Java recursion odd of sum

Question

We have to find sum of array and then return if the value is odd (boolean) using recursion

A = {87, 31, 15, 25, 10, 15, 21, 75)
methodname(A, pos) //position is 0 at the begining

i did this so far, but i'm way off since i can't sum and return boolean in the same line

if (pos == array.length-1) {
    return A[pos] % 2 != 0
} else {
    if (pos < A.length - 1)
        return A[pos] + methodname(A, pos + 1) % 2 == 1;
}

@Rab: he's supposed to use recursion (in place of a loop), but he does not post a complete method, making the question very difficult to answer. — Hovercraft Full Of Eels, Oct 17 '18 at 01:29
@HovercraftFullOfEels oh alright, Didn't really look at the title. I thought it would be something else because his returning in both statements — SamHoque, Oct 17 '18 at 01:30
int[] A = {1,5,7,8}; System.out.println(Rec(A, 0)); //0 is the position — Asrar, Oct 17 '18 at 02:02

Kartik · Answer 1 · 2018-10-17T02:43:14.003

2

public static void main(String[] args) {
    System.out.println(isSumOdd(new int[]{3, 3, 4}, 0));
}

private static boolean isSumOdd(int[] arr, int pos) {
    return pos == arr.length - 1
            ? arr[pos] % 2 != 0
            : isSumOdd(arr, pos + 1) ^ arr[pos] % 2 != 0; //Sum of 2 numbers can be odd iff exactly one of them is odd.
}

edited Oct 17 '18 at 02:43

answered Oct 17 '18 at 01:39

Kartik

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@Asrar Why not? – Scary Wombat Oct 17 '18 at 02:08
I see ? and : in the code. What is that for? am i missing something – Asrar Oct 17 '18 at 02:12
@Asrar it's called ternary operator. This is equivalent to writing `if (pos == arr.length - 1) { return arr[pos] % 2 != 0; } else { return isSumOdd(arr, pos + 1) ^ arr[pos] % 2 != 0; }` – Kartik Oct 17 '18 at 02:15
2

@Asrar *This solution is not even close* is a pretty rude comment to make when you do not even understand the code. I think you should apologize to Kartlik who has gone to the trouble to help you. – Scary Wombat Oct 17 '18 at 02:30
@ScaryWombat I said "2 numbers", you gave example of 3 numbers :). Anyway, would you like me to remove that line? No apologies needed, non-native speakers can sometimes sound rude while they don't actually mean it (me being one of them). – Kartik Oct 17 '18 at 02:33
My bad if i said something wrong. But i never came across ^ in java yet. May be i have to learn a lot yet – Asrar Oct 17 '18 at 02:37
Moved that line to the right place to avoid confusion. @Asrar ^ is [XOR operator](https://stackoverflow.com/a/726665/1039555). – Kartik Oct 17 '18 at 02:45

mettleap · Answer 2 · 2018-10-17T02:05:58.860

0

You can try using the following approach (syntax might be wrong, treat the following as pseudo code):

  // gives whether the sum from i to end 
  // is even or odd 
  Boolean sumIsEven(int[] arr, int i ){

      //base case
      if(arr.length-1==i){
           return arr[i]%2 ==0;
      }

      if(arr[i]%2 == 0){
           return sumIsEven(arr, i +1);
      }else{
           // arr[i] is odd
           return !sumIsEven(arr, i +1);
      }
}

But this will give only if the whole sum is odd/even

edited Oct 17 '18 at 02:05

answered Oct 17 '18 at 01:37

mettleap

1,390
8
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Why do you have int return type. I want to first sum all the numbers in the array and then return a boolean if that is odd or not. – Asrar Oct 17 '18 at 02:05
Sorry return type should be boolean... will fix that – mettleap Oct 17 '18 at 02:06

Java recursion odd of sum

2 Answers2