added day

codenitros · codenitros · commit dec80a2e32c1 · 2020-05-18T21:19:40.000+05:30
diff --git a/presence_of_permuted_substring_day_18.cpp b/presence_of_permuted_substring_day_18.cpp
@@ -0,0 +1,62 @@
+/*
+
+Given two strings s1 and s2, write a function to return true if s2 contains 
+the permutation of s1. In other words, one of the first string's permutations 
+is the substring of the second string.
+
+ 
+
+Example 1:
+
+Input: s1 = "ab" s2 = "eidbaooo"
+Output: True
+Explanation: s2 contains one permutation of s1 ("ba").
+
+*/
+
+//solution
+
+class Solution {
+public:
+    
+    bool compare(char ar1[],char ar2[])
+    {
+        for(int i=0;i<26;i++)
+        {
+            if(ar1[i]!=ar2[i])return false;
+        }
+        return true;
+    }
+    bool checkInclusion(string s1, string s2) {
+        int s1len = s1.length();
+        int s2len = s2.length();
+        int flag=0;
+        //corners
+        if(s1len == 0 || s2len < s1len)return false;
+    
+        char letters[26]={0};
+        char stri[26]={0};
+        for(int i=0;i<s1len;i++)
+        {
+            letters[s1[i]-'a']++;
+            stri[s2[i]-'a']++;
+        }
+        
+        for(int j=s1len;j<s2len;j++)
+        {
+            if(compare(letters,stri))
+            {
+                return true;
+            }
+            stri[s2[j]-'a']++;
+            stri[s2[j-s1len]-'a']--;
+        }
+        if(compare(letters,stri))
+        {
+            return true;
+        }
+        return false;
+        
+        
+    }
+};
