Generate all binary strings from given pattern
Last Updated :
20 Oct, 2022
Given a string containing of '0', '1' and '?' wildcard characters, generate all binary strings that can be formed by replacing each wildcard character by '0' or '1'.
Example :
Input: str = "1??0?101"
Output:
10000101
10001101
10100101
10101101
11000101
11001101
11100101
11101101
Method 1 (Using Recursion)
We pass index of next character to the recursive function. If the current character is a wildcard character '?', we replace it with '0' or '1' and do the same for all remaining characters. We print the string if we reach at its end.
Algorithm:
Step 1: Initialize the string first with some wildcard characters in it.
Step 2: Check if index position is equals to the size of string, If it is return.
Step 3: If wildcard character is present at index location, replace it by 0 or 1 accordingly.
Step 4: Print the output
Below is the implementation of the above approach:
C++
// Recursive C++ program to generate all binary strings
// formed by replacing each wildcard character by 0 or 1
#include <iostream>
using namespace std;
// Recursive function to generate all binary strings
// formed by replacing each wildcard character by 0 or 1
void print(string str, int index)
{
if (index == str.size())
{
cout << str << endl;
return;
}
if (str[index] == '?')
{
// replace '?' by '0' and recurse
str[index] = '0';
print(str, index + 1);
// replace '?' by '1' and recurse
str[index] = '1';
print(str, index + 1);
// No need to backtrack as string is passed
// by value to the function
}
else
print(str, index + 1);
}
// Driver code to test above function
int main()
{
string str = "1??0?101";
print(str, 0);
return 0;
}
// Recursive Java program to generate all
// binary strings formed by replacing
// each wildcard character by 0 or 1
import java.util.*;
import java.lang.*;
import java.io.*;
class binStr
{
// Recursive function to generate all binary
// strings formed by replacing each wildcard
// character by 0 or 1
public static void print(char str[], int index)
{
if (index == str.length)
{
System.out.println(str);
return;
}
if (str[index] == '?')
{
// replace '?' by '0' and recurse
str[index] = '0';
print(str, index + 1);
// replace '?' by '1' and recurse
str[index] = '1';
print(str, index + 1);
// NOTE: Need to backtrack as string
// is passed by reference to the
// function
str[index] = '?';
}
else
print(str, index + 1);
}
// driver code
public static void main (String[] args)
{
String input = "1??0?101";
char[] str = input.toCharArray();
print(str, 0);
}
}
// This code is contributed by Chhavi
# Recursive Python program to generate all
# binary strings formed by replacing
# each wildcard character by 0 or 1
# Recursive function to generate all binary
# strings formed by replacing each wildcard
# character by 0 or 1
def _print(string, index):
if index == len(string):
print(''.join(string))
return
if string[index] == "?":
# replace '?' by '0' and recurse
string[index] = '0'
_print(string, index + 1)
# replace '?' by '1' and recurse
string[index] = '1'
_print(string, index + 1)
# NOTE: Need to backtrack as string
# is passed by reference to the
# function
string[index] = '?'
else:
_print(string, index + 1)
# Driver code
if __name__ == "__main__":
string = "1??0?101"
string = list(string)
_print(string, 0)
# This code is contributed by
# sanjeev2552
# Note: function name _print is used because
# print is already a predefined function in Python
// Recursive C# program to generate all
// binary strings formed by replacing
// each wildcard character by 0 or 1
using System;
class GFG
{
// Recursive function to generate
// all binary strings formed by
// replacing each wildcard character
// by 0 or 1
public static void print(char []str,
int index)
{
if (index == str.Length)
{
Console.WriteLine(str);
return;
}
if (str[index] == '?')
{
// replace '?' by
// '0' and recurse
str[index] = '0';
print(str, index + 1);
// replace '?' by
// '1' and recurse
str[index] = '1';
print(str, index + 1);
// NOTE: Need to backtrack
// as string is passed by
// reference to the function
str[index] = '?';
}
else
print(str, index + 1);
}
// Driver Code
public static void Main ()
{
string input = "1??0?101";
char []str = input.ToCharArray();
print(str, 0);
}
}
// This code is contributed by nitin mittal.
