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Check if a string can be split into even length palindromic substrings

  • Difficulty Level : Easy
  • Last Updated : 22 Jul, 2021
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Given a string str, the task is to check if it is possible to split the given string into even length palindromic substrings
Examples: 
 

Input: str = “abbacc” 
Output: Yes 
Explanation: 
Strings “abba” and “cc” are the even length palindromic substrings.
Input: str = “abcde” 
Output: No 
Explanation: 
No even length palindromic substrings possible. 
 

 

Approach: The idea is to use Stack Data Structure. Below are the steps: 
 

  1. Initialise an empty stack.
  2. Traverse the given string str.
  3. For each character in the given string, do the following: 
    • If the character is equal to the character at the top of the stack then pop the top element from the stack.
    • Else push the current character into the stack.
  4. If stack is empty after the above steps then the given string can be break into a palindromic substring of even length.
  5. Else the given string cannot be break into palindromic substring of even length.

Below is the implementation of the above approach:
 



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check string str can be
// split a string into even length
// palindromic substrings
bool check(string s, int n)
{
 
    // Initialize a stack
    stack<char> st;
 
    // Iterate the string
    for (int i = 0; i < n; i++) {
 
        // If the i-th character is same
        // as that at the top of the stack
        // then pop the top element
        if (!st.empty() && st.top() == s[i])
            st.pop();
 
        // Else push the current charactor
        // into the stack
        else
            st.push(s[i]);
    }
 
    // If the stack is empty, then even
    // palindromic substrings are possible
    if (st.empty()) {
        return true;
    }
 
    // Else not-possible
    else {
        return false;
    }
}
 
// Driver Code
int main()
{
    // Given string
    string str = "aanncddc";
 
    int n = str.length();
 
    // Function Call
    if (check(str, n)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
 
    return 0;
}

Java




// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to check String str can be
// split a String into even length
// palindromic subStrings
static boolean check(String s, int n)
{
     
    // Initialize a stack
    Stack<Character> st = new Stack<Character>();
 
    // Iterate the String
    for(int i = 0; i < n; i++)
    {
         
       // If the i-th character is same
       // as that at the top of the stack
       // then pop the top element
       if (!st.isEmpty() &&
               st.peek() == s.charAt(i))
           st.pop();
            
       // Else push the current charactor
       // into the stack
       else
           st.add(s.charAt(i));
    }
     
    // If the stack is empty, then even
    // palindromic subStrings are possible
    if (st.isEmpty())
    {
        return true;
    }
     
    // Else not-possible
    else
    {
        return false;
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given String
    String str = "aanncddc";
 
    int n = str.length();
 
    // Function Call
    if (check(str, n))
    {
        System.out.print("Yes" + "\n");
    }
    else
    {
        System.out.print("No" + "\n");
    }
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 program for the above approach
 
# Function to check string str can be
# split a string into even length
# palindromic substrings
def check(s, n):
 
    st = []
 
    # Iterate the string
    for i in range(n):
 
        # If the i-th character is same
        # as that at the top of the stack
        # then pop the top element
        if (len(st) != 0 and
         st[len(st) - 1] == s[i]):
            st.pop();
 
        # Else push the current charactor
        # into the stack
        else:
            st.append(s[i]);
     
    # If the stack is empty, then even
    # palindromic substrings are possible
    if (len(st) == 0):
        return True;
     
    # Else not-possible
    else:
        return False;
     
# Driver Code
 
# Given string
str = "aanncddc";
n = len(str)
 
# Function Call
if (check(str, n)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by grand_master   

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to check String str can be
// split a String into even length
// palindromic subStrings
static bool check(String s, int n)
{
     
    // Initialize a stack
    Stack<int> st = new Stack<int>();
 
    // Iterate the String
    for(int i = 0; i < n; i++)
    {
         
        // If the i-th character is same
        // as that at the top of the stack
        // then pop the top element
        if (st.Count != 0 &&
            st.Peek() == s[i])
            st.Pop();
                 
        // Else push the current charactor
        // into the stack
        else
            st.Push(s[i]);
    }
     
    // If the stack is empty, then even
    // palindromic subStrings are possible
    if (st.Count == 0)
    {
        return true;
    }
     
    // Else not-possible
    else
    {
        return false;
    }
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given String
    String str = "aanncddc";
 
    int n = str.Length;
 
    // Function call
    if (check(str, n))
    {
        Console.Write("Yes" + "\n");
    }
    else
    {
        Console.Write("No" + "\n");
    }
}
}
 
// This code is contributed by sapnasingh4991

Javascript




<script>
 
// JavaScript program for the above approach
 
// Function to check string str can be
// split a string into even length
// palindromic substrings
function check(s, n)
{
 
    // Initialize a stack
    var st = [];
 
    // Iterate the string
    for (var i = 0; i < n; i++) {
 
        // If the i-th character is same
        // as that at the top of the stack
        // then pop the top element
        if (st.length!=0 && st[st.length-1] == s[i])
            st.pop();
 
        // Else push the current charactor
        // into the stack
        else
            st.push(s[i]);
    }
 
    // If the stack is empty, then even
    // palindromic substrings are possible
    if (st.length==0) {
        return true;
    }
 
    // Else not-possible
    else {
        return false;
    }
}
 
// Driver Code
 
// Given string
var str = "aanncddc";
var n = str.length;
 
// Function Call
if (check(str, n)) {
    document.write( "Yes" );
}
else {
    document.write( "No" );
}
 
</script>
Output: 
Yes

 

Time Complexity: O(N), where N is the length of the given string.
 




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