Minimum number whose binary form is not a subsequence of given binary string
Given a binary string S of size N, the task is to find the minimum non-negative integer which is not a subsequence of the given string S in its binary form.
Examples:
Input: S = “0000”
Output:1
Explanation: 1 whose binary representation is “1” is the smallest non-negative integer which is not a subsequence of the given string in its binary form.Input: S = “10101”
Output: 8
Approach: The idea is to convert the given string into its decimal representation, say R. Then iterate in the range [0, R] to check for each integer whether it exists or not as a subsequence in its binary form in the given string, S. If not, then break the loop and print the required result.
Follow the steps below to solve the problem:
- Store the decimal number of the binary string, S in a variable R.
- Initialize a variable ans as R+1 to store the required result.
- Iterate in the range [0, R] using the variable i
- Convert the given integer i into its binary string, and store it in string P.
- Check if string P is a subsequence of string S or not. If it is not, then update ans to i and break out of the loop.
- Print the value of ans as the result.
Below is the implementation of the above approach:
- C++
- Java
- Python3
- C#
- Javascript
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if string str1 is a// subsequence of string str2bool isSubsequence(string str1, string str2, int m, int n){ // Store index of str1 int j = 0; // Traverse str2 and str1, and compare // current character of str2 with first // unmatched char of str1 for (int i = 0; i < n && j < m; i++) // If matched, move ahead in str1 if (str1[j] == str2[i]) j++; // If all characters of str1 were // found in str2 return (j == m);}// Function to find the minimum number which // is not a subsequence of the given binary // string in its binary formvoid findMinimumNumber(string s){ // Store the decimal number of string, S int r = stoi(s, 0, 2); // Initialize the required result int ans = r + 1; // Iterate in the range [0, R] for (int i = 0; i <= r; i++) { // Convert integer i to binary string string p = ""; int j = i; while (j > 0) { p += to_string(j % 2); j = j / 2; } int m = p.length(); int n = s.length(); reverse(p.begin(), p.end()); // Check if the string is not a subsequence if (!isSubsequence(p, s, m, n)) { // Update ans and break ans = i; break; } } // Print the required result cout << ans;}// Driver Codeint main(){ // Function Call string s = "10101"; // Function Call findMinimumNumber(s); return 0;} |
Java
Python3
C#
Javascript
8
Time Complexity: O(N*R), where R is the decimal representation of the given binary string, S
Auxiliary Space: O(N)
