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Form binary palindromic String with X 0s and Y 1s

Last Updated : 16 Aug, 2022
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Given two integers M and N the task is to construct a binary palindromic string consisting of M occurrences of 0s and N occurrences of 1s.

Examples:

Input: M = 4, N = 3
Output: 0011100
Explanation: The string 0011100 is palindrome containing 
4 occurrences of '0' and 3 occurrences of '1'.

Input: M = 3, N = 3
Output: -1 

 

Approach: 

For the string to be a palindrome, atleast one among M and N should be even, otherwise if both M and N are odd then string can not be a palindrome.

Follow the below steps to solve the problem:

  • If both M and N are odd then print -1.
  • If N is Even, print '1' N/2 times, then print '0' M times, then again print '1' N/2 times.
  • If M is Even, print '0' M/2 times, then print '1' N times, then again print '0' M/2 times.

Below is the implementation of the above approach: 

C++ 
// C++ code for above approach
#include <iostream>
using namespace std;

// Function to build the palindromic
// string if possible
string buildString(int M, int N, char P, char Q)
{
    string res = "";
    for (int i = 0; i < M / 2; i++)
        res += P;
    for (int i = 0; i < N; i++)
        res += Q;
    for (int i = 0; i < M / 2; i++)
        res += P;
    return res;
}

// Function to check if it is possible
// to make the string palindrome
string makePalin(int M, int N)
{
    // If both M and N are odd then it is
    // impossible to make the
    // string palindrome
    if ((M & 1) && (N & 1)) {
        return "-1";
    }
    else {
        string ans = "";
        if (N & 1) {
            ans = buildString(M, N, '0', '1');
        }
        else {
            ans = buildString(N, M, '1', '0');
        }
        return ans;
    }
}

// Driver Code
int main()
{
    int M = 4;
    int N = 3;

    // Function Call
    cout << makePalin(M, N);
    return 0;
}
Java 
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG
{
  
  // Java code for above approach
  // Function to build the palindromic
  // string if possible
  static String buildString(int M, int N, char P, char Q)
  {
    String res = "";
    for (int i = 0; i < M / 2; i++)
      res += P;
    for (int i = 0; i < N; i++)
      res += Q;
    for (int i = 0; i < M / 2; i++)
      res += P;
    return res;
  }

  // Function to check if it is possible
  // to make the string palindrome
  static String makePalin(int M, int N)
  {
    // If both M and N are odd then it is
    // impossible to make the
    // string palindrome
    if ((M % 2==1) && (N %2==1)) {
      return "-1";
    }
    else {
      String ans = "";
      if (N %2==1) {
        ans = buildString(M, N, '0', '1');
      }
      else {
        ans = buildString(N, M, '1', '0');
      }
      return ans;
    }
  }

  // Driver Code
  public static void main (String[] args)
  {
    int M = 4;
    int N = 3;

    // Function Call
    System.out.println(makePalin(M, N));
  }
}

// This code is contributed by satwik4409.
Python3 
# Python code for above approach

# Function to build the palindromic
# string if possible
def buildString(M, N, P, Q):
    res = ""
    for i in range(M//2):
        res += P
    for i in range(N):
        res += Q
    for i in range(M//2):
        res += P
    return res

# Function to check if it is possible
# to make the string palindrome
def makePalin(M, N):
  
    # If both M and N are odd then it is
    # impossible to make the
    # string palindrome
    if (((M & 1) != 0) and ((N & 1) != 0)):
        return "-1"
    else:
        ans = ""
        if ((N & 1) != 0):
            ans = buildString(M, N, '0', '1')
        else:
            ans = buildString(N, M, '1', '0')

        return ans

# Driver Code
if __name__ == "__main__":
    M = 4
    N = 3
    # Function Call
    print(makePalin(M, N))

# This code is contributed by Rohit Pradhan
C# 
using System;

public class GFG{

  // Function to build the palindromic
  // string if possible
  public static string buildString(int M, int N, char P, char Q)
  {
    string res = "";
    for (int i = 0; i < M / 2; i++)
      res += P;
    for (int i = 0; i < N; i++)
      res += Q;
    for (int i = 0; i < M / 2; i++)
      res += P;
    return res;
  }

  // Function to check if it is possible
  // to make the string palindrome
  public static string makePalin(int M, int N)
  {
    // If both M and N are odd then it is
    // impossible to make the
    // string palindrome
    if ( Convert.ToBoolean(M & 1) && Convert.ToBoolean(N & 1)) {
      return "-1";
    }
    else {
      string ans = "";
      if (Convert.ToBoolean(N & 1)) {
        ans = buildString(M, N, '0', '1');
      }
      else {
        ans = buildString(N, M, '1', '0');
      }
      return ans;
    }
  }

  static public void Main (){
    int M = 4;
    int N = 3;

    // Function Call
    Console.WriteLine(makePalin(M, N));

  }
}

// This code is contributed by akashish__
JavaScript 
<script>
    // JavaScript code for the above approach

  // Function to build the palindromic
  // string if possible
  function buildString(M, N, P, Q)
  {
    let res = "";
    for (let i = 0; i < M / 2; i++)
      res += P;
    for (let i = 0; i < N; i++)
      res += Q;
    for (let i = 0; i < M / 2; i++)
      res += P;
    return res;
  }

  // Function to check if it is possible
  // to make the string palindrome
  function makePalin(M, N)
  {
    // If both M and N are odd then it is
    // impossible to make the
    // string palindrome
    if ((M % 2==1) && (N %2==1)) {
      return "-1";
    }
    else {
      let ans = "";
      if (N %2==1) {
        ans = buildString(M, N, '0', '1');
      }
      else {
        ans = buildString(N, M, '1', '0');
      }
      return ans;
    }
  }

    // Driver Code

    let M = 4;
    let N = 3;

    // Function Call
    document.write(makePalin(M, N));
    
    // This code is contributed by sanjoy_62.
</script>

Output
0011100

Time Complexity: O(M+N)
Auxiliary Space: O(1)


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