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.


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 

We strongly recommend to minimize the browser and try this yourself first.

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.

// 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 << "  ";

    // 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;


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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