<?php
// Recursive PHP program to generate all binary strings
// formed by replacing each wildcard character by 0 or 1
// Recursive function to generate all binary strings
// formed by replacing each wildcard character by 0 or 1
function print1($str, $index)
{
if ($index == strlen($str))
{
echo $str."\n";
return;
}
if ($str[$index] == '?')
{
// replace '?' by '0' and recurse
$str[$index] = '0';
print1($str, $index + 1);
// replace '?' by '1' and recurse
$str[$index] = '1';
print1($str, $index + 1);
// No need to backtrack as string is passed
// by value to the function
}
else
print1($str, $index + 1);
}
// Driver code
$str = "1??0?101";
print1($str, 0);
// This code is contributed by chandan_jnu
?>
<script>
// Recursive JavaScript program to generate all
// binary strings formed by replacing
// each wildcard character by 0 or 1
// Recursive function to generate
// all binary strings formed by
// replacing each wildcard character
// by 0 or 1
function print(str, index) {
if (index === str.length) {
document.write(str.join("") + "<br>");
return;
}
if (str[index] === "?") {
// replace '?' by
// '0' and recurse
str[index] = "0";
print(str, index + 1);
// replace '?' by
// '1' and recurse
str[index] = "1";
print(str, index + 1);
// NOTE: Need to backtrack
// as string is passed by
// reference to the function
str[index] = "?";
} else print(str, index + 1);
}
// Driver Code
var input = "1??0?101";
var str = input.split("");
print(str, 0);
</script>
Output: 10000101
10001101
10100101
10101101
11000101
11001101
11100101
11101101
Time Complexity: O(2N), where N is the length of the given string and there are 2 possibilities.
Auxiliary Space: O(N2), as a copy of the string is created in every recursive call.
Method 2 (Using Queue)
We can also achieve this by using iteration. The idea is to use queue, We find position of first occurrence of wildcard character in the input string and replace it by '0' , then '1' and push both strings into the queue. Then we pop next string from the queue, and repeat the process till queue is empty. If no wildcard characters are left, we simply print the string.
Iterative C++ implementation using queue.
C++
// Iterative C++ program to generate all binary
// strings formed by replacing each wildcard
// character by 0 or 1
#include <iostream>
#include <queue>
using namespace std;
// Iterative function to generate all binary strings
// formed by replacing each wildcard character by 0
// or 1
void print(string str)
{
queue<string> q;
q.push(str);
while (!q.empty())
{
string str = q.front();
// find position of first occurrence of wildcard
size_t index = str.find('?');
// If no matches were found,
// find returns string::npos
if(index != string::npos)
{
// replace '?' by '0' and push string into queue
str[index] = '0';
q.push(str);
// replace '?' by '1' and push string into queue
str[index] = '1';
q.push(str);
}
else
// If no wildcard characters are left,
// print the string.
cout << str << endl;
q.pop();
}
}
// Driver code to test above function
int main()
{
string str = "1??0?101";
print(str);
return 0;
}
/*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
class GFG {
// Iterative Java program to generate all binary
// strings formed by replacing each wildcard
// character by 0 or 1
// Iterative function to generate all binary strings
// formed by replacing each wildcard character by 0
// or 1
static void print(String str)
{
Queue<String> q = new LinkedList<>();
q.add(str);
while (!q.isEmpty())
{
str = q.remove();
// find position of first occurrence of wildcard
int index = str.indexOf('?');
// If no matches were found,
// find returns string::npos
if(index != -1)
{
// replace '?' by '0' and add string into queue
str = str.substring(0,index) + '0' + str.substring(index+1);
q.add(str);
// replace '?' by '1' and add string into queue
str = str.substring(0,index) + '1' + str.substring(index+1);
q.add(str);
}
else
// If no wildcard characters are left,
// print the string.
