CODEWITHSUNDEEP
javarecursion
PoohSan
PoohSan

Reputation: 45

counting the times a character appears in string java using recursion

I'm doing some recursion exercises and one has quite confused me. The problem stated that I should count the times "yo" has appeared in a string, but if the letter 'o' appears before the "yo", I'm not supposed to count it.

"yoyo" is counted as two, but "yooyo" is counted as one. I have done a code but it doesn't follow the condition of not counting the "yo" that has an 'o' before it. Thanks for the help!

My code:

import java.util.*;

public class Mp3
{
    static int oui(String arr, int index)
    {
      int count = 0;
      if(index >= arr.length())
        return 0;
  
      if(count == 0)
      {
        if(arr.charAt(index) == 'y')
        {
          if(arr.charAt(index + 1) == 'o')
            count++;
          else if(arr.charAt(index - 1) == 'o' && arr.charAt(index - 2) != 'y')
            --count;
          else
            return count + oui(arr, index + 1);
        }
     }
     return count + oui(arr, index + 1);
  }

  public static void main (String[] args)
  {
     String inp3 = "yoyooyoxhadjiohioyooyoyoxxyoyo";

     int res3 = oui(inp3, 0);

     System.out.println(inp3 + ":" + res3);
  }
}

Upvotes: 1

Views: 123

Answers (2)

Papai from BEKOAIL
Papai from BEKOAIL

Reputation: 1529

Make this simple:

Algo

let's assume index represents 0-th index initially.

  1. check the index-ed and index+1-ed character , if it is y and o increment count by +1 and index-ed value by +2;
  2. if it is not, simply ignore the count and increment index-ed value by +1.
  3. do it repetitively until you reach at the end of the string.

Code

caller :: getCountYo("yoyooyoxhadjiohioyooyoyoxxyoyo", 0);

private int getCountYo(String input, int index) {
    int count = 0;
    /* when you reach at the end of the string */
    if(index == input.length() || index+1 == input.length()) {
        return 0;
    }else {
        /* when "yo" found as a sub-string */
        if(input.charAt(index) == 'y' && input.charAt(index+1) == 'o') {
            count += getCountYo(input, index+2)+1;
        }else if(input.charAt(index) != 'y') {
            count += getCountYo(input, index+1);
        }
    }
    return count;
   }

Upvotes: 0

Saso
Saso

Reputation: 74

import java.util.*;

public class Mp3
{
  static int oui(String arr, int index)
  {
    int count = 0;
    if(index >= arr.length())
      return 0;
    if(arr.charAt(index)=='o' && arr.charAt(index+1)=='y'&&arr.charAt(index+2)=='o') 
      
       return count + oui(arr, index + 3);
      

    if(arr.charAt(index) == 'y' && arr.charAt(index + 1) == 'o')
       
       return count+ 1 +oui(arr, index + 2);
     
    
    return count + oui(arr, index + 1);
 }


  public static void main (String[] args)
 {
  String inp3 = "yoyooyoxhadjiohioyooyoyoxxyoyo";

  int res3 = oui(inp3, 0);

  System.out.println(inp3 + ":" + res3);
 }

}

**I tried not to change a lot from your code :

  1. I check if index= o then if followed by yo I skip them and check afterward
  2. I check if index= y and followed by o then I take them and check afterwards cause I always check for o first
  3. Simple :) If you have any questions post a comment and I will try to explain better**

Upvotes: 3

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