All possible Strings of 1 and 0

Question

if i wanted to create all possible strings of n 1's and n 0's where there are never more 0's than 1's using 2 for loops and recursion how would i go about this? (the first for loop controlls how many 1's are in the string the second how many 0's up to the number of 1's) What would the recursive alogrithim have to look like inorder for me to create every possible string of n 1's and n 0's where there are never more 0's than 1's ?

Answer 1

void spew(int c0, int c1, int bal, String soFar){
  if(c0 == 0 && c1 == 0)System.out.println soFar;
  if(c0 > 0 && bal >= 0) spew(c0 - 1, c1, bal - 1, soFar+"0");
  if(c1 > 0) spew(c0, c1 - 1, bal + 1, soFar+"1");
void spew(int n){spew(n,n,0,"");}And for those of you that are interested, the ones and zeros were not really ones and zeros, they were open and close parenthesis. The reason for not allowing the zeros to ever outnumber the ones was so avoid a mis-balanced set of parenthesis in the initial portion of the string, and the reason to have the same number of ones and zeros was so that final string would have a balanced set of parenthesis.
Once you cast the problem in it's proper form, you can see that the real question was how to emit Strings with exactly n balanced pairs of parens. The total number of strings of this form are known as the Catalan numbers and you can look them up in Sloan's list of integer sequences. And of course the Catalan numbers are also the way that you count the distinct number of binary trees that you can create with exactly n nodes.
Of course, I don't know any of that, I am just guessing the intent from the otherwise peculiar clarification of the problem and from the other posts by the OP asking about binary trees. It is a pity that the OP felt a need to obscure the original problem, making it both harder to solve and less likely that anyone googling for similar help would ever find it here.