vault backup: 2022-07-14 11:03:17

OJ notes/pages/Leetcode Permutation-In-String.md

@@ -53,6 +53,47 @@ In other words, return `true` if one of `s1`'s permutations is the substring of
 
 I tried to use kadane's algorithm, but the problem is a premature string, not set. So I gave up.
 
-
+Then I found out this is sliding window problem, similar to
+[[Leetcode Longest-Substring-Without-Repeating-Characters]]
 
 ### Solution
+
+```cpp
+class Solution {
+public:
+  bool checkInclusion(string s1, string s2) {
+    // Sliding windows problem.
+
+    // First, register the freqs of s1
+    vector<int> freq(26, 0);
+    for (char ch : s1) {
+      freq[ch - 'a']++;
+    }
+
+    // Second, start constructing window and window freq;
+    int len = s1.length();
+    int l = 0, r = 0;
+    vector<int> window(26, 0);
+
+    // Start expanding and sliding!
+    while (r < s2.length()) {
+      // I add value to window here, to avoid r OOB when r = s2.length
+      window[s2[r] - 'a']++;
+      if (r - l + 1 < len) {
+        // Expand right
+        r++;
+      } else if (window == freq) { // note that I use else if, to make sure
+                                   // every time r is incremented once.
+        return true;
+      } else {
+        // Remove l from the window, slide right
+        window[s2[l] - 'a']--;
+        l++;
+        r++;
+      }
+    }
+
+    return false;
+  }
+};
+```