Construct a binary string following the given constraints

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)