feat: add solutions to lc problem: No.1100

No.1100.Find K-Length Substrings With No Repeated Characters

Author: yanglbme

solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/README.md

@@ -39,7 +39,17 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
-固定大小的滑动窗口。
+**方法一：双指针 + 计数器**
+
+我们观察发现，字符均为小写字母，也即最多有 $26$ 种不同的字符。因此，如果 $k \gt 26$ 或者 $k \gt n$，则无法找到任何长度为 $k$ 且不含重复字符的子串，直接返回 $0$ 即可。
+
+接下来，我们用双指针 $j$ 和 $i$ 维护一个滑动窗口，其中 $j$ 是滑动窗口的左端点，$i$ 是滑动窗口的右端点，用一个计数器 $cnt$ 统计滑动窗口中每个字符出现的次数。
+
+遍历字符串 $s$，每次将 $s[i]$ 加入滑动窗口，即 $cnt[s[i]]++$，如果此时 $cnt[s[i]] \gt 1$ 或者 $i - j + 1 \gt k$，则循环将 $s[j]$ 从滑动窗口中移除，即 $cnt[s[j]]--$，并将 $j$ 右移。如果 $j$ 右移结束后，窗口大小 $i - j + 1$ 恰好等于 $k$，则说明滑动窗口中的字符串是一个符合题意的子串，将结果加一。
+
+遍历结束后，即可得到所有符合题意的子串的个数。
+
+时间复杂度 $O(n)$，空间复杂度 $O(C)$。其中 $n$ 为字符串 $s$ 的长度；而 $C$ 为字符集的大小，本题中 $C = 26$。
 
 <!-- tabs:start -->
 
@@ -50,14 +60,17 @@
 ```python
 class Solution:
     def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
+        n = len(s)
+        if k > n or k > 26:
+            return 0
         ans = j = 0
-        mp = {}
+        cnt = Counter()
         for i, c in enumerate(s):
-            if c in mp:
-                j = max(j, mp[c] + 1)
-            mp[c] = i
-            if i - j + 1 >= k:
-                ans += 1
+            cnt[c] += 1
+            while cnt[c] > 1 or i - j + 1 > k:
+                cnt[s[j]] -= 1
+                j += 1
+            ans += i - j + 1 == k
         return ans
 ```
 
@@ -68,17 +81,18 @@ class Solution:
 ```java
 class Solution {
     public int numKLenSubstrNoRepeats(String s, int k) {
+        int n = s.length();
+        if (k > n || k > 26) {
+            return 0;
+        }
+        int[] cnt = new int[128];
         int ans = 0;
-        Map<Character, Integer> mp = new HashMap<>();
-        for (int i = 0, j = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            if (mp.containsKey(c)) {
-                j = Math.max(j, mp.get(c) + 1);
-            }
-            mp.put(c, i);
-            if (i - j + 1 >= k) {
-                ++ans;
+        for (int i = 0, j = 0; i < n; ++i) {
+            ++cnt[s.charAt(i)];
+            while (cnt[s.charAt(i)] > 1 || i - j + 1 > k) {
+                cnt[s.charAt(j++)]--;
             }
+            ans += i - j + 1 == k ? 1 : 0;
         }
         return ans;
     }
@@ -91,13 +105,18 @@ class Solution {
 class Solution {
 public:
     int numKLenSubstrNoRepeats(string s, int k) {
+        int n = s.size();
+        if (k > n || k > 26) {
+            return 0;
+        }
+        int cnt[128]{};
         int ans = 0;
-        unordered_map<int, int> mp;
-        for (int i = 0, j = 0; i < s.size(); ++i) {
-            char c = s[i];
-            if (mp.count(c)) j = max(j, mp[c] + 1);
-            mp[c] = i;
-            if (i - j + 1 >= k) ++ans;
+        for (int i = 0, j = 0; i < n; ++i) {
+            ++cnt[s[i]];
+            while (cnt[s[i]] > 1 || i - j + 1 > k) {
+                --cnt[s[j++]];
+            }
+            ans += i - j + 1 == k;
         }
         return ans;
     }
@@ -107,26 +126,22 @@ public:
 ### **Go**
 
