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Recursion Interview Questions

Character Combine

Prepbytes

CONCEPTS USED:

Recursion

DIFFICULTY LEVEL:

Medium

PROBLEM STATEMENT(SIMPLIFIED):

Given two characters C1, C2 and an integer K, print all the possible
K length combinations of both the characters. Print the combinations lexicographically.

NOTE:

  1. Characters can be repeated.

  2. Consecutive C2 characters are not allowed.

For Example:

Can we use Recursion here ?

Yes, here we need to find combinations that satisfy the given conditions. Recursion is a good choice to find all such combinations.

SOLVING APPROACH:

  1. Initialize an empty string str and recursively keep adding characters C1 and C2 to it based on the following cases :

    • If str is empty, you can add either C1 or C2.

    • If str is not empty, check if the last character is C2 or not. If it is C2, you can add only C1 as consecutive C2's are not allowed. If it is C1, you can add either C1 or C2.

    • Keep adding characters till length of str becomes equal to K and then print str.

STATE SPACE TREE:

SOLUTIONS:

#include <stdio.h>
#include <string.h>

/* function for appending a char to a char array */
void append(char* s, char c) {
        int len = strlen(s);
        s[len] = c;
        s[len+1] = '\0';
}

void solve(char *s, char a, char b, int n){

  if(n == 0){
    printf("%s\n", s);
    return;
  }
  else if(n > 0){
    int size = strlen(s);
    /* creating temporary char arrays */
      char temp_a[10];
      char temp_b[10];

      /* copying original char array to temp arrays */
      strcpy(temp_a, s);
      strcpy(temp_b, s);

      /* append char a and b to char arrays */
      append(temp_a, a);
      append(temp_b, b);

    char empty[2] = "";
    if(strcmp(s, empty) == 0){                  //If string is empty start with any character
      solve(temp_a, a, b, n-1);
      solve(temp_b, a, b, n-1);
    }
    else{                         //If string is not empty check if the last char was b or not
      if(s[size - 1] == b){       // If it is b then go solving with a else solve with both
        solve(temp_a, a, b, n-1);
      }
      else{
        solve(temp_a, a, b, n-1);
        solve(temp_b, a, b, n-1);
      }
    }
  }
}
int main(){
  int t; scanf("%d", &t);
  while(t--){

    char a,b; scanf(" %c %c", &a, &b);

    int n; scanf("%d", &n);

    char s[10] = "";

    solve(s, a, b, n);
  }
  return 0;
}

#include <bits/stdc++.h>
using namespace std;

void solve(string s,char a,char b,int n){
  if(n > 0){
    if(s == ""){                  //If string is empty start with any character
      solve(s+ a,a,b,n-1);
      solve(s+ b,a,b,n-1);
    }
    else{                         //If string is not empty check if the last char was b or not
      if(s[s.size()-1]==b){       // If it is b then go solving with a else solve with both
        solve(s + a,a,b,n-1);
      }
      else{
        solve(s + a,a,b,n-1);
        solve(s + b,a,b,n-1);
      }
    }
  }
  else if(n == 0)
    cout<<s<<"\n";
}
int main(){
  int t;cin>>t;
  while(t--){
    char a,b;cin>>a>>b;
    int n;cin>>n;
    string s = "";
    solve(s,a,b,n);
  }
  return 0;
}

import java.util.*;
import java.io.*;

public class Main {
  static void combine(String s,char a,char b,int n){
  if(n > 0){
    if(s == ""){           //If string is empty start with any character
      combine(s+ a,a,b,n-1);
      combine(s+ b,a,b,n-1);
    }
    else{                   //If string is not empty check if the last char was b or not
      if(s.charAt(s.length()-1)==b){       // If it is b then go solving with a else solve with both
        combine(s + a,a,b,n-1);
      }
      else{
        combine(s + a,a,b,n-1);
        combine(s + b,a,b,n-1);
      }
    }
  }
  else if(n == 0)
    System.out.println(s);
}
  public static void main(String args[]) throws IOException {
    Scanner sc = new Scanner(System.in);
    int t = sc.nextInt();
    while(t!=0){
      char a = sc.next().charAt(0);
      char b = sc.next().charAt(0);
      int n = sc.nextInt();
      String s = "";
      combine(s,a,b,n);
      t--;
    }
  }
}

def solve(s, a, b, n):
	if(n > 0):
		
		if(s == ""):
		
			solve(s + a, a, b, n-1)
			solve(s + b, a, b, n-1)

		else:

			if(s[len(s)-1] == b):
			
				solve(s + a, a, b, n-1)
		
			else:
				
				solve(s + a, a, b, n-1)
				solve(s + b, a, b, n-1)

	elif(n == 0):
		print(s)

for _ in range(int(input())):
	
	a, b, n = input().split()
	n = int(n)
	s = ""

	solve(s, a, b, n)

[forminator_quiz id="1545"]

So, in this blog, we have tried to explain the concept of Recursion. If you want to solve more questions on Recursion, which are curated by our expert mentors at PrepBytes, you can follow this link Recursion.

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