Given an integer N, the task is to generate all the binary strings with equal 0s and 1s. If no strings are possible, print -1
Examples:
Input: N = 2
Output: “01”, “10”
Explanation: All possible binary strings of length 2 are: 01, 10, 11, 00. Out of these, only 2 have equal number of 0s and 1sInput: 4
Output: “0011”, “0101”, “0110”, “1100”, “1010”, “1001”
Approach: The task can be solved by using recursion. If N is odd, then the answer is -1, else, we can use recursion to generate all the binary strings with equal 0s and 1s. Follow the below steps to solve the problem:
- Variable ones keep track of the number of 1’s and variable zeros keeps a track of the number of 0’s in the string.
- Both ones and zeros should have frequency N/2.
- Base condition: The string s stores the output string. So, when the length of s reaches N we stop recursive calls and print the output string s.
- If the frequency of 1’s is less than N/2 then add 1 to the string and increment ones.
- If the frequency of 0’s is less than N/2 then add 0 to the string and increment zeros.
Below is the implementation of the above code:
C++
// C++ code for the above approach#include <bits/stdc++.h>using namespace std;// Recursive function that prints// all strings of N length with equal 1's and 0'svoid binaryNum(int n, string s, int ones, int zeros){ // String s contains the output to be printed // ones stores the frequency of 1's // zeros stores the frequency of 0's // Base Condition: When the length of string s // becomes N if (s.length() == n) { cout << (s) << endl; return; } // If frequency of 1's is less than N/2 then // add 1 to the string and increment ones if (ones < n / 2) binaryNum(n, s + "1", ones + 1, zeros); // If frequency of 0's is less than N/2 then // add 0 to the string and increment zeros if (zeros < n / 2) binaryNum(n, s + "0", ones, zeros + 1);}// Driver Codeint main(){ string s = ""; binaryNum(4, s, 0, 0); return 0;}// This code is contributed by Potta Lokesh |
Java
// Java program for the above approachimport java.io.*;class GFG { // Recursive function that prints // all strings of N length with equal 1's and 0's static void binaryNum(int n, String s, int ones, int zeros) { // String s contains the output to be printed // ones stores the frequency of 1's // zeros stores the frequency of 0's // Base Condition: When the length of string s // becomes N if (s.length() == n) { System.out.println(s); return; } // If frequency of 1's is less than N/2 then // add 1 to the string and increment ones if (ones < n / 2) binaryNum(n, s + "1", ones + 1, zeros); // If frequency of 0's is less than N/2 then // add 0 to the string and increment zeros if (zeros < n / 2) binaryNum(n, s + "0", ones, zeros + 1); } // Driver Code public static void main(String[] args) { String s = ""; binaryNum(4, s, 0, 0); }} |
Python3
# python code for the above approach# Recursive function that prints# all strings of N length with equal 1's and 0'sdef binaryNum(n, s, ones, zeros): # String s contains the output to be printed # ones stores the frequency of 1's # zeros stores the frequency of 0's # Base Condition: When the length of string s # becomes N if (len(s) == n): print(s) return # If frequency of 1's is less than N/2 then # add 1 to the string and increment ones if (ones < n / 2): binaryNum(n, s + "1", ones + 1, zeros) # If frequency of 0's is less than N/2 then # add 0 to the string and increment zeros if (zeros < n / 2): binaryNum(n, s + "0", ones, zeros + 1)# Driver Codeif __name__ == "__main__": s = "" binaryNum(4, s, 0, 0)# This code is contributed by rakeshsahni |
C#
// C# program for the above approachusing System;class GFG { // Recursive function that prints // all strings of N length with equal 1's and 0's static void binaryNum(int n, string s, int ones, int zeros) { // String s contains the output to be printed // ones stores the frequency of 1's // zeros stores the frequency of 0's // Base Condition: When the length of string s // becomes N if (s.Length == n) { Console.WriteLine(s); return; } // If frequency of 1's is less than N/2 then // add 1 to the string and increment ones if (ones < n / 2) binaryNum(n, s + "1", ones + 1, zeros); // If frequency of 0's is less than N/2 then // add 0 to the string and increment zeros if (zeros < n / 2) binaryNum(n, s + "0", ones, zeros + 1); } // Driver Code public static void Main(string[] args) { string s = ""; binaryNum(4, s, 0, 0); }}// This code is contributed by ukasp. |
Javascript
<script>// javascript program for the above approach // Recursive function that prints // all strings of N length with equal 1's and 0's function binaryNum(n, s, ones, zeros) { // String s contains the output to be printed // ones stores the frequency of 1's // zeros stores the frequency of 0's // Base Condition: When the length of string s // becomes N if (s.length == n) { document.write(s+"<br>"); return; } // If frequency of 1's is less than N/2 then // add 1 to the string and increment ones if (ones < n / 2) binaryNum(n, s + "1", ones + 1, zeros); // If frequency of 0's is less than N/2 then // add 0 to the string and increment zeros if (zeros < n / 2) binaryNum(n, s + "0", ones, zeros + 1); }// Driver Codevar s = "";binaryNum(4, s, 0, 0);// This code is contributed by 29AjayKumar </script> |
Output
1100 1010 1001 0110 0101 0011
Time Complexity: O(2N)
Auxiliary Space: O(1)
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