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I'm working on a coding problem where we have to mutate an existing array into a new array. The rules are that an element at the arrays index is equal to, a[i - 1] + a[i] + a[i + 1]. The catch is that if the a[i - 1] or a[i + 1] doesn't exist, they should be counted as zero.

My test cases are passing for every value up to the last value, and I can't understand why it isn't being calculated, when every other value is,

int[] mutateTheArray(int n, int[] a) 
{
    int b[] = new int[a.length + 1];
    if(a.length == 1) 
    {
        return a;
    }
    for(int i = 0; i < a.length - 1; i++) 
    {
        if(i == 0) 
        {
            b[0] = 0 + a[i] + a[i + 1];
        }
        if(i == a.length)
        {
            b[a.length] = a[i - 1] + a[i] + 0;
        }
        else if(i != a.length && i != 0)
        {
            b[i] = a[i - 1] + a[i] + a[i + 1];
        }
    }

    return b;
}

The output should be for an array a = [4, 0, 1, -2, 3], output should be: [4, 5, -1, 2, 1]. I'm getting the answer except for the last value, which is calculating to 0. I know the issue is in accessing the array index - 1, but I don't know how to get each element without it going out of bounds. Any help is appreciated, thanks :)

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  • if(i == a.length) should be if(i == a.length - 1). The condition of the last else if can just be dropped. Commented Feb 26, 2020 at 3:53
  • Note that you never reach i == a.length, as the condition of your for-loop is i < a.length - 1. Commented Feb 26, 2020 at 3:56

2 Answers 2

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2
  1. Your array b should have the size a.length, not a.length + 1.
  2. You should change the condition of your for-loop to i < a.length, as the last element should be included.
  3. Your second if-condition should be i == a.length - 1. (Note: Array.length returns the amount of elements in the Array, which is the last index + 1.)
    Also, you can use else if, as this condition can not be met if the first one was.
  4. The last if-condition is unnecessary as the condition is always false now. Just use else.

Btw.: You may want to look into the ternary operator.

All in all, your code would look like this:

int[] mutateTheArray(int n, int[] a) 
{
    int b[] = new int[a.length];
    if(a.length == 1)
        return a;
    for(int i = 0; i < a.length; i++) 
    {
        if(i == 0) 
        {
            b[0] = 0 + a[i] + a[i + 1];
        }
        else if(i == a.length-1)
        {
            b[a.length-1] = a[i - 1] + a[i] + 0;
        }
        else
        {
            b[i] = a[i - 1] + a[i] + a[i + 1];
        }
    }
    return b;
}

... or when using the ternary operator:

int[] mutateTheArray(int n, int[] a) 
{
    int b[] = new int[a.length];
    if(a.length == 1)
        return a;
    for(int i = 0; i < a.length; i++) 
    {
            b[i] = (i == 0 ? 0 : a[i - 1]) + a[i] + (i == a.length-1 ? 0 : a[i + 1]);
    }
    return b;
}

PS: Edited from b[a.length] = ... to b[a.length-1] = ... at line 15.

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4 Comments

Btw.: I see no need for the first argument int n. I have left it in for compatibily, but you may want to remove it if you're going to ignore it anyway.
Joja, have you tested this code? It is breaking here
Edited from b[a.length] = a[i - 1] + a[i] + 0; to b[a.length-1] = a[i - 1] + a[i] + 0;
@PedroCoelho Indeed, I have not seen this problem. Thanks.
2 
int[] mutateTheArray(int n, int[] a) {
    int b[] = new int[a.length];
    for(int i = 0; i < a.length; i++) {
        b[i] = a[i];
        b[i] += i > 0 ? a[i-1] : 0;
        b[i] += i < a.length-1 ? a[i+1] : 0;
    }

    return b;
}
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