Open In App

Construct a binary string following the given constraints

Last Updated : 12 Sep, 2022
Comments
Improve
Suggest changes
Like Article
Like
Report

Given three integers A, B and X. The task is to construct a binary string str which has exactly A number of 0's and B number of 1's provided there has to be at least X indices such that str[i] != str[i+1]. Inputs are such that there's always a valid solution.

Examples: 

Input: A = 2, B = 2, X = 1 
Output: 1100 
There are two 0's and two 1's and one (=X) index such that s[i] != s[i+1] (i.e. i = 1)

Input: A = 4, B = 3, X = 2 
Output: 0111000 

Approach: 

  • Divide x by 2 and store it in a variable d.
  • Check if d is even and d / 2 != a, if the condition is true then print 0 and decrement d and a by 1.
  • Loop from 1 to d and print 10 and in the end update a = a - d and b = b - d.
  • Finally print the remaining 0's and 1's depending on the values of a and b.

Below is the implementation of the above approach:

C++ 
// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to print a binary string which has
// 'a' number of 0's, 'b' number of 1's and there are
// at least 'x' indices such that s[i] != s[i+1]
int constructBinString(int a, int b, int x)
{
    int d, i;

    // Divide index value by 2 and store
    // it into d
    d = x / 2;

    // If index value x is even and
    // x/2 is not equal to a
    if (x % 2 == 0 && x / 2 != a) {
        d--;
        cout << 0;
        a--;
    }

    // Loop for d for each d print 10
    for (i = 0; i < d; i++)
        cout << "10";

    // subtract d from a and b
    a = a - d;
    b = b - d;

    // Loop for b to print remaining 1's
    for (i = 0; i < b; i++) {
        cout << "1";
    }

    // Loop for a to print remaining 0's
    for (i = 0; i < a; i++) {
        cout << "0";
    }
}

// Driver code
int main()
{
    int a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
    return 0;
}
Java 
// Java implementation of the approach
class GFG
{
// Function to print a binary string which has
// 'a' number of 0's, 'b' number of 1's and there are
// at least 'x' indices such that s[i] != s[i+1]
static void constructBinString(int a, int b, int x)
{
    int d, i;

    // Divide index value by 2 and store
    // it into d
    d = x / 2;

    // If index value x is even and
    // x/2 is not equal to a
    if (x % 2 == 0 && x / 2 != a) 
    {
        d--;
        System.out.print("0");
        a--;
    }

    // Loop for d for each d print 10
    for (i = 0; i < d; i++)
        System.out.print("10");

    // subtract d from a and b
    a = a - d;
    b = b - d;

    // Loop for b to print remaining 1's
    for (i = 0; i < b; i++) 
    {
        System.out.print("1");
    }

    // Loop for a to print remaining 0's
    for (i = 0; i < a; i++) 
    {
        System.out.print("0");
    }
}

// Driver code
public static void main(String[] args)
{
    int a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
}
}

// This code is contributed
// by Mukul Singh
Python3 
# Python3 implementation of the above approach 

# Function to print a binary string which 
# has 'a' number of 0's, 'b' number of 1's 
# and there are at least 'x' indices such
# that s[i] != s[i+1] 
def constructBinString(a, b, x): 

    # Divide index value by 2 and 
    # store it into d 
    d = x // 2

    # If index value x is even and 
    # x/2 is not equal to a 
    if x % 2 == 0 and x // 2 != a: 
        d -= 1
        print("0", end = "") 
        a -= 1

    # Loop for d for each d print 10 
    for i in range(d): 
        print("10", end = "") 

    # subtract d from a and b 
    a = a - d 
    b = b - d 

    # Loop for b to print remaining 1's 
    for i in range(b): 
        print("1", end = "")
    
    # Loop for a to print remaining 0's 
    for i in range(a): 
        print("0", end = "")

# Driver Code
if __name__ == "__main__": 

    a, b, x = 4, 3, 2
    constructBinString(a, b, x) 

# This code is contributed by Rituraj_Jain
C# 
// C# implementation of the approach
using System;

class GFG
{
// Function to print a binary string which has
// 'a' number of 0's, 'b' number of 1's and there are
// at least 'x' indices such that s[i] != s[i+1]
static void constructBinString(int a, int b, int x)
{
    int d, i;

    // Divide index value by 2 and store
    // it into d
    d = x / 2;

    // If index value x is even and
    // x/2 is not equal to a
    if (x % 2 == 0 && x / 2 != a) 
    {
        d--;
        Console.Write("0");
        a--;
    }

    // Loop for d for each d print 10
    for (i = 0; i < d; i++)
        Console.Write("10");

    // subtract d from a and b
    a = a - d;
    b = b - d;

    // Loop for b to print remaining 1's
    for (i = 0; i < b; i++) 
    {
        Console.Write("1");
    }

    // Loop for a to print remaining 0's
    for (i = 0; i < a; i++) 
    {
        Console.Write("0");
    }
}

// Driver code
public static void Main()
{
    int a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
}
}

// This code is contributed
// by Akanksha Rai
PHP 
<?php
// PHP implementation of the 
// above approach

// Function to print a binary string 
// which has 'a' number of 0's, 'b' 
// number of 1's and there are at least
// 'x' indices such that s[i] != s[i+1]
function constructBinString($a, $b, $x)
{
    $d; $i;

    // Divide index value by 2 
    // and store it into d
    $d = $x / 2;

    // If index value x is even and
    // x/2 is not equal to a
    if ($x % 2 == 0 && $x / 2 != $a) 
    {
        $d--;
        echo 0;
        $a--;
    }

    // Loop for d for each d print 10
    for ($i = 0; $i < $d; $i++)
        echo "10";

    // subtract d from a and b
    $a = $a - $d;
    $b = $b - $d;

    // Loop for b to print remaining 1's
    for ($i = 0; $i < $b; $i++)
    {
        echo "1";
    }

    // Loop for a to print remaining 0's
    for ($i = 0; $i < $a; $i++) 
    {
        echo "0";
    }
}

// Driver code
$a = 4;
$b = 3;
$x = 2;
constructBinString($a, $b, $x);

// This code is contributed by ajit
?>
JavaScript 
<script>

    // Javascript implementation of the approach
    
    // Function to print a binary string which has
    // 'a' number of 0's, 'b' number of 1's and there are
    // at least 'x' indices such that s[i] != s[i+1]
    function constructBinString(a, b, x)
    {
        let d, i;

        // Divide index value by 2 and store
        // it into d
        d = parseInt(x / 2, 10);

        // If index value x is even and
        // x/2 is not equal to a
        if (x % 2 == 0 && parseInt(x / 2, 10) != a) 
        {
            d--;
            document.write("0");
            a--;
        }

        // Loop for d for each d print 10
        for (i = 0; i < d; i++)
            document.write("10");

        // subtract d from a and b
        a = a - d;
        b = b - d;

        // Loop for b to print remaining 1's
        for (i = 0; i < b; i++) 
        {
            document.write("1");
        }

        // Loop for a to print remaining 0's
        for (i = 0; i < a; i++) 
        {
            document.write("0");
        }
    }
    
    let a = 4, b = 3, x = 2;
    constructBinString(a, b, x);
    
</script>

Output
0111000

Complexity Analysis:

  • Time Complexity: O(max(a,b,x))
  • Auxiliary Space: O(1)

Next Article
Construct a binary string following the given constraints

S
Samdare B
Improve
Article Tags :
Practice Tags :

Similar Reads

We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox