CODEWITHSUNDEEP
javaarrays
Get Off My Lawn
Get Off My Lawn

Reputation: 36299

get closest value to a number in array

I have an array of positive/negative ints

int[] numbers = new int[10];
numbers[0] = 100;
numbers[1] = -34200;
numbers[2] = 3040;
numbers[3] = 400433;
numbers[4] = 500;
numbers[5] = -100;
numbers[6] = -200;
numbers[7] = 532;
numbers[8] = 6584;
numbers[9] = -945;

Now, I would like to test another int against this array, and return the number that is closest to the int.

For example if I used the number 490 i would get back item #4 from numbers 500 what is the best way to do something like this?

int myNumber = 490;
int distance = 0;
int idx = 0;
for(int c = 0; c < numbers.length; c++){
    int cdistance = numbers[c] - myNumber;
    if(cdistance < distance){
        idx = c;
        distance = cdistance;
    }
}
int theNumber = numbers[idx];

That doesn't work. Any suggestions on a good method to do this?

Upvotes: 37

Views: 125950

Answers (13)

Anoop M Maddasseri
Anoop M Maddasseri

Reputation: 10529

Kotlin - TreeSet lower method returns the greatest element in this set strictly less than the given element, or null if there is no such element.

import java.util.*

fun printLowest(number: Int) {
    val numbers = listOf(100, 90, 50, -100, -200, 532, 6584, -945)
    val lower = TreeSet(numbers).lower(number)
    println(lower)
}

printLowest(100) // Prints 90

Upvotes: -2

Islam Ahmed
Islam Ahmed

Reputation: 736

Kotlin is so helpful

fun List<Int>.closestValue(value: Int) = minBy { abs(value - it) }

val values = listOf(1, 8, 4, -6)

println(values.closestValue(-7)) // -6
println(values.closestValue(2)) // 1
println(values.closestValue(7)) // 8

List doesn't need to be sorted BTW

Upvotes: 1

Petronella
Petronella

Reputation: 2535

int valueToFind = 490;

Map<Integer, Integer> map = new HashMap();

for (int i = 0, i < numbers.length; i++){
    map.put(Math.abs(numbers[i] - valueToFind), numbers[i]);
}

List<Integer> keys = new ArrayList(map.keySet());
Collections.sort(keys);

return map.get(keys.get(0));

Upvotes: 3

Raj
Raj

Reputation: 21

You can tweak the good old binary search and implement this efficiently. 

Arrays.sort(numbers);
nearestNumber = nearestNumberBinarySearch(numbers, 0, numbers.length - 1, myNumber);

private static int nearestNumberBinarySearch(int[] numbers, int start, int end, int myNumber) {
    int mid = (start + end) / 2;
    if (numbers[mid] == myNumber)
        return numbers[mid];
    if (start == end - 1)
        if (Math.abs(numbers[end] - myNumber) >= Math.abs(numbers[start] - myNumber))
            return numbers[start];
        else
            return numbers[end];
     if(numbers[mid]> myNumber)
        return nearestNumberBinarySearch(numbers, start,mid, myNumber);
     else
         return nearestNumberBinarySearch(numbers,mid, end, myNumber);

}

Upvotes: 2

Oleksandr Pyrohov
Oleksandr Pyrohov

Reputation: 16216

In Java 8:

List<Integer> list = Arrays.stream(numbers).boxed().collect(Collectors.toList());

int n = 490;

int c = list.stream()
            .min(Comparator.comparingInt(i -> Math.abs(i - n)))
            .orElseThrow(() -> new NoSuchElementException("No value present"));

Initially, you can use a List instead of an Array (lists have much more functionality).

Upvotes: 36

Rivean
Rivean

Reputation: 1

public class Main    
{
    public static void main(String[] args)
    {   
        int[] numbers = {6,5,10,1,3,4,2,14,11,12};

        for(int i =0; i<numbers.length; i++)
        {
            sum(numbers, i, numbers[i], 12, String.valueOf(numbers[i]));
        }
    }

    static void sum(int[] arr, int i, int sum, int target, String s)
    {

        int flag = 0;

        for(int j = i+1; j<arr.length; j++)
        {

            if(arr[i] == target && flag==0)
            {
                System.out.println(String.valueOf(arr[i]));
                flag =1;

            }
            else if(sum+arr[j] == target)
            { 
                System.out.println(s+" "+String.valueOf(arr[j]));

            }
            else
            {
                sum(arr, j, sum+arr[j], target, s+" "+String.valueOf(arr[j]));
            }
        }        
    }
}

