Generate all binary permutations such that there are more or equal 1’s than 0’s before every point in all permutations
Generate all permutations of given length such that every permutation has more or equal 1’s than 0’s in all prefixes of the permutation.
Examples:
Input: len = 4 Output: 1111 1110 1101 1100 1011 1010 Note that a permutation like 0101 can not be in output because there are more 0's from index 0 to 2 in this permutation. Input: len = 3 Output: 111 110 101 Input: len = 2 Output: 11 10
Like permutation generation problems, recursion is the simplest approach to solve this. We start with an empty string, attach 1 to it and recur. While recurring, if we find more 1’s at any point, we append a 0 and make one more recursive call.
Below is the implementation:
C++
// C++ program to generate all permutations of 1's and 0's such that // every permutation has more 1's than 0's at all indexes. #include <iostream> #include <cstring> using namespace std; // ones & zeroes --> counts of 1's and 0's in current string 'str' // len ---> desired length of every permutation void generate(int ones, int zeroes, string str, int len) { // If length of current string becomes same as desired length if (len == str.length()) { cout << str << " "; return; } // Append a 1 and recur generate(ones+1, zeroes, str+"1", len); // If there are more 1's, append a 0 as well, and recur if (ones > zeroes) generate(ones, zeroes+1, str+"0", len); } // Driver program to test above function int main() { string str = ""; generate(0, 0, str, 4); return 0; } |
Java
// Java program to generate all permutations of 1's and 0's such that // every permutation has more 1's than 0's at all indexes. class GFG { // ones & zeroes --> counts of 1's and 0's in current string 'str' // len ---> desired length of every permutation static void generate(int ones, int zeroes, String str, int len) { // If length of current string becomes same as desired length if (len == str.length()) { System.out.print(str + " "); return; } // Append a 1 and recur generate(ones + 1, zeroes, str + "1", len); // If there are more 1's, append a 0 as well, and recur if (ones > zeroes) { generate(ones, zeroes + 1, str + "0", len); } } // Driver program to test above function public static void main(String[] args) { String str = ""; generate(0, 0, str, 4); } } /* This Java code is contributed by PrinciRaj1992*/ |
Python3
# Python 3 program to generate all permutations of 1's and 0's such that # every permutation has more 1's than 0's at all indexes. # ones & zeroes --> counts of 1's and 0's in current string 'str' # len ---> desired length of every permutation def generate(ones, zeroes, str, len1): # If length of current string becomes same as desired length if (len1 == len(str)): print(str,end =" ") return # Append a 1 and recur generate(ones+1, zeroes, str+"1", len1) # If there are more 1's, append a 0 as well, and recur if (ones > zeroes): generate(ones, zeroes+1, str+"0", len1) # Driver program to test above function if __name__ == '__main__': str = "" generate(0, 0, str, 4) # This code is contributed by # Surendra_Gangwar |
C#
// C# program to generate all permutations of 1's and 0's such that // every permutation has more 1's than 0's at all indexes. using System; public class GFG { // ones & zeroes --> counts of 1's and 0's in current string 'str' // len ---> desired length of every permutation static void generate(int ones, int zeroes, String str, int len) { // If length of current string becomes same as desired length if (len == str.Length) { Console.Write(str + " "); return; } // Append a 1 and recur generate(ones + 1, zeroes, str + "1", len); // If there are more 1's, append a 0 as well, and recur if (ones > zeroes) { generate(ones, zeroes + 1, str + "0", len); } } // Driver program to test above function public static void Main() { String str = ""; generate(0, 0, str, 4); } } /* This Java code is contributed by 29AjayKumar*/ |
Output:
1111 1110 1101 1100 1011 1010
This article is contributed by Sachin. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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