Unable to solve

#1

please tell base case and recursive case condition

#2

Logic:
first try to place 0 or 1 at starting and call recursion to slove

now in recursion

// base case:
if (n == K) 
	{ 
		// terminate binary string 
		str[n] = '\0' ; 
		cout << str << "\n"; 
		return ; 
	}

in recursive case you have to place 0 or 1 and call the recursion again

if previous character is '1' then we put 
only 0 at end of string 
example str = "01" then new string be "010" 

if previous character is '0' than we put 
both '1' and '0' at end of string 
example str = "00" then new string "001" and "000"

complete code

i hope this helps
if yes hit a like and don’t forgot to mark doubt as resolved :grinning:
if you have more doubts regarding this feel free to ask

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#3

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