Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException in Finding Max Element

Question

Whenever I am trying to run this code, it gives me out of bound exception. Can anyone point me out what's wrong with it.

package com.programs.interview;
import java.util.Scanner;

public class FindMaxNumInArray {

public static void main (String[] args) 
{ 
    Scanner scan = new Scanner (System.in); 
    System.out.print("Enter the size of the array: "); 
    int arraySize = scan.nextInt(); 
    int[] myArray = new int[arraySize]; 
    System.out.print("Enter the " + arraySize + " values of the array: "); 
    for (int i = 0; i < arraySize; i++)
        myArray[i] = scan.nextInt(); 
    for (int j = 0; j < arraySize; j++) 
        System.out.println(myArray[j]); 
    System.out.println("In the array entered, the larget value is "+ maximum(myArray,arraySize) + "."); 
} 

public static int maximum(int[] arr, int Arraylength){
    int tmp;
    if (Arraylength == 0)
        return arr[Arraylength];
    tmp = maximum(arr, Arraylength -1);
    if (arr[Arraylength] > tmp)
        return arr[Arraylength];
    return tmp;
  }
}

Output

Enter the size of the array: 5 Enter the 5 values of the array: 1 2 3 4 5 1 2 3 4 5 Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 5 at com.programs.interview.FindMaxNumInArray.maximum(FindMaxNumInArray.java:26) at com.programs.interview.FindMaxNumInArray.main(FindMaxNumInArray.java:17)

Answer 1

This is the problem:

if (arr[Arraylength] > tmp)

Valid array indexes go from 0 to length-1 inclusive. array[array.length] is always invalid, and on the initial call, ArrayLength is equal to arr.length.

It's not clear why you're using recursion at all, to be honest. An iterative solution would be much simpler - but you'll need to work out what you want to do if the array is empty.

EDIT: If you really want how I would write the recursive form, it would be something like this:

/** Returns the maximum value in the array. */
private static int maximum(int[] array) {
    if (array.length == 0) {
        // You need to decide what to do here... throw an exception,
        // return some constant, whatever.
    }
    // Okay, so the length will definitely be at least 1...
    return maximumRecursive(array, array.length);
}

/** Returns the maximum value in array in the range [0, upperBoundExclusive) */
private static int maximumRecursive(int[] array, int upperBoundExclusive) {
    // We know that upperBoundExclusive cannot be lower than 1, due to the
    // way that this is called. You could add a precondition if you really
    // wanted.
    if (upperBoundExclusive == 1) {
        return array[0];
    }
    int earlierMax = maximumRecursive(array, upperBoundExclusive - 1);
    int topValue = array[upperBoundExclusive - 1];
    return Math.max(topValue, earlierMax);

    // Or if you don't want to use Math.max
    // return earlierMax > topValue ? earlierMax : topValue;
}

Answer 2

you can't access

arr[Arraylength]

the last element would be at

arr[Arraylength -1]

for example if you have

int arr[] = new int[5];

then the elements would be at 4, because index starts from 0

arr[0], arr[1], arr[2], arr[3], arr[4]

Answer 3

Your issue is in the following piece of code:

if (arr[Arraylength] > tmp)
            return arr[Arraylength];

Indexes start at 0, so you will be out of bound for an array with 5 elements [1,2,3,4,5] indexes: [0,1,2,3,4].

Answer 4

I would use a plain loop. Java doesn't do recursion particularly well.

public static int maximum(int[] arr) {
    int max = Integer.MIN_VALUE;
    for(int i : arr) if (i > max) max = i;
    return max;
}

Answer 5

here

System.out.println("In the array entered, the larget value is "+ maximum(myArray,arraySize) + ".");

you are passing the arraysize where in maximum method you are returning arr[Arraylength] which giving ArrayIndexOutOfBound so change either in calling maximum(yArray,arraySize-1) or return arr[Arraylength-1] statement.