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++ 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 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. |
# 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 |
<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# 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);
} } |
10
Time Complexity: O(2N)
Auxiliary Space: O(1)