Upvotes: -1

Hamada Ibrahim
Hamada Ibrahim

Reputation: 103

public int getClosestToTarget(int target, int[] values) {

    if (values.length < 1)
        throw new IllegalArgumentException("The values should be at least one element");
    if (values.length == 1) {
        return values[0];
    }
    int closestValue = values[0];
    int leastDistance = Math.abs(values[0] - target);
    for (int i = 0; i < values.length; i++) {
        int currentDistance = Math.abs(values[i] - target);
        if (currentDistance < leastDistance) {
            closestValue = values[i];
            leastDistance = currentDistance;
        }
    }
    return closestValue;
}

Upvotes: 1

farid_z
farid_z

Reputation: 1673

One statement block to initialize and set the closest match. Also, return -1 if no closest match is found (empty array). 

 protected int getClosestIndex(final int[] values, int value) {
    class Closest {
        Integer dif;
        int index = -1;
    };
    Closest closest = new Closest();
    for (int i = 0; i < values.length; ++i) {
        final int dif = Math.abs(value - values[i]);
        if (closest.dif == null || dif < closest.dif) {
            closest.index = i;
            closest.dif = dif;
        }
    }
    return closest.index;
}

Upvotes: 1

DeadlySnipa999
DeadlySnipa999

Reputation: 1

I did this as an assignment for my course, and I programmed it in Ready to Program Java, so sorry if its a bit confusing.

// The "Ass_1_B_3" class.
import java.awt.*;
import hsa.Console;

public class Ass_1_B_3
{
    static Console c;           // The output console

    public static void main (String[] args)
    {
        c = new Console ();

        int [] data = {3, 1, 5, 7, 4, 12, -3, 8, -2};
        int nearZero = 0;
        int temp = 0;
        int temp2 = data[0];

        for (int i = 0; i < data.length; i++)
        {
            temp = Math.abs (data[i]);
            nearZero = temp2;   
            if (temp < temp2)
            {
                temp2 = temp;
                nearZero = data[i];
            }


        }

        c.println ("The number closest to zero is: " + nearZero);

        // Place your program here.  'c' is the output console
    } // main method
} // Ass_1_B_3 class

Upvotes: 0

Chris Hayes
Chris Hayes

Reputation: 12020

int myNumber = 490;
int distance = Math.abs(numbers[0] - myNumber);
int idx = 0;
for(int c = 1; c < numbers.length; c++){
    int cdistance = Math.abs(numbers[c] - myNumber);
    if(cdistance < distance){
        idx = c;
        distance = cdistance;
    }
}
int theNumber = numbers[idx];

Always initialize your min/max functions with the first element you're considering. Using things like Integer.MAX_VALUE or Integer.MIN_VALUE is a naive way of getting your answer; it doesn't hold up well if you change datatypes later (whoops, MAX_LONG and MAX_INT are very different!) or if you, in the future, want to write a generic min/max method for any datatype.

Upvotes: 51

ntstha
ntstha

Reputation: 1173

Here is something that i did...

import javax.swing.JOptionPane;

public class NearestNumber {

public static void main(String[] arg)
{
    int[] array={100,-3420,3040,400433,500,-100,-200,532,6584,-945};

    String myNumberString =JOptionPane.showInputDialog(null,"Enter the number to test:");
    int myNumber = Integer.parseInt(myNumberString);

    int nearestNumber = findNearestNumber(array,myNumber);

    JOptionPane.showMessageDialog(null,"The nearest number is "+nearestNumber);
}

public static int findNearestNumber(int[] array,int myNumber)
{

    int min=0,max=0,nearestNumber;

    for(int i=0;i<array.length;i++)
    {
        if(array[i]<myNumber)
        {
            if(min==0)
            {
                min=array[i];
            }
            else if(array[i]>min)
            {
                min=array[i];
            }
        }
        else if(array[i]>myNumber)
        {
            if(max==0)
            {
                max=array[i];
            }
            else if(array[i]<max)
            {
                max=array[i];
            }
        }
        else
        {
            return array[i];
        }
    }

    if(Math.abs(myNumber-min)<Math.abs(myNumber-max))
    {
        nearestNumber=min;
    }
    else
    {
        nearestNumber=max;
    }

    return nearestNumber;
}

}

Upvotes: -8

irrelephant
irrelephant

Reputation: 4111

cdistance = numbers[c] - myNumber. You're not taking the absolute value of the difference. If myNumber is a lot greater than numbers[c] or if numbers[c] is negative, the comparison will register as the "minimum difference".

Take for example the case where numbers[c] = -34200. numbers[c] - myNumber would then be -34690, a lot less than the distance.

Also, you should initialize distance to a large value, as no solution has been found at the start.

Upvotes: 3

Don
Don

Reputation: 555

you are very close. I think the initial value of 'distance' should be a big number instead of 0. And use the absolute value for the cdistance.

Upvotes: 4

Related Questions