CareerCup

void generate(int n) {
		if (n < 1) {
			System.out.println(Arrays.toString(A));
			return;
		}
		A[n - 1] = 0;
		generate(n - 1);
		A[n - 1] = 1;
		generate(n - 1);
	}

this is kind of wrong..coz in the above code it goes wrong when n numbers are all zeroes..

we have to permute the n numbers in such a way that there are no repetitions..

basically in this question we have suppose a ->zeroes and b->ones then we need to permute then such that,possible arrangements are n!/(a!*b!)..
so can anyone help me how to code that thing..as i want to print non-repeated permutations..i m unable to find its solution..

increment a n bit unsigned number until the value overflows back to 0
print the values alone the way

The following code may avoid repetitions:

public static void printAll(int[] data)
    {
        if(data == null) return;
        if(data.length == 0) return;
        
        int numOne = 0;
        for(int i = 0; i < data.length; ++i)
        {
            numOne += data[i];
        }
        int numZero = data.length - numOne;
        String str = "";
        printHelper(numZero, numOne, 0, str, 0, 0);
    }
    
    private static void printHelper(int numZero, int numOne, int level, String str, int numZeroUsed, int numOneUsed)
    {
        if(level == numZero + numOne)
        {
            System.out.println(str);
            return;
        }
        if(numZeroUsed < numZero)
        {
            String strNew = str;
            strNew += "0";
            printHelper(numZero, numOne, level+1, strNew , numZeroUsed+1, numOneUsed);
        }
        if(numOneUsed < numOne)
        {
            String strNew = str;
            strNew += "1";
            printHelper(numZero, numOne, level+1, strNew, numZeroUsed, numOneUsed+1);
        }
    }

1. count the number of ones
2. put a 1 at the lowest index, recursively put the rest ones in the rest n - 1 positions
3. repeat step 2 for 2, 3, ...

here comes the codes in Python:

import itertools
m, n = 3, 2
for i in itertools.product(*((range(n), )*m)):
    print i

#include <stdio.h>
#define N 3
main(x){
	int i;  
	for (i = 0; i<N;printf("%d",x >> i &1), i++); 
	printf("\n");
	(1 << N)-x && main(x+1); 
	return 0; 
}

or use this function:

void foo(unsigned int n)
{
	int i, j; 
	for (i = 0; i < 1 << n; printf("\n"), i++)
		for (j = 0; j < n; printf("%d", i >> j&1), j++); 
}