Find Binary permutations of given size not present in the Array
Last Updated :
20 Sep, 2021
Given a positive integer N and an array arr[] of size K consisting of binary string where each string is of size N, the task is to find all the binary strings of size N that are not present in the array arr[].
Examples:
Input: N = 3, arr[] = {"101", "111", "001", "010", "011", "100", "110"}
Output: 000
Explanation: Only 000 is absent in the given array.
Input: N = 2, arr[] = {"00", "01"}
Output: 10 11
Approach: The given problem can be solved by finding all binary strings of size N using Backtracking and then print those strings that are not present in the array arr[]. Follow the steps below to solve the problem:
- Initialize an unordered_set of strings S that stores all the strings in the array arr[].
- Initialize a variable, say total and set it as the power of 2N where N is the size of the string.
- Iterate over the range [0, total) using the variable i and perform the following steps:
- Initialize an empty string variable num as "".
- Iterate over the range [N - 1, 0] using the variable j and if the value of the Bitwise AND of i and 2j is true, then push the character 1 into the string num. Otherwise, push the character 0.
- If num is not present in the set S then print the string num as the one of the resultant string.
- After completing the above steps, if all the possible binary strings are present in the array then print "-1".
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find a Binary String of
// same length other than the Strings
// present in the array
void findMissingBinaryString(vector<string>& nums, int N)
{
unordered_set<string> s;
int counter = 0;
// Map all the strings present in
// the array
for (string str : nums) {
s.insert(str);
}
int total = (int)pow(2, N);
string ans = "";
// Find all the substring that
// can be made
for (int i = 0; i < total; i++) {
string num = "";
for (int j = N - 1; j >= 0; j--) {
if (i & (1 << j)) {
num += '1';
}
else {
num += '0';
}
}
// If num already exists then
// increase counter
if (s.find(num) != s.end()) {
continue;
counter++;
}
// If not found print
else {
cout << num << ", ";
}
}
// If all the substrings are present
// then print -1
if (counter == total) {
cout << "-1";
}
}
// Driver Code
int main()
{
int N = 3;
vector<string> arr
= { "101", "111", "001", "011", "100", "110" };
findMissingBinaryString(arr, N);
return 0;
}
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find a Binary String of
// same length other than the Strings
// present in the array
static void findMissingBinaryString(String[] nums, int N)
{
HashSet<String> s = new HashSet<String>();
int counter = 0;
// Map all the Strings present in
// the array
for (String str : nums) {
s.add(str);
}
int total = (int)Math.pow(2, N);
String ans = "";
// Find all the subString that
// can be made
for (int i = 0; i < total; i++) {
String num = "";
for (int j = N - 1; j >= 0; j--) {
if ((i & (1 << j))>0) {
num += '1';
}
else {
num += '0';
}
}
// If num already exists then
// increase counter
if (s.contains(num)) {
counter++;
continue;
}
// If not found print
else {
System.out.print(num+ ", ");
}
}
// If all the subStrings are present
// then print -1
if (counter == total) {
System.out.print("-1");
}
}
// Driver Code
public static void main(String[] args)
{
int N = 3;
String []arr
= { "101", "111", "001", "011", "100", "110" };
findMissingBinaryString(arr, N);
}
}
// This code is contributed by Princi Singh
# Python 3 program for the above approach
from math import pow
# Function to find a Binary String of
# same length other than the Strings
# present in the array
def findMissingBinaryString(nums, N):
s = set()
counter = 0
# Map all the strings present in
# the array
for x in nums:
s.add(x)
total = int(pow(2, N))
ans = ""
# Find all the substring that
# can be made
for i in range(total):
num = ""
j = N - 1
while(j >= 0):
if (i & (1 << j)):
num += '1'
else:
num += '0'
j -= 1
# If num already exists then
# increase counter
if (num in s):
continue
counter += 1
# If not found print
else:
print(num,end = ", ")
# If all the substrings are present
# then print -1
if (counter == total):
print("-1")
# Driver Code
if __name__ == '__main__':
N = 3
arr = ["101", "111", "001", "011", "100", "110"]
findMissingBinaryString(arr, N)
# This code is contributed by SURENDRA_GANGWAR.
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to find a Binary String of
// same length other than the Strings
// present in the array
static void findMissingBinaryString(string[] nums,
int N)
{
HashSet<string> s = new HashSet<string>();
int counter = 0;
// Map all the Strings present in
// the array
foreach(string str in nums) { s.Add(str); }
int total = (int)Math.Pow(2, N);
// string ans = "";
// Find all the subString that
// can be made
for (int i = 0; i < total; i++) {
string num = "";
for (int j = N - 1; j >= 0; j--) {
if ((i & (1 << j)) > 0) {
num += '1';
}
else {
num += '0';
}
}
// If num already exists then
// increase counter
if (s.Contains(num)) {
counter++;
continue;
}
// If not found print
else {
Console.Write(num + ", ");
}
}
// If all the subStrings are present
// then print -1
if (counter == total) {
Console.Write("-1");
}
}
// Driver Code
public static void Main(string[] args)
{
int N = 3;
string[] arr
= { "101", "111", "001", "011", "100", "110" };
findMissingBinaryString(arr, N);
}
}
// This code is contributed by ukasp.
<script>
// Javascript program for the above approach
// Function to find a Binary String of
// same length other than the Strings
// present in the array
function findMissingBinaryString(nums, N) {
let s = new Set();
let counter = 0;
// Map all the strings present in
// the array
for (let str of nums) {
s.add(str);
}
let total = Math.pow(2, N);
let ans = "";
// Find all the substring that
// can be made
for (let i = 0; i < total; i++) {
let num = "";
for (let j = N - 1; j >= 0; j--) {
if (i & (1 << j)) {
num += "1";
} else {
num += "0";
}
}
// If num already exists then
// increase counter
if (s.has(num)) {
continue;
counter++;
}
// If not found print
else {
document.write(num + ", ");
}
}
// If all the substrings are present
// then print -1
if (counter == total) {
document.write("-1");
}
}
// Driver Code
let N = 3;
let arr = ["101", "111", "001", "011", "100", "110"];
findMissingBinaryString(arr, N);
// This code is contributed by _saurabh_jaiswal.
</script>
Time Complexity: O(N*2N)
Auxiliary Space: O(K), where K is the size of the array
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