Home  >  Article  >  Java  >  In Java, add odd numbers at odd index positions and even numbers at even index positions in an array, and the sum is divisible by the array length.

In Java, add odd numbers at odd index positions and even numbers at even index positions in an array, and the sum is divisible by the array length.

WBOY
WBOYforward
2023-09-12 22:41:071101browse

In Java, add odd numbers at odd index positions and even numbers at even index positions in an array, and the sum is divisible by the array length.

Understanding how arrays work is fundamental for any developer, and Java is no exception. In Java, an array is an object that stores multiple variables of the same type. However, arrays can often be used in more complex ways. One such example is when you need to calculate the sum of an array, considering only the even numbers at odd indices and the odd numbers at even indices, whether they are evenly divisible by the size of the array. In this article, we will guide you step by step on how to solve this problem.

Problem Statement

Given an array of integers, write a function in Java to determine whether the sum of even numbers at odd indices and the sum of odd numbers at even indices are divisible by the size of the array.

method

The solution involves looping through the array and selectively adding numbers to the sum. We will iterate through each index. For an even index, we will check if the number is odd and if so, add it to our sum. For odd indexes we will check if the number is even and if so add it to our sum. Finally, we will check if the sum is divisible by the size of the array.

The Chinese translation of

Example

is:

Example

The following is a simple implementation of the above method in Java:

public class Main {
   public static boolean isSumDivisible(int[] array) {
      int sum = 0;
      for (int i = 0; i < array.length; i++) {
         if (i % 2 == 0 && array[i] % 2 != 0) {
               sum += array[i];
         } else if (i % 2 != 0 && array[i] % 2 == 0) {
               sum += array[i];
         }
      }
      return sum % array.length == 0;
   }
   public static void main(String[] args) {
      int[] array = {1, 2, 3, 4, 5, 6};
      System.out.println(isSumDivisible(array));
   }
}

Output

false
The Chinese translation of

Explanation

is:

Explanation

Let's take a look at the example array {1, 2, 3, 4, 5, 6}. In this case, we have −

  • At index 0 (an even index), we have the number 1, which is odd.

  • At index 1 (odd index) we have even 2.

  • At index 2 (an even index), we have the number 3, which is odd.

  • On index 3 (an odd index), we have even 4.

  • At index 4 (an even index), we have the number 5, which is odd.

  • At index 5 (odd index) we have even 6.

So, we add these numbers to the total and we get 1 2 3 4 5 6 = 21. The size of the array is 6. Since 21 is not divisible by 6, the output of the function isSumDivisible(array) will be "false".

This question demonstrates a good understanding of arrays, iteration and conditional logic

in conclusion

Understanding how to manipulate arrays and use conditional logic in Java is crucial to solving many problems in programming. The specific problem of checking whether the sum of even numbers at odd indices and the sum of odd numbers at even indices is divisible by the size of the array is a good example of how to apply these concepts. Practicing solving problems like this can enhance your understanding of Java and improve your problem-solving skills.

The above is the detailed content of In Java, add odd numbers at odd index positions and even numbers at even index positions in an array, and the sum is divisible by the array length.. For more information, please follow other related articles on the PHP Chinese website!

Statement:
This article is reproduced at:tutorialspoint.com. If there is any infringement, please contact admin@php.cn delete