Java Creating a Method to find Max Difference in an Array.

[Java] Creating a Method to find Max Difference in an Array. - Printable Version

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Creating a Method to find Max Difference in an Array. - SIM-810273 - Nov 18, 2015

Hi all, I have to create a function that will look at an array of 5 numbers and find the difference between adjacent numbers and from them differences then return the largest difference.

Honestly I'm not sure where to start I know using loops will be my best bet but I'm unsure.

Arrays

Code 
int [] numbers = {12, 8, 34, 10, 59};

int [] numbers2 = {-50, 100, 20, -40};

So for the first one the maximum difference would be 49 (10 to 59)
Second one the maximum difference would be 150 (-50 to 100)

Remember it has to be between numbers next to each other.

Thanks in advance!

* - Sparx710 - Nov 18, 2015

If the array always contains 5 numbers, then you can hard code the comparison of the numbers.
Calculate Math.abs(numbers[0]-numbers[1]), Math.abs(numbers[1]-numbers[2]) and so on.
Then print/return the largest one.

Or loop through the array until numbers.length-1 and calculate Math.abs(numbers[i]-numbers[i+1])

As I cannot write PMs yet, here is my response to your message^^
You can compare a value in an array to the next one like i wrote above (array[index] and array[index+1])

PHP Code 


<?php 
public static int maxDifference(int[] a) {
        int maxDiff = 0;
        //Loop until the second last value, so you can compare it with the last one at the end
        for (int index = 0; index < a.length - 1; index++) {
            int diff = a[index] - a[index + 1]; //Calculate the difference of two numbers, that are next to each other
            if (diff < 0) diff = diff * -1; //Get a postive value
            if (diff > maxDiff) maxDiff = diff; //Set new maxDiff
        }
        return maxDiff;
    } 

* - WainWright - Nov 19, 2015

I can imagine many ways to do this but the simplest one I have is to simply sort the array in ascending order and then subtract the last element from the first.

Code 
Arrays.sort(numbers);//Arrays.parallelSort(numbers);

return numbers[4] - numbers[0];//return numbers[numbers.length - 1] - numbers[0];

I don't know if performance would be better to manually find the largest and smallest or not. Sometimes the methods Oracle provides are more efficient.

* - FocalBanger - Nov 20, 2015

Iterate through the array and keep track of the largest number (ex. if(biggestNumber < array[i]){biggestNumber = array[i]) then keep track of the biggest difference by subtracting the biggest number from each element in the array.

* - Enchanter - Nov 20, 2015

In Java8 you can do

Code 
public static int getMaxAd(int[] arr){

        return Arrays.stream(IntStream.range(0, arr.length - 1).toArray())

                  .map(x -> arr[x + 1] - arr[x])

                  .max().getAsInt();

    }

Code 
int max = getMaxAd(numbers);