31/10/22

Author: jaygajera17

.vscode/settings.json

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+{
+    "files.associations": {
+        "string": "cpp"
+    }
+}

Palindrome_String.cpp

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+/*
+
+https://practice.geeksforgeeks.org/problems/palindrome-string0817/1
+
+
+Input: S = "abba"
+Output: 1
+Explanation: S is a palindrome
+Example 2:
+
+Input: S = "abc" 
+Output: 0
+Explanation: S is not a palindrome
+
+*/
+
+int isPalindrome(string s)
+	{
+	    // Your code goes here
+	    int n=s.length();
+	    int f=0;
+	    for(int i=0;i<n/2;i++)
+	    {
+	        if(s[i]!=s[n-i-1])
+	        {
+	            f=1;
+	        }
+	    }
+	    if(f==1)
+	    {
+	        return 0;
+	    }else
+	    return 1;
+	}

Reverse_Word_in_String.cpp

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+/*
+
+https://practice.geeksforgeeks.org/problems/reverse-words-in-a-given-string5459/1
+
+Input:
+S = i.like.this.program.very.much
+Output: much.very.program.this.like.i
+Explanation: After reversing the whole
+string(not individual words), the input
+string becomes
+much.very.program.this.like.
+
+
+Input:
+S = pqr.mno
+Output: mno.pqr
+Explanation: After reversing the whole
+string , the input string becomes
+mno.pqr
+
+*/
+
+ string reverseWords(string s) 
+    {
+        // code here 
+      reverse(s.begin(),s.end());
+      int i=0;
+      string st;
+      vector<string>ans;
+      
+      string ans1="";
+      int j=s.length();
+      string t;
+      while(s[j]!='.')  // only push when "." occur so 1st word Not pushed .
+      {
+          t=t+s[j];
+          j--;
+      }
+      while(i!=s.length()+1)
+      {
+          if(s[i]=='.')
+          {
+              reverse(st.begin(),st.end()); //reverse of reverse 
+              ans.push_back(st);
+              ans.push_back('.');
+              st="";
+          }
+          else
+          {
+             st=st+s[i];
+          }
+          i++;
+      }
+      for(auto i:ans)
+      {
+          ans1=ans1+i;
+      }
+     return ans1+t;
+    }

paranthesis_Checker.cpp

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+
+https://practice.geeksforgeeks.org/problems/parenthesis-checker2744/1
+
+Given an expression string x. Examine whether the pairs and the orders of “{“,”}”,”(“,”)”,”[“,”]” are correct in exp.
+For example, the function should return 'true' for exp = “[()]{}{[()()]()}” and 'false' for exp = “[(])”.
+
+Input:
+{([])}
+Output: 
+true
+Explanation: 
+{ ( [ ] ) }. Same colored brackets can form 
+balaced pairs, with 0 number of 
+unbalanced bracket.
+
+Input: 
+([]
+Output: 
+false
+Explanation: 
+([]. Here square bracket is balanced but 
+the small bracket is not balanced and 
+Hence , the output will be unbalanced.
+
+
+  bool ispar(string x)
+    {
+        // Your code here
+      stack<char> s;
+       for(int i=0;i<x.length();i++)
+       {
+           if(s.empty())
+           {
+               s.push(x[i]);
+           }
+           else if((s.top()=='('&& x[i]==')')  ||  (s.top()=='{' && x[i]=='}')  ||  (s.top()=='[' && x[i]==']'))
+            {
+                s.pop();
+            }
+           else
+           {
+               s.push(x[i]);
+           }
+       }
+       if(s.empty())
+       {
+           return true;
+       }
+       return false;
+    }