All possible combinations of a 3 digit binary number are 2c1 * 2c1 * 2c1 = 8 (000, 001, 010, 011, 100, 101, 110, 111). Each digit can be 0 or 1 so 2c1. Below is the code for it.
Starting with 000, we get the next combination by incrementing the LSbs and carrying out the ripple effect with the for loop.
`
void printCombinations(int digits):
char[] out = new char[digits]
for (int i = 0; i < out.length; i++):
out[i] = ‘0’
while (true):
print(out)
int i
for(i = out.length -1; i >= 0; i–):
if (out[i] == ‘0’):
out[i] = ‘1’
else if (out[i] == ‘1’):
out[i] = ‘0’
continue
break
if (i == -1):
break
`
If each digit can have more values (for decimal 0 to 9, or characters A B C … Z), the if-else would change (better with a mapping function). There is a (intuitive, simple) recursive solution too but recursion is a overhead for large combinations.
`
printCombinations_Recursive (char[] out, int index):
if (index == out.length):
print(out)
return
for (int bit = 0; bit <= 1; bit++):
out[index] = bit //(char) (bit +'0’) or a mapping function
printCombinations_Recursive(out, index + 1)
printRecursiveWrapper (int digits):
char[] out = new char[digits]
printCombinations_Recursive(out, 0)`