CODEWITHSUNDEEP
javaarrays
BayAreaLearner
BayAreaLearner

Reputation: 15

How to search odd index elements in an array

int modeOdd = 0;

System.out.println("Occurence of all existing ODD digits --");

  for (int i = 1;  i < ary.length; i += 2) {  // This prints the array element at the odd indices 
      if (ary[i] > 0) {
      System.out.println("Digit " + i + " : "  + ary[i]);
      }
    }

  System.out.println( "\nthe odd digit (s) that has/have the "
          + "higest occurence-");



for (int i = 1; i < ary.length; i += 2){
    int newNumber = ary[i];
    if (newNumber > ary[modeOdd] ) && ( i % 2 != 0 )){
        modeOdd = i;
    }
} 
System.out.println(modeOdd);


}

The first portion of the code works and prints the array elements at my odd indices. However, the second portion of the code is finding the mode of all my array elements. I don't understand why it is doing this since I start it at index i and increment it by 2. I tried i modulus 2 can't equal 0 and don't see any changes.

What do I need to change? Thank you.

Upvotes: 0

Views: 2836

Answers (2)

Pham Trung
Pham Trung

Reputation: 11284

The bug is this line:

int modeOdd = 0;

You should declare modeOdd = 1 , as currently you declared it to 0, which is not odd, so if ary[0] contains value greater than any values in odd indices, it will never change.

Note: Be careful if length of the array is less than 2

Upvotes: 3

Ankur Singhal
Ankur Singhal

Reputation: 26077

public class Test {
    public static void main(String[] args) {
        int modeOdd = 1;

        int[] ary = new int[] { 2, 3, 4, 5, 6, 78, 9, 3 };
        System.out.println("Occurence of all existing ODD digits --");

        for (int i = 1 ; i < ary.length ; i += 2) {  // This prints the array element at the odd indices
            if (ary[i] > 0) {
                System.out.println("Digit " + i + " : " + ary[i]);
            }
        }

        System.out.println("\nthe odd digit (s) that has/have the " + "higest occurence-");

        for (int i = 1 ; i < ary.length ; i += 2) {
            int newNumber = ary[i];
            if (newNumber > ary[modeOdd] && (i % 2 != 0)) {
                modeOdd = i;
            }
        }
        System.out.println(modeOdd);

    }
}

output

Occurence of all existing ODD digits --
Digit 1 : 3
Digit 3 : 5
Digit 5 : 78
Digit 7 : 3

the odd digit (s) that has/have the higest occurence-
5

Upvotes: 0

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