System.out.println(str);
}
}
// Driver Code
public static void main(String args[])
{
String str = "1??0?101";
print(str);
}
}
// This code is contributed by shinjanpatra
# Iterative Python program to generate all binary
# strings formed by replacing each wildcard
# character by 0 or 1
# Iterative function to generate all binary strings
# formed by replacing each wildcard character by 0
# or 1
def Print(Str):
q = []
q.append(Str)
while(len(q) > 0):
Str = q[0]
# find position of first occurrence of wildcard
try:
index = Str.index('?')
except ValueError:
index = -1
# If no matches were found,
# find returns -1
if(index != -1):
# replace '?' by '0' and push string into queue
s1=Str.replace('?','0',1)
q.append(s1)
# replace '?' by '1' and push string into queue
s2=Str.replace('?','1',1)
q.append(s2)
else:
# If no wildcard characters are left,
# print the string.
print(Str)
q.pop(0)
# Driver code
Str = "1??0?101"
Print(Str)
# This code is contributed by Pushpesh Raj
// C# program to implement the approach
using System;
using System.Collections.Generic;
public class GFG
{
// Iterative C# program to generate all binary
// strings formed by replacing each wildcard
// character by 0 or 1
// Iterative function to generate all binary strings
// formed by replacing each wildcard character by 0
// or 1
static void print(string Str)
{
var q = new List<string>();
q.Add(Str);
while (q.Count > 0) {
Str = q[0];
// find position of first occurrence of wildcard
int index = Str.IndexOf('?');
// If no matches were found,
// find returns -1
if (index != -1) {
// replace '?' by '0' and push string into
// queue
Str = Str.Substring(0, index) + '0'
+ Str.Substring(index + 1);
q.Add(Str);
// replace '?' by '1' and push string into
// queue
Str = Str.Substring(0, index) + '1'
+ Str.Substring(index + 1);
q.Add(Str);
}
else {
// If no wildcard characters are left,
// print the string.
Console.WriteLine(Str);
}
q.RemoveAt(0);
}
}
// Driver Code
public static void Main(string[] args)
{
string str = "1??0?101";
// Function call
print(str);
}
}
// This code is contributed by phasing17
<script>
// Iterative JavaScript program to generate all binary
// strings formed by replacing each wildcard
// character by 0 or 1
// Iterative function to generate all binary strings
// formed by replacing each wildcard character by 0
// or 1
function Print(Str){
let q = []
q.push(Str)
while (q.length > 0){
let Str = q[0]
// find position of first occurrence of wildcard
let index = Str.indexOf('?')
// If no matches were found,
// find returns -1
if(index != -1)
{
// replace '?' by '0' and push string into queue
Str = Str.replace(Str[index] , '0')
q.push(Str)
// replace '?' by '1' and push string into queue
Str = Str.replace(Str[index] , '1')
q.push(Str)
}
else
{
// If no wildcard characters are left,
// print the string.
document.write(Str,"</br>")
}
q.shift()
}
}
// Driver code to test above function
let Str = "1??0?101"
Print(Str)
// This code is contributed by shinjanpatra
</script>
Output: 10000101
10001101
10100101
10101101
11000101
11001101
11100101
11101101
Time Complexity: O(N*2N), where N is the size of the string.
Auxiliary Space: O(2N)
Method 3 (Using str and Recursion)
C++
// C++ program to implement the approach
#include <bits/stdc++.h>
using namespace std;
/* we store processed strings in all (array)
we see if string as "?", if so, replace it with 0 and 1
and send it back to recursive func until base case is
reached which is no wildcard left */
vector<string> res;
void genBin(string s)
{
auto pos = s.find('?');
if (pos != string::npos) {
// copying s to s1
string s1 = s;
// replacing first occurrence of ?
// with 0
s1.replace(pos, 1, "0");
// copying s to s2
string s2 = s;
// replacing first occurrence of ?
