CODEWITHSUNDEEP
javaarrays
user1979029
user1979029

Reputation: 11

Find the first occurrence of an odd and count them

import java.util.*;

public class FirstOddOccurrence {
    public static void main(String[] args) {
        int[] x = {2, 4, 6, 7, 4, 3, 2, 7, 6, 7, 7};
        int i;

        display(x);

        System.out.printf("# Occurrences of first odd = %3d\n", firstOddCount(x));
    }

    private static void display(int[] x) {
        int i;

        System.out.print("Array: ");
        for (i = 0; i < x.length; i++) {
            if (i < x.length - 1)
                System.out.printf("%3d, ", x[i]);
            else
                System.out.printf("%3d\n", x[i]);
        }
    }

    public static int odd(int[] x) {
        int i;
        int y;
        for (i = 0; i < x.length; i++) {
            y = x[i] % 2;
            if (y == 1) {
                return x[i];
            } else {
                return 0;
            }
        }
        return x[i];
    }

    public static int firstOddCount(int x[]) {
        int i;
        int c = 0;
        for (i = 0; i < x.length; i++) {
            if (x[i] == odd(x))
                c++;

        }
        return c;
    }
}

I'm trying to find the first occurrence of an odd number in the array that has been provided. What is wrong with my program? I can't seem to get the program to count the first odd occurrences.

Upvotes: 0

Views: 677

Answers (4)

paxdiablo
paxdiablo

Reputation: 881383

Your particular problem is that you return 0 if you find an even number. That means that the list {2, 4, 6, 8, 1} will give you 0, rather than 1, as the first odd number.

What you should do is ignore leading even numbers and continue to process the list.

However, the way you've organised your program, you're processing the first all-even part of the list twice, once in odd() to find the first odd number, then again in firstOddCount() to count how many of that number there are - that's totally unnecessary.

Once you find the first odd number, I think you can be reasonably certain that number (or any other odd number for that matter) does not exist in the space you've already searched. Otherwise it would have been the first odd number. Hence it makes little sense to go back and look at that initial part of the list again.

A way in which you can easily just process the list once is as follows:

public static int firstOddCount (int numbers[]) {
    // Find first odd number or end of list.

    int idx = 0, len = numbers.length;
    while ((idx < len) && ((numbers[idx] % 2) == 0)
        idx++;

    // If at end of list, everything is even => count(first(odd)) is 0.

    if (idx == len)
        return 0;

    // Otherwise, start counting from current position.

    int count = 1, oddnum = numbers[idx];
    while (++idx < len)
        if (numbers[idx] == oddnum)
            count++;

    return count;
}

Upvotes: 0

Trasvi
Trasvi

Reputation: 1247

Andr's solution fixes your issue; odd(x) will return 0 if x[0] is even, and x[0] if it is odd.

You could also improve firstOddCount like so: odd(x) will always return the same value, so only calculate it once.

public static int firstOddCount(int x[]) {
   int firstOdd = odd(x);
   int c=0;
   for(int i=0; i < x.length; i++) {
       if (x[i]==firstOdd)
            c++;
   }
   return c;

}

Upvotes: 1

Kanagaraj M
Kanagaraj M

Reputation: 966

If you are trying to get one element from group you should use 'break' when your condition matched first time else it will give all...

Upvotes: -1

andr
andr

Reputation: 16054

Your code here:

if (y == 1) {
    return x[i];
} else {
    return 0;
}

does not work - if a tested number is even, you immediately return 0. Instead you want to skip these even numbers and wait until an odd number comes up. In the end, if you don't find any odd number, you return 0. Here is the corrected version of odd():

int i;
int y;
for (i = 0; i < x.length; i++) {
    y = x[i] % 2;
    if (y == 1) {
        return x[i];
    }
}
return 0;

Upvotes: 2

Related Questions