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 codeint main(){ int a = 4, b = 3, x = 2; constructBinString(a, b, x); return 0;} |
Java
// Java implementation of the approachclass 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 codepublic 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 Codeif __name__ == "__main__": a, b, x = 4, 3, 2 constructBinString(a, b, x) # This code is contributed by Rituraj_Jain |
C#
// C# implementation of the approachusing 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 codepublic 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)
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