 ```go
-func numKLenSubstrNoRepeats(s string, k int) int {
-	mp := make(map[rune]int)
-	ans, j := 0, 0
-	for i, c := range s {
-		if v, ok := mp[c]; ok {
-			j = max(j, v+1)
+func numKLenSubstrNoRepeats(s string, k int) (ans int) {
+	if k > len(s) || k > 26 {
+		return 0
+	}
+	cnt := [128]int{}
+	for i, j := 0, 0; i < len(s); i++ {
+		cnt[s[i]]++
+		for cnt[s[i]] > 1 || i-j+1 > k {
+			cnt[s[j]]--
+			j++
 		}
-		mp[c] = i
-		if i-j+1 >= k {
+		if i-j+1 == k {
 			ans++
 		}
 	}
-	return ans
-}
-
-func max(a, b int) int {
-	if a > b {
-		return a
-	}
-	return b
+	return
 }
 ```
 

solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/README_EN.md

@@ -41,14 +41,17 @@
 ```python
 class Solution:
     def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
+        n = len(s)
+        if k > n or k > 26:
+            return 0
         ans = j = 0
-        mp = {}
+        cnt = Counter()
         for i, c in enumerate(s):
-            if c in mp:
-                j = max(j, mp[c] + 1)
-            mp[c] = i
-            if i - j + 1 >= k:
-                ans += 1
+            cnt[c] += 1
+            while cnt[c] > 1 or i - j + 1 > k:
+                cnt[s[j]] -= 1
+                j += 1
+            ans += i - j + 1 == k
         return ans
 ```
 
@@ -57,17 +60,18 @@ class Solution:
 ```java
 class Solution {
     public int numKLenSubstrNoRepeats(String s, int k) {
+        int n = s.length();
+        if (k > n || k > 26) {
+            return 0;
+        }
+        int[] cnt = new int[128];
         int ans = 0;
-        Map<Character, Integer> mp = new HashMap<>();
-        for (int i = 0, j = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            if (mp.containsKey(c)) {
-                j = Math.max(j, mp.get(c) + 1);
-            }
-            mp.put(c, i);
-            if (i - j + 1 >= k) {
-                ++ans;
+        for (int i = 0, j = 0; i < n; ++i) {
+            ++cnt[s.charAt(i)];
+            while (cnt[s.charAt(i)] > 1 || i - j + 1 > k) {
+                cnt[s.charAt(j++)]--;
             }
+            ans += i - j + 1 == k ? 1 : 0;
         }
         return ans;
     }
@@ -80,13 +84,18 @@ class Solution {
 class Solution {
 public:
     int numKLenSubstrNoRepeats(string s, int k) {
+        int n = s.size();
+        if (k > n || k > 26) {
+            return 0;
+        }
+        int cnt[128]{};
         int ans = 0;
-        unordered_map<int, int> mp;
-        for (int i = 0, j = 0; i < s.size(); ++i) {
-            char c = s[i];
-            if (mp.count(c)) j = max(j, mp[c] + 1);
-            mp[c] = i;
-            if (i - j + 1 >= k) ++ans;
+        for (int i = 0, j = 0; i < n; ++i) {
+            ++cnt[s[i]];
+            while (cnt[s[i]] > 1 || i - j + 1 > k) {
+                --cnt[s[j++]];
+            }
+            ans += i - j + 1 == k;
         }
         return ans;
     }
@@ -96,26 +105,22 @@ public:
 ### **Go**
 
 ```go
-func numKLenSubstrNoRepeats(s string, k int) int {
-	mp := make(map[rune]int)
-	ans, j := 0, 0
-	for i, c := range s {
-		if v, ok := mp[c]; ok {
-			j = max(j, v+1)
+func numKLenSubstrNoRepeats(s string, k int) (ans int) {
+	if k > len(s) || k > 26 {
+		return 0
+	}
+	cnt := [128]int{}
+	for i, j := 0, 0; i < len(s); i++ {
+		cnt[s[i]]++
+		for cnt[s[i]] > 1 || i-j+1 > k {
+			cnt[s[j]]--
+			j++
 		}
-		mp[c] = i
-		if i-j+1 >= k {
+		if i-j+1 == k {
 			ans++
 		}
 	}
-	return ans
-}
-
-func max(a, b int) int {
-	if a > b {
-		return a
-	}
-	return b
+	return
 }
 ```
 

solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.cpp

@@ -1,13 +1,18 @@
 class Solution {
 public:
     int numKLenSubstrNoRepeats(string s, int k) {
+        int n = s.size();
+        if (k > n || k > 26) {
+            return 0;
+        }
+        int cnt[128]{};
         int ans = 0;
-        unordered_map<int, int> mp;
-        for (int i = 0, j = 0; i < s.size(); ++i) {
-            char c = s[i];
-            if (mp.count(c)) j = max(j, mp[c] + 1);
-            mp[c] = i;
-            if (i - j + 1 >= k) ++ans;
+        for (int i = 0, j = 0; i < n; ++i) {
+            ++cnt[s[i]];
+            while (cnt[s[i]] > 1 || i - j + 1 > k) {
+                --cnt[s[j++]];
+            }
+            ans += i - j + 1 == k;
         }
         return ans;
     }

solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.go

@@ -1,21 +1,17 @@
-func numKLenSubstrNoRepeats(s string, k int) int {
-	mp := make(map[rune]int)
-	ans, j := 0, 0
-	for i, c := range s {
-		if v, ok := mp[c]; ok {
-			j = max(j, v+1)
+func numKLenSubstrNoRepeats(s string, k int) (ans int) {
+	if k > len(s) || k > 26 {
+		return 0
+	}
+	cnt := [128]int{}
+	for i, j := 0, 0; i < len(s); i++ {
+		cnt[s[i]]++
+		for cnt[s[i]] > 1 || i-j+1 > k {
+			cnt[s[j]]--
+			j++
 		}
-		mp[c] = i
-		if i-j+1 >= k {
+		if i-j+1 == k {
 			ans++
 		}
 	}
-	return ans
-}
-
-func max(a, b int) int {
-	if a > b {
-		return a
-	}
-	return b
+	return
 }

solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.java

@@ -1,16 +1,17 @@
 class Solution {
     public int numKLenSubstrNoRepeats(String s, int k) {
+        int n = s.length();
+        if (k > n || k > 26) {
+            return 0;
+        }
+        int[] cnt = new int[128];
         int ans = 0;
-        Map<Character, Integer> mp = new HashMap<>();
-        for (int i = 0, j = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            if (mp.containsKey(c)) {
-                j = Math.max(j, mp.get(c) + 1);
-            }
-            mp.put(c, i);
-            if (i - j + 1 >= k) {
-                ++ans;
+        for (int i = 0, j = 0; i < n; ++i) {
+            ++cnt[s.charAt(i)];
+            while (cnt[s.charAt(i)] > 1 || i - j + 1 > k) {
+                cnt[s.charAt(j++)]--;
             }
+            ans += i - j + 1 == k ? 1 : 0;
         }
         return ans;
     }

solution/1100-1199/1100.Find K-Length Substrings With No Repeated Characters/Solution.py

@@ -1,11 +1,14 @@
 class Solution:
     def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
+        n = len(s)
+        if k > n or k > 26:
+            return 0
         ans = j = 0
-        mp = {}
+        cnt = Counter()
         for i, c in enumerate(s):
-            if c in mp:
-                j = max(j, mp[c] + 1)
-            mp[c] = i
-            if i - j + 1 >= k:
-                ans += 1
+            cnt[c] += 1
+            while cnt[c] > 1 or i - j + 1 > k:
+                cnt[s[j]] -= 1
+                j += 1
+            ans += i - j + 1 == k
         return ans