Find Mth lexicographically smallest Binary String with no two adjacent 1
Given two integers N and M, the task is to find the Mth lexicographically smallest binary string (have only characters 1 and 0) of length N where there cannot be two consecutive 1s.
Examples:
Input: N = 2, M = 3.
Output: 10
Explanation: The only strings that can be made of size 2 are [“00”, “01”, “10”] and the 3rd string is “10”.Input: N = 3, M = 2.
Output: 001
Approach: The problem can be solved based on the following approach:
Form all the N sized strings and find the Mth smallest among them.
Follow the steps mentioned below to implement the idea.
- For each character, there are two choices:
- Make the character 0.
- If the last character of the string formed till now is not 1, then the current character can also be 1.
- To implement this use recursion.
- As the target is to find the Mth smallest, for any character call the recursive function for 0 first and for 1 after that (if 1 can be used).
- Each time a string of length N is formed increase the count of strings
- If count = M, that string is the required lexicographically Mth smallest string.
Below is the implementation of above approach:
C++
// C++ code to implement the approach #include <bits/stdc++.h> using namespace std; // Declared 2 global variable // one is the answer string and // the other is the count of created string string ans = "" ; int Count = 0; // Function to find the mth string. void findString( int idx, int n, int m, string curr) { // When size of string is equal to n if (idx == n) { // If count of strings created // is equal to m-1 if (Count == m - 1) { ans = curr; } else { Count += 1; } return ; } // Call the function to recurse for // currentstring + "0" curr += "0" ; findString(idx + 1, n, m, curr); curr.pop_back(); // If the last character of curr is not 1 // then similarly recurse for "1". if (curr[curr.length() - 1] != '1' ) { curr += "1" ; findString(idx + 1, n, m, curr); curr.pop_back(); } } // Driver Code int main() { int N = 2, M = 3; // Function call findString(0, N, M, "" ); cout << ans << endl; return 0; } |
Java
// Java code to implement the approach import java.io.*; class GFG { // Declared 2 global variable // one is the answer string and // the other is the count of created string static String ans = "" ; public static int Count = 0 ; // Function to find the mth string. public static void findString( int idx, int n, int m, String curr) { // When size of string is equal to n if (idx == n) { // If count of strings created // is equal to m-1 if (Count == m - 1 ) { ans = curr; } else { Count += 1 ; } return ; } // Call the function to recurse for // currentstring + "0" curr += "0" ; findString(idx + 1 , n, m, curr); curr=curr.substring( 0 ,curr.length()- 1 ); // If the last character of curr is not 1 // then similarly recurse for "1". if (curr.length()== 0 || curr.charAt(curr.length() - 1 ) != '1' ) { curr += "1" ; findString(idx + 1 , n, m, curr); curr=curr.substring( 0 ,curr.length()- 1 ); } } // Driver Code public static void main(String[] args) { int N = 2 , M = 3 ; // Function call findString( 0 , N, M, "" ); System.out.println(ans); } } // This code is contributed by jana_sayantan. |
Python3
# Python code to implement the approach # Declared 2 global variable # one is the answer string and # the other is the count of created string ans = "" Count = 0 # Function to find the mth string. def findString(idx, n, m, curr): global ans,Count # When size of string is equal to n if (idx = = n): # If count of strings created # is equal to m-1 if (Count = = m - 1 ): ans = curr else : Count + = 1 return # Call the function to recurse for # currentstring + "0" curr + = "0" findString(idx + 1 , n, m, curr) curr = curr[ 0 : len (curr) - 1 ] # If the last character of curr is not 1 # then similarly recurse for "1". if ( len (curr) = = 0 or curr[ len (curr) - 1 ] ! = '1' ): curr + = "1" findString(idx + 1 , n, m, curr) curr = curr[ 0 : len (curr) - 1 ] # Driver Code N,M = 2 , 3 # Function call findString( 0 , N, M, "") print (ans) # This code is contributed by shinjanpatra |
Javascript
<script> // JavaScript code to implement the approach // Declared 2 global variable // one is the answer string and // the other is the count of created string let ans = "" let Count = 0 // Function to find the mth string. function findString(idx, n, m, curr){ // When size of string is equal to n if (idx == n){ // If count of strings created // is equal to m-1 if (Count == m - 1) ans = curr else Count += 1 return } // Call the function to recurse for // currentstring + "0" curr += "0" findString(idx + 1, n, m, curr) curr = curr.substring(0,curr.length - 1) // If the last character of curr is not 1 // then similarly recurse for "1". if (curr.length == 0 || curr[curr.length - 1] != '1' ){ curr += "1" findString(idx + 1, n, m, curr) curr = curr.substring(0,curr.length - 1) } } // Driver Code let N = 2,M = 3 // Function call findString(0, N, M, "" ) document.write(ans) // This code is contributed by shinjanpatra </script> |
C#
// C# code for the above approach using System; using System.Collections.Generic; class GFG { // Declared 2 global variable // one is the answer string and // the other is the count of created string static String ans = "" ; public static int Count = 0; // Function to find the mth string. public static void findString( int idx, int n, int m, string curr) { // When size of string is equal to n if (idx == n) { // If count of strings created // is equal to m-1 if (Count == m - 1) { ans = curr; } else { Count += 1; } return ; } // Call the function to recurse for // currentstring + "0" curr += "0" ; findString(idx + 1, n, m, curr); curr=curr.Substring(0,curr.Length-1); // If the last character of curr is not 1 // then similarly recurse for "1". if (curr.Length==0|| curr[(curr.Length - 1)] != '1' ) { curr += "1" ; findString(idx + 1, n, m, curr); curr=curr.Substring(0,curr.Length-1); } } // Driver Code public static void Main() { int N = 2, M = 3; // Function call findString(0, N, M, "" ); Console.WriteLine(ans); } } |
Output
10
Time Complexity: O(2N)
Auxiliary Space: O(1)
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