Algorithm to find the largest integer in array

Question

I am trying to create a method which returns an int - the value of the largest integer in the sent array. The way I want this method to work, is to check the first and the last element of the array in a for-loop, and work their way to the middle. So i = first integer, k = last integer. When i = 0, k = n-1 (indexes), when i = 1, k = n-2 if you catch my drift. In every loop it needs to check if a[i]>a[k]. Then they switch places. Then I know that the largest number is in the leading half of the array, and then I want it to check that half, so ultimately the largest int is at index 0.

I tried like this:

public static int maxOfArray(int[] a)
{
    int length = a.length;

    if(length<1)
        throw new NoSuchElementException("Not at least one integer in array");

    while (length > 1)
    {
        int k = length;

        for(int i = 0; i < length/2; i++)
        {
            k--;

            if(a[i]<a[k])
            {
                int j = a[i];
                a[i] = a[k];
                a[k] = j;
            }
        }
        length /=2;
    }
    return a[0];
}

..but I don't really get it.. I'm having a hard time "picturing" what's happening here.. But it's not always working.. (though sometimes).

EDIT Also: The array {6,15,2,5,8,14,10,16,11,17,13,7,1,18,3,4,9,12}; will spit out 17 as the largest number. I realize I have to fix the odd-length bug, but I would like to solve this even-length array first..

Answer 1

A bug is when encountering length is odd.

In these cases, you "miss" the middle element.

Example: for input int[] arr = { 8, 1, 5, 4, 9, 4, 3, 7, 2 }; - the element 9 will be compared and checked against itself, but then you reduce the size of length, you exclude 9 from the array you are going to iterate next.

I believe it can be solved by reducing the problem to ceil(length/2) instead of length/2 (and handling special case of length==1)

The other issue as was mentioned in comments is: you need to iterate up to length/2 rather then up to length, otherwise you are overriding yourself.

Lastly - the sign is wrong.

if(a[i]>a[k])

should be

if(a[i]<a[k])

Remember - you are trying to swap the elements if the first is smaller the the second in order to push the larger elements to the head of your array.

Answer 2

but I don't really get it.. I'm having a hard time "picturing" what's happening here.. But it's not always working.. (though sometimes).

In that case you should use a debugger to step through the code to get a picture of what each line of code does.

---

What I would do is:

public static int maxOfArray(int[] a) {
    int max = a[0];
    for (int i : a)
        if (max < i)
            max = i;
    return max;
}

public static int findMaxTheHardWay(int[] array) {
    for (int length = array.length; length > 1; length = (length + 1) / 2) {
        for (int i = 0; i < length / 2; i++) {
            if (array[i] < array[length - i - 1])
                array[i] = array[length - i - 1]; // don't need to swap.
        }
    }
    return array[0];
}

public static void main(String... args) {
    Random rand = new Random(1);
    for (int i = 1; i <= 1000; i++) {
        int[] a = new int[i];
        for (int j = 0; j < i; j++) a[j] = rand.nextInt();
        int max = maxOfArray(a);
        int max2 = findMaxTheHardWay(a);
        if (max != max2)
            throw new AssertionError(i + ": " + max + " != " + max2);
    }
}

Answer 3

This is rather a crazy way to solve the problem, but I'll play along.

The problem is in the inner loop.

• You start out with i = 0 and k = length - 1.
• If a[i] > a[k] you swap them.
• ...
• You end up with k = 0 and i = length - 1
• If a[i] > a[k] you swap them.

If you look at that carefully you will notice that if we swapped the elements in the first swap, we will also swap them in the last swap; i.e. we will UNDO the effects of the first swap. And the same applies pair-wise through the entire array slice.

See?

What you need to do is to stop the inner loop half way ... and then take account of the case where length is odd.

---

By the way, the reason I called this "rather crazy", because the obvious and simple way is much faster: O(N) versus O(NlogN)

Answer 4

int a[] = {1,7,3};   
List<Integer> list = Arrays.asList(a);  
Integer largest = Collections.max(list);

This will give you Largest number in Array.

Answer 5

Here is a solution that fits the specifications that you want (unlike many other here, humm, humm):

    final Integer[] input = {1, 2, 6, 32, 4, 44 ,12, 42, 3, 7, 17, 22, 57, 23, 102, 103 };

    int half = (input.length / 2);
    int mod = input.length % 2;
    while (half >= 0) {
        for (int i = 0, j = (half * 2) + mod - 1; i <= half && j >= half; i++, j--) {
            if (input[i] < input[j]) {
                final int tmp = input[i];
                input[i] = input[j];
                input[j] = tmp;
            }
        }
        if (half == 0) break;
        half = half / 2;
        mod = half % 2;
    }
    //Here, input[0] = the biggest number in the original input.

Edit: Added mod, so it works if the last element is the largest..

Answer 6

I think your code is working, you just have to ceil the length / 2 in case of odd array but my tests return proper result:

package org.devince.largestinteger;

import java.util.NoSuchElementException;

public class LargestInteger {
    final static int[] input = {1, 2, 6, 32, 4, 44 ,12, 42, 3, 7, 17, 22, 57, 23, 102, 103 };
//  final static int[] input = { 8, 1, 5, 4, 9, 4, 3, 7, 2 };
//  final static int[] input = {1,3,7};

    /**
     * @param args
     */
    public static void main(String[] args) {
        System.out.println(String.valueOf(maxOfArray(input)));
    }

    public static int maxOfArray(int[] a)
    {
        int length = a.length;

        if(length<1)
            throw new NoSuchElementException("Not at least one integer in array");

        while (length > 1)
        {
            int k = length;

            for(int i = 0; i < length; i++)
            {
                k--;

                if(a[i]>a[k])
                {
                    int j = a[i];
                    a[i] = a[k];
                    a[k] = j;
                }
            }
            length = (int) Math.ceil(length / 2f);
        }
        return a[0];
    }

}

Answer 7

Why not just store the first value of the array to a variable max.

After that just loop through the array starting from second position till the last , in the loop just check if the current value is greater than max or not.If it is greater just assign max that value.

Return max and you have the largest number.

public int FindLargest()
{
    int[] num = { 1, 2, 5, 12, 13, 56, 16, 4 };
    int max = num[0];

    for (int i = 1; i <num.length; i++)
    {
        if (num[i] > max)
        {
            max = num[i];
        }
    }

    return max;
}

Answer 8

As the same u can approach like also,

int length = a.length;
   while (length > 1)
    {
        int k = length;
        for(int i = 0; i < length; i++)
        { 
            for(int y = k-1; y >= i; y--) 
            {         
                if(a[i]<a[y])
                {
                    int j = a[i];
                    a[i] = a[y];
                    a[y] = j;               
                }
            }       
        }
        length /=2;
    }

Answer 9

         final int validSampleRates[] = new int[]{
                    5644800, 2822400, 352800, 192000, 176400, 96000,
                    88200, 50400, 50000, 4800,47250, 44100, 44056, 37800, 32000, 22050, 16000, 11025, 4800, 8000};
          ArrayList <Integer> YourArray = new ArrayList <Integer> ():
 for (int smaple : validSampleRates){

                    YourArray.add(smaple);
                }
            Integer largest = Collections.max(YourArray);
            System.out.println("Largest   " + String.valueOf(largest));

The best way is to use Array that extends List Collection as ArrayList