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);
}
}
Output
10
Time Complexity: O(2N)
Auxiliary Space: O(1)
