mask ^= 1 << y + } + mask |= 1 << x + if r-l+1 >= 3 { + ans++ + } + } + return +} diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.java b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.java index e0bb30cee667f..5f25fdcac1abe 100644 --- a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.java +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.java @@ -1,12 +1,16 @@ class Solution { public int countGoodSubstrings(String s) { - int count = 0, n = s.length(); - for (int i = 0; i < n - 2; ++i) { - char a = s.charAt(i), b = s.charAt(i + 1), c = s.charAt(i + 2); - if (a != b && a != c && b != c) { - ++count; + int ans = 0; + int n = s.length(); + for (int l = 0, r = 0, mask = 0; r < n; ++r) { + int x = s.charAt(r) - 'a'; + while ((mask >> x & 1) == 1) { + int y = s.charAt(l++) - 'a'; + mask ^= 1 << y; } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; } - return count; + return ans; } -} \ No newline at end of file +} diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.php b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.php index 4677592fedbcf..ebbdb06379db4 100644 --- a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.php +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.php @@ -4,12 +4,25 @@ class Solution { * @return Integer */ function countGoodSubstrings($s) { - $cnt = 0; - for ($i = 0; $i < strlen($s) - 2; $i++) { - if ($s[$i] != $s[$i + 1] && $s[$i] != $s[$i + 2] && $s[$i + 1] != $s[$i + 2]) { - $cnt++; + $ans = 0; + $n = strlen($s); + $l = 0; + $r = 0; + $mask = 0; + + while ($r < $n) { + $x = ord($s[$r]) - ord('a'); + while (($mask >> $x) & 1) { + $y = ord($s[$l++]) - ord('a'); + $mask ^= 1 << $y; } + $mask |= 1 << $x; + if ($r - $l + 1 >= 3) { + $ans++; + } + $r++; } - return $cnt++; + + return $ans; } } diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.py b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.py index 603ec32710f21..c0c1d5957b308 100644 --- a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.py +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.py @@ -1,6 +1,11 @@ class Solution: def countGoodSubstrings(self, s: str) -> int: - count, n = 0, len(s) - for i in range(n - 2): - count += s[i] != s[i + 1] and s[i] != s[i + 2] and s[i + 1] != s[i + 2] - return count + ans = mask = l = 0 + for r, x in enumerate(map(lambda c: ord(c) - 97, s)): + while mask >> x & 1: + y = ord(s[l]) - 97 + mask ^= 1 << y + l += 1 + mask |= 1 << x + ans += int(r - l + 1 >= 3) + return ans diff --git a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.ts b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.ts index e4c044a1243c2..90cfa1e13065c 100644 --- a/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.ts +++ b/solution/1800-1899/1876.Substrings of Size Three with Distinct Characters/Solution.ts @@ -1,13 +1,14 @@ function countGoodSubstrings(s: string): number { - const n: number = s.length; - let count: number = 0; - for (let i: number = 0; i < n - 2; ++i) { - let a: string = s.charAt(i), - b: string = s.charAt(i + 1), - c: string = s.charAt(i + 2); - if (a != b && a != c && b != c) { - ++count; + let ans = 0; + const n = s.length; + for (let l = 0, r = 0, mask = 0; r < n; ++r) { + const x = s.charCodeAt(r) - 'a'.charCodeAt(0); + while ((mask >> x) & 1) { + const y = s.charCodeAt(l++) - 'a'.charCodeAt(0); + mask ^= 1 << y; } + mask |= 1 << x; + ans += r - l + 1 >= 3 ? 1 : 0; } - return count; + return ans; }