Construct binary palindrome by repeated appending and trimming

Given n and k, Construct a palindrome of size n using a binary number of size k repeating itself to wrap into the palindrome. The palindrome must always begin with 1 and contains maximum number of zeros.

Examples :

Input : n = 5,  k = 3
Output : 11011 
Explanation : the 3 sized substring is
110 combined twice and trimming the extra 
0 in the end to give 11011.

Input : n = 2,  k = 8
Output : 11 
Explanation : the 8 sized substring is 11...... 
wrapped to two places to give 11.

The naive approach would be to try every palindrome of size k starting with 1 such that a palindrome of size n is formed. This approach has an exponential complexity.

A better way to do this is to initialize the k sized binary number with the index and connect the palindrome in the way it should be. Like last character of palindrome should match to first, find which indexes will be present at those locations and link them. Set every character linked with 0th index to 1 and the string is ready. This approach will have a linear complexity.

In this approach, first lay the index of the k sized binary to hold into an array, for example if n = 7, k = 3 arr becomes [0, 1, 2, 0, 1, 2, 0]. Following that in the connect chars graph, connect the indices of the k sized binary which should be same by going through the property of palindrome which is kth and (n – k – 1)th variable should be same, such that 0 is linked to 1(and vice versa), 1 is linked to 2(and vice versa) and so on. After that, check what is linked with 0 in connect chars array and make all of the associated indices one (because the first number should be non-zero) by using dfs approach. In the dfs, pass 0, the final answer string and the graph. 

Begin by making the parent 1 and checking if its children are zero, if they are make them and their children 1. This makes only the required indices of the k sized string one, others are left zero. Finally, the answer contains the 0 to k – 1 indexes and corresponding to arr the digits are printed. 

Implementation:

C++

// CPP code to form binary palindrome
#include <iostream>
#include <vector>
using namespace std;

// function to apply DFS
void dfs(int parent, int ans[], vector<int> connectchars[])
{
    // set the parent marked
    ans[parent] = 1;

    // if the node has not been visited set it and
    // its children marked
    for (int i = 0; i < connectchars[parent].size(); i++) {
        if (!ans[connectchars[parent][i]])
            dfs(connectchars[parent][i], ans, connectchars);
    }
}

void printBinaryPalindrome(int n, int k)
{
    int arr[n], ans[n] = { 0 };

    // link which digits must be equal
    vector<int> connectchars[k];

    for (int i = 0; i < n; i++)
        arr[i] = i % k;

    // connect the two indices
    for (int i = 0; i < n / 2; i++) {
        connectchars[arr[i]].push_back(arr[n - i - 1]);
        connectchars[arr[n - i - 1]].push_back(arr[i]);
    }

    // set everything connected to
    // first character as 1
    dfs(0, ans, connectchars);

    for (int i = 0; i < n; i++)
        cout << ans[arr[i]];
}

// driver code
int main()
{
    int n = 10, k = 4;
    printBinaryPalindrome(n, k);
    return 0;
}

Java

// JAVA code to form binary palindrome
import java.util.*;

class GFG
{

// function to apply DFS
static void dfs(int parent, int ans[],
                Vector<Integer> connectchars[])
{
    // set the parent marked
    ans[parent] = 1;

    // if the node has not been visited set it and
    // its children marked
    for (int i = 0; i < connectchars[parent].size(); i++)
    {
        if (ans[connectchars[parent].get(i)] != 1)
            dfs(connectchars[parent].get(i), ans, connectchars);
    }
}

static void printBinaryPalindrome(int n, int k)
{
    int []arr = new int[n];
    int []ans = new int[n];

    // link which digits must be equal
    Vector<Integer> []connectchars = new Vector[k];
    for (int i = 0; i < k; i++)
        connectchars[i] = new Vector<Integer>();
    for (int i = 0; i < n; i++)
        arr[i] = i % k;

    // connect the two indices
    for (int i = 0; i < n / 2; i++)
    {
        connectchars[arr[i]].add(arr[n - i - 1]);
        connectchars[arr[n - i - 1]].add(arr[i]);
    }

