21/10

Author: jaygajera17

Remove_Character.cpp

@@ -0,0 +1,33 @@
+/*
+Link:- https://practice.geeksforgeeks.org/problems/remove-character3815/1
+Problem:- Given two strings string1 and string2, remove those characters from first string(string1) which are present in second string(string2). Both the strings are different and contain only lowercase characters.
+
+Input:
+string1 = "computer"
+string2 = "cat"
+Output: "ompuer"
+Explanation: After removing characters(c, a, t)
+from string1 we get "ompuer".
+
+Input:
+string1 = "occurrence"
+string2 = "car"
+Output: "ouene"
+Explanation: After removing characters
+(c, a, r) from string1 we get "ouene".
+
+*/
+  string removeChars(string s1, string s2) {
+        // code here
+        string ans="";
+        
+        for(int i=0;i<s1.length();i++)
+        {
+            // npos -> not found
+            if(s2.find(s1[i])==string::npos)
+            {
+                ans=ans+s1[i];
+            }  
+        }
+        return ans;
+    }

StrStr.cpp

@@ -0,0 +1,60 @@
+/*
+
+Link:- https://practice.geeksforgeeks.org/problems/implement-strstr/1
+
+Problem:- Your task is to implement the function strstr. The function takes two strings as arguments (s,x) and  locates the occurrence of the string x in the string s. The function returns and integer denoting the first occurrence of the string x in s (0 based indexing).
+
+Note: You are not allowed to use inbuilt function.
+
+Input:
+s = GeeksForGeeks, x = Fr
+Output: -1
+Explanation: Fr is not present in the
+string GeeksForGeeks as substring.
+
+Input:
+s = GeeksForGeeks, x = For
+Output: 5
+Explanation: For is present as substring
+in GeeksForGeeks from index 5 (0 based
+indexing).
+
+*/
+
+/******  Using  String Matching Naive Algo  */
+ if(x.length()>s.length())
+    {
+        return -1;
+    }
+    
+    for(int i=0;i<=s.length()-x.length();i++)
+    {
+        for(int j=0;j<x.length();j++)
+        {
+          if(s[i+j]!=x[j]) 
+          {
+              break;
+          }
+          else if(j==x.length()-1)
+            {
+                return i;
+            }
+           
+        }
+    }
+    return -1;
+
+
+
+// ******** Using inbuild Function find() ********
+   
+// The function returns the index of the first occurrence of sub-string, if the sub-string is not found it returns string::npos
+ if(s.find(x)!=string::npos)
+     {
+         return s.find(x);
+     }
+     else
+     {
+         return -1;
+     }
+

check_for_SUBSEQUENCE.cpp

@@ -0,0 +1,43 @@
+/*
+
+problem:- Given two strings A and B, find if A is a subsequence of B.
+
+Link:- https://practice.geeksforgeeks.org/problems/check-for-subsequence4930/1
+
+
+Input:
+A = AXY 
+B = YADXCP
+Output: 0 
+Explanation: A is not a subsequence of B
+as 'Y' appears before 'A'.
+
+Input:
+A = gksrek
+B = geeksforgeeks
+Output: 1
+Explanation: A is a subsequence of B.
+
+*/
+
+  bool isSubSequence(string s1, string s2) 
+    {
+        // code here
+        int i,j=0;
+        while(i<s1.length() && j<s2.length())
+        {
+            if(s1[i]==s2[j])
+            {
+                i++;
+            }
+                j++;
+        }
+        if(i==s1.length())
+        {
+            return 1;
+        }
+        else
+        {
+            return 0;
+        }
+    }

find_occurence_of_s2_in_s1.cpp

@@ -0,0 +1,41 @@
+/*
+
+Link:- https://practice.geeksforgeeks.org/problems/maximum-occuring-character-1587115620/1
+
+problem:- Given a string str. The task is to find the maximum occurring character in the string str. If more than one character occurs the maximum number of time then print the lexicographically smaller character.
+
+Input:
+str = testsample
+Output: e
+Explanation: e is the character which
+is having the highest frequency.
+
+Input:
+str = output
+Output: t
+Explanation:  t and u are the characters
+with the same frequency, but t is
+lexicographically smaller.
+
+*/
+
+ char getMaxOccuringChar(string str)
+    {
+        // Your code here
+        map<char,int>m;
+        for(int i=0;i<str.length();i++)
+        {
+            m[str[i]]++;
+        }
+        int max=0;
+        char c;
+        for(auto i=m.begin();i!=m.end();i++)
+        {
+            if(max<i->second)
+            {
+                max=i->second;
+                c=i->first;
+            }
+        }
+        return c;
+    }

maximum_occuring_char.cpp

@@ -36,4 +36,23 @@ class Solution
      }
      return '#'; // '#' = -1 in ascii
     }
-};
+};
+
+/* 
+  if given find FIRST repeated character 
+
+   string ans="";
+    unordered_map<char,int>m;
+    for(int i=0;i<s.length();i++)
+    {
+        m[s[i]]++;
+       
+        if(m[s[i]]==2)
+        {
+            ans=ans+s[i];
+            return  ans;
+        }
+    }
+
+
+*/