// with 1
s2.replace(pos, 1, "1");
genBin(s1);
genBin(s2);
}
else {
res.push_back(s);
}
}
// Driver code
int main()
{
genBin("1??0?101");
for (string x : res) {
cout << x << " ";
}
}
// This code is contributed by phasing17
import java.io.*;
import java.util.*;
class GFG
{
// we store processed strings in all (array)
// we see if string as "?", if so, replace it with 0 and 1
// and send it back to recursive func until base case is reached
// which is no wildcard left
static ArrayList<String>res = new ArrayList<String>();
static void genBin(String s) {
if (s.indexOf('?') != -1) {
String s1 = s.replaceFirst("\\?", "0"); // only replace once
String s2 = s.replaceFirst("\\?", "1"); // only replace once
genBin(s1);
genBin(s2);
}
else res.add(s);
}
public static void main(String[] args) {
genBin("1??0?101");
System.out.println(res);
}
}
#we store processed strings in all (array)
#we see if string as "?", if so, replace it with 0 and 1
#and send it back to recursive func until base case is reached
#which is no wildcard left
res = []
def genBin(s):
if '?' in s:
s1 = s.replace('?','0',1) #only replace once
s2 = s.replace('?','1',1) #only replace once
genBin(s1)
genBin(s2)
else: res.append(s)
# Driver code
genBin("1??0?101")
print(res)
# This code is contributed by
# divay pandey
// C# code to implement the approach
using System;
using System.Collections.Generic;
class GFG {
// we store processed strings in all (array)
// we see if string as "?", if so, replace it with 0 and
// 1 and send it back to recursive func until base case
// is reached which is no wildcard left
static List<string> res = new List<string>();
static void genBin(string s)
{
int ind = s.IndexOf("?");
if (ind != -1) {
string s1 = s.Remove(ind, 1).Insert(
ind, "0"); // only replace once
string s2 = s.Remove(ind, 1).Insert(
ind, "1"); // only replace once
genBin(s1);
genBin(s2);
}
else
res.Add(s);
}
// Driver code
public static void Main(string[] args)
{
genBin("1??0?101");
foreach(var ele in res) Console.Write(ele + " ");
}
}
// This code is contributed by phasing17
<script>
/* we store processed strings in all (array)
we see if string as "?", if so, replace it with 0 and 1
and send it back to recursive func until base case is reached
which is no wildcard left */
let res = []
function genBin(s){
if(s.includes('?')){
let s1 = s.replace(/\?/,'0') //only replace once
let s2 = s.replace(/\?/,'1') //only replace once
genBin(s1)
genBin(s2)
}else{
res.push(s)
}
}
// Driver code
genBin("1??0?101")
document.write(res)
// This code is contributed by
// Santanu Panda
</script>
Output: ['10000101', '10001101', '10100101', '10101101', '11000101', '11001101', '11100101', '11101101']
Time Complexity: O(N*2N), where N is the size of the string.
Auxiliary Space: O(2N)
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Print all subsequences of a stringGiven a string, we have to find out all its subsequences of it. A String is said to be a subsequence of another String, if it can be obtained by deleting 0 or more character without changing its order.Examples: Input : abOutput : "", "a", "b", "ab"Input : abcOutput : "", "a", "b", "c", "ab", "ac", "
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Count Distinct SubsequencesGiven a string str of length n, your task is to find the count of distinct subsequences of it.Examples: Input: str = "gfg"Output: 7Explanation: The seven distinct subsequences are "", "g", "f", "gf", "fg", "gg" and "gfg" Input: str = "ggg"Output: 4Explanation: The four distinct subsequences are "",
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Count distinct occurrences as a subsequenceGiven two strings pat and txt, where pat is always shorter than txt, count the distinct occurrences of pat as a subsequence in txt.Examples: Input: txt = abba, pat = abaOutput: 2Explanation: pat appears in txt as below two subsequences.[abba], [abba]Input: txt = banana, pat = banOutput: 3Explanation
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Longest Common Subsequence (LCS)Given two strings, s1 and s2, the task is to find the length of the Longest Common Subsequence. If there is no common subsequence, return 0. A subsequence is a string generated from the original string by deleting 0 or more characters, without changing the relative order of the remaining characters.