    // set everything connected to
    // first character as 1
    dfs(0, ans, connectchars);

    for (int i = 0; i < n; i++)
        System.out.print(ans[arr[i]]);
}

// Driver code
public static void main(String[] args)
{
    int n = 10, k = 4;
    printBinaryPalindrome(n, k);
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 code to form binary palindrome

# function to apply DFS
def dfs(parent, ans, connectchars):

    # set the parent marked
    ans[parent] = 1

    # if the node has not been visited
    # set it and its children marked
    for i in range(len(connectchars[parent])):
        if (not ans[connectchars[parent][i]]):
            dfs(connectchars[parent][i], ans,
                             connectchars)

def printBinaryPalindrome(n, k):
    arr = [0] * n
    ans = [0] * n

    # link which digits must be equal
    connectchars = [[] for i in range(k)]

    for i in range(n):
        arr[i] = i % k

    # connect the two indices
    for i in range(int(n / 2)):
        connectchars[arr[i]].append(arr[n - i - 1])
        connectchars[arr[n - i - 1]].append(arr[i])

    # set everything connected to
    # first character as 1
    dfs(0, ans, connectchars)

    for i in range(n):
        print(ans[arr[i]], end = "")

# Driver Code
if __name__ == '__main__':

    n = 10
    k = 4
    printBinaryPalindrome(n, k)

# This code is contributed by PranchalK

C#

// C# code to form binary palindrome
using System;
using System.Collections.Generic;

class GFG
{

// function to apply DFS
static void dfs(int parent, int []ans,
                List<int> []connectchars)
{
    // set the parent marked
    ans[parent] = 1;

    // if the node has not been visited set it and
    // its children marked
    for (int i = 0; i < connectchars[parent].Count; i++)
    {
        if (ans[connectchars[parent][i]] != 1)
            dfs(connectchars[parent][i], ans, connectchars);
    }
}

static void printBinaryPalindrome(int n, int k)
{
    int []arr = new int[n];
    int []ans = new int[n];

    // link which digits must be equal
    List<int> []connectchars = new List<int>[k];
    for (int i = 0; i < k; i++)
        connectchars[i] = new List<int>();
    for (int i = 0; i < n; i++)
        arr[i] = i % k;

    // connect the two indices
    for (int i = 0; i < n / 2; i++)
    {
        connectchars[arr[i]].Add(arr[n - i - 1]);
        connectchars[arr[n - i - 1]].Add(arr[i]);
    }

    // set everything connected to
    // first character as 1
    dfs(0, ans, connectchars);

    for (int i = 0; i < n; i++)
        Console.Write(ans[arr[i]]);
}

// Driver code
public static void Main(String[] args)
{
    int n = 10, k = 4;
    printBinaryPalindrome(n, k);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript code to form binary palindrome


// function to apply DFS
function dfs(parent, ans, connectchars) {
    // set the parent marked
    ans[parent] = 1;

    // if the node has not been visited set it and
    // its children marked
    for (let i = 0; i < connectchars[parent].length; i++) {
        if (!ans[connectchars[parent][i]])
            dfs(connectchars[parent][i], ans, connectchars);
    }
}

function printBinaryPalindrome(n, k) {
    let arr = new Array(n), ans = new Array(n).fill(0);

    // link which digits must be equal
    let connectchars = new Array(k).fill(0).map(() => new Array(k).fill(0));

    for (let i = 0; i < n; i++)
        arr[i] = i % k;

    // connect the two indices
    for (let i = 0; i < n / 2; i++) {
        connectchars[arr[i]].push(arr[n - i - 1]);
        connectchars[arr[n - i - 1]].push(arr[i]);
    }

    // set everything connected to
    // first character as 1
    dfs(0, ans, connectchars);

    for (let i = 0; i < n; i++)
        document.write(ans[arr[i]]);
}

// driver code

let n = 10, k = 4;
printBinaryPalindrome(n, k);

// This code is contributed by gfgking.
</script>

Output

1100110011

Time Complexity: O(n)