Find the Last Multiple of 3 in an Array using Java

Question

Write a method that returns the index of the last value in the array that is a multiple of 3.

For example, in the array

[4, 7, 9, 7, 12] the last multiple of three is ‘12’, which occurs at index 4.

This is my code so far. Could anyone help me revising my code?

public int findMultipleOfThree(int[] arr)
{
    int result = 0;
    for (int i = arr.length-1; i > 0; i--)
    {
        if(i%3==0)
        {
            result = arr[i];
        }
    }
    return result;
}

I know this is far from the right answer. Please help

Answer 1

Problem

You have switched the element for comparison and result, you need to

• check the value if (arr[i] % 3 == 0)
• return the indice result = i

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Solution

• best: you can do it from the end, and return the first match, you do the minimum iteration

    public static int findMultipleOfThreeBackward(int[] arr) {
        for (int i = arr.length - 1; i > 0; i--) {
            if (arr[i] % 3 == 0) {
                return i;
            }
        }
        return -1;
    }
  

• You can do it from the beginning and override result to return the last one

   public static int findMultipleOfThreeForward(int[] arr) {
        int result = 0;
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] % 3 == 0) {
                result = i;
            }
        }
        return result;
    }
  

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Note

I'd also suggest to use -1, as it's an unvalid index, it shows you did not find any value that matches the condition