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In this basic array problem, I need to add the sum or the value of the previous index of the array to the current index value using recursion.

For example, {5,3,1,2} becomes {5,8,9,11}.

Since the 0th index is the first index so it remains as it is.

I think my code is correct but instruction pointer is not flowing according to my dry run and the return statement (for the base condition) is not executing. Since I have my code in Java, after adding try-catch I'm getting output as expected. But adding try-catch is not the solution.

class kr1
{
    public static void main(String[] args) {
        int a[]= {5,1,3,9,5};
        a=addr (a,0,a.length);
        for (int i:a)
            System.out.print (i +" ");
    }

    static int[] addr(int a[],int i,int n)
    {
        if (i==0)
            addr(a,i+1,n);
        if (i==n-1)
        {
            a[i] = a[i]+a[i-1];
            return a ; //this return statement is not executing
        }
             //this below part is executing even if i has reached to n-1
        a[i] = a[i] + a[i-1];
        addr(a,i+1,n);
        return a;
    }
   }
ggorlen
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vrnvav97
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3 Answers3

2
static int[] addr(int a[],int i,int n)
{
    if (i==0)
        addr(a,i+1,n); // buggy
    if (i==n-1)
    {
        a[i] = a[i]+a[i-1];
        return a;
    }

    // code executed after if (i==0) block returns
    a[i] = a[i] + a[i-1];
    addr(a,i+1,n);
    return a;
}

As there is no return a; inside the if (i==0) block, the subsequent code block will be executed. Since the value of i is 0, a[i-1] is equivalent to a[-1], which results in an ArrayOutOfBoundsException being thrown.

You can resolve this by:

  1. Adding a return a; inside the if (i==0) block.

  2. Having the main function to call this recursive function with i = 1. That way, you can remove the if (i==0) block.

Since Java arrays are pass-by-reference, you actually don't need to return the array.

class kr1
{
    public static void main(String[] args) {
        int a[]= {5,1,3,9,5};
        addr (a,0,a.length);
        for (int i:a)
            System.out.print (i +" ");
    }

    static void addr(int a[],int i,int n)
    {
        if (i==0)
            addr(a,i+1,n);
            return;
        if (i==n-1)
        {
            a[i] = a[i]+a[i-1];
            return;
        }
        a[i] = a[i] + a[i-1];
        addr(a,i+1,n);
    }
   }

That would make the code more readable.

Zhiyuan-Amos
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  • that was a silly mistake done by me..and for method returning value part.. you are correct..i was initially using void..but when i was not getting the expected result..i was trying by returning the array – vrnvav97 Apr 20 '19 at 13:02
0
static int[] addr(int a[],int i,int n)
{
    if (i==0)
        return addr(a,i+1,n);
    if (i==n-1)
    {
        a[i] = a[i]+a[i-1];
        return a ; //this return statement is not executing
    }
    //this below part is executing even if i has reached to n-1
    a[i] = a[i] + a[i-1];
   return addr(a,i+1,n);
}

In recursion when recalling the function, you should return the value of the function call, you didn't do that. So I just added a return to the recalling functions and it worked. Also, check your algorithm, it fails when the length if the array is one, you could add a simple check to ensure that when the array size is 1 or zero the array is returned back and no manipulation is done to the array. Cheers.

oziomajnr
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0

just change this part of your code

if (i==0)
    addr(a,i+1,n);

to this one

if (i==0){
     addr(a,i+1,n);
     return a;
}
Pythoscorpion
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