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Shortest Superstring ProblemGiven a set of n strings arr[], find the smallest string that contains each string in the given set as substring. We may assume that no string in arr[] is substring of another string.Examples: Input: arr[] = {"geeks", "quiz", "for"}Output: geeksquizforExplanation: "geeksquizfor" contains all the thr
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Printing Shortest Common SupersequenceGiven two strings s1 and s2, find the shortest string which has both s1 and s2 as its sub-sequences. If multiple shortest super-sequence exists, print any one of them.Examples:Input: s1 = "geek", s2 = "eke"Output: geekeExplanation: String "geeke" has both string "geek" and "eke" as subsequences.Inpu
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Shortest Common SupersequenceGiven two strings s1 and s2, the task is to find the length of the shortest string that has both s1 and s2 as subsequences.Examples: Input: s1 = "geek", s2 = "eke"Output: 5Explanation: String "geeke" has both string "geek" and "eke" as subsequences.Input: s1 = "AGGTAB", s2 = "GXTXAYB"Output: 9Explan
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Longest Repeating SubsequenceGiven a string s, the task is to find the length of the longest repeating subsequence, such that the two subsequences don't have the same string character at the same position, i.e. any ith character in the two subsequences shouldn't have the same index in the original string. Examples:Input: s= "ab
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Longest Palindromic Subsequence (LPS)Given a string s, find the length of the Longest Palindromic Subsequence in it. Note: The Longest Palindromic Subsequence (LPS) is the maximum-length subsequence of a given string that is also a Palindrome. Longest Palindromic SubsequenceExamples:Input: s = "bbabcbcab"Output: 7Explanation: Subsequen
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Longest Palindromic SubstringGiven a string s, the task is to find the longest substring which is a palindrome. If there are multiple answers, then return the first appearing substring.Examples:Input: s = "forgeeksskeegfor" Output: "geeksskeeg"Explanation: There are several possible palindromic substrings like "kssk", "ss", "ee
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Palindrome
C Program to Check for Palindrome StringA string is said to be palindrome if the reverse of the string is the same as the string. In this article, we will learn how to check whether the given string is palindrome or not using C program.The simplest method to check for palindrome string is to reverse the given string and store it in a temp
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Check if a given string is a rotation of a palindromeGiven a string, check if it is a rotation of a palindrome. For example your function should return true for "aab" as it is a rotation of "aba". Examples: Input: str = "aaaad" Output: 1 // "aaaad" is a rotation of a palindrome "aadaa" Input: str = "abcd" Output: 0 // "abcd" is not a rotation of any p
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Check if characters of a given string can be rearranged to form a palindromeGiven a string, Check if the characters of the given string can be rearranged to form a palindrome. For example characters of "geeksogeeks" can be rearranged to form a palindrome "geeksoskeeg", but characters of "geeksforgeeks" cannot be rearranged to form a palindrome. Recommended PracticeAnagram P
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Online algorithm for checking palindrome in a streamGiven a stream of characters (characters are received one by one), write a function that prints 'Yes' if a character makes the complete string palindrome, else prints 'No'. Examples:Input: str[] = "abcba"Output: a Yes // "a" is palindrome b No // "ab" is not palindrome c No // "abc" is not palindrom
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Print all Palindromic Partitions of a String using Bit ManipulationGiven a string, find all possible palindromic partitions of a given string. Note that this problem is different from Palindrome Partitioning Problem, there the task was to find the partitioning with minimum cuts in input string. Here we need to print all possible partitions. Example: Input: nitinOut
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Minimum Characters to Add at Front for PalindromeGiven a string s, the task is to find the minimum number of characters to be added to the front of s to make it palindrome. A palindrome string is a sequence of characters that reads the same forward and backward. Examples: Input: s = "abc"Output: 2Explanation: We can make above string palindrome as
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Make largest palindrome by changing at most K-digitsYou are given a string s consisting of digits (0-9) and an integer k. Convert the string into a palindrome by changing at most k digits. If multiple palindromes are possible, return the lexicographically largest one. If it's impossible to form a palindrome with k changes, return "Not Possible".Examp
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Minimum Deletions to Make a String PalindromeGiven a string s of length n, the task is to remove or delete the minimum number of characters from the string so that the resultant string is a palindrome. Note: The order of characters should be maintained. Examples : Input : s = "aebcbda"Output : 2Explanation: Remove characters 'e' and 'd'. Resul
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Minimum insertions to form a palindrome with permutations allowedGiven a string of lowercase letters. Find minimum characters to be inserted in the string so that it can become palindrome. We can change the positions of characters in the string.Examples: Input: geeksforgeeksOutput: 2Explanation: geeksforgeeks can be changed as: geeksroforskeeg or geeksorfroskeeg
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