Java BinarySearch

Question

Can I get some help please? I have tried many methods to get this to work i got the array sorted and to print but after that my binary search function doesnt want to run and give me right results. It always gives me -1. Any help?

public class BinarySearch {
public static final int NOT_FOUND = -1;
public static int binarySearch(double[] a, double key) {
    int low = 0;
    int high = a.length -1;
    int mid;
    while (low<=high) {
        mid = (low+high) /2;
        if (mid > key) 
            high = mid -1;
        else if (mid < key) 
            low = mid +1;
        else 
            return mid;
    }
    return NOT_FOUND;
}
public static void main(String[] args) {
    double key = 10.5, index;
    double a[] ={10,5,4,10.5,30.5};
    int i;
    int l = a.length;
    int j;
    System.out.println("The array currently looks like");
    for (i=0; i<a.length; i++)
        System.out.println(a[i]);
    System.out.println("The array after sorting looks like");
    for (j=1; j < l; j++) {
        for (i=0; i < l-j; i++) {
            if (a[i] > a[i+1]) {
                double temp = a[i];
                a[i] = a[i+1];
                a[i+1] = temp;
            }
        }
    }
    for (i=0;i < l;i++) {
        System.out.println(a[i]);
    }
    System.out.println("Found " + key + " at " + binarySearch(double a[], key));
}   
}

I would step in a debugger through the code and see why it not behaving the way it should. btw. A bug you are unlikely to find on your own is the mid should be mid = (low+high) >>> 1; I would also compare it to the code for the Arrays.binarySearch in the JDK as it works and is almost the same. ;) — Peter Lawrey
– Peter Lawrey, Commented Sep 20, 2012 at 17:32

Shades88 · Accepted Answer · 2012-09-20 17:44:35Z

12

you are not actually comparing with the array values. in

while (low <= high) {
      mid = (low + high) / 2;
      if (mid > key) {
          high = mid - 1;
      } else if (mid < key) {
          low = mid + 1;
      } else {
          return mid;
      }
}

Instead use this section

    while (low <= high) {
        mid = (low + high) / 2;
        if (a[mid] > key) {
            high = mid - 1;
        } else if (a[mid] < key) {
            low = mid + 1;
        } else {
            return mid;
        }
    }

You were correct to find the indexes, but what you were doing is that you were just comparing index number with your key, which is obviously incorrect. When you write a[mid] you will actually compare your key with the number which is at index mid.

Also the last line of code is giving compile error, it should be

System.out.println("Found " + key + " at " + binarySearch(a, key));

answered Sep 20, 2012 at 17:44

Shades88

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2 Comments

John Over a year ago

Note that this is technically not correct implementation for relatively large arrays, and this subtle bug is often found in literature. Low + high may overflow int range even if both are within range on their own. Instead mid should be calculated as low + (high - low)/2.

TinkerTenorSoftwareGuy Over a year ago

I've seen this elsewhere. Now I know why. Thanks so much!

nullpotent · Accepted Answer · 2012-09-20 17:38:11Z

1

Here

    if (mid > key) 
        high = mid -1;
    else if (mid < key) 
        low = mid +1;
    else 
        return mid;

You're comparing index to a value (key) in array. You should instead compare it to a[mid]

And,

System.out.println("Found " + key + " at " + binarySearch(double a[], key));

Should be

System.out.println("Found " + key + " at " + binarySearch(a, key));

answered Sep 20, 2012 at 17:38

nullpotent

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Comments

Abhijit Gaikwad · Accepted Answer · 2013-03-14 04:44:30Z

1

public static double binarySearch(double[] a, double key) {

    if (a.length == 0) {
      return -1;
    }
    int low = 0;
    int high = a.length-1;

    while(low <= high) {
      int middle = (low+high) /2; 
      if (b> a[middle]){
        low = middle +1;
      } else if (b< a[middle]){
        high = middle -1;
      } else { // The element has been found
        return a[middle]; 
      }
    }
    return -1;
  }

answered Mar 14, 2013 at 4:44

Abhijit Gaikwad

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Comments

Ragulan · Accepted Answer · 2018-07-10 00:55:19Z

int binarySearch(int list[], int lowIndex, int highIndex, int find)
    {
        if (highIndex>=lowIndex)
        {
            int mid = lowIndex + (highIndex - lowIndex)/2;

            // If the element is present at the
            // middle itself
            if (list[mid] == find)
                return mid;

            // If element is smaller than mid, then
            // it can only be present in left subarray
            if (list[mid] > find)
                return binarySearch(list, lowIndex, mid-1, find);

            // Else the element can only be present
            // in right subarray
            return binarySearch(list, mid+1, highIndex, find);
        }

        // We reach here when element is not present
        //  in array
        return -1;
    }

PoweredByRice · Accepted Answer · 2013-08-21 21:00:21Z

I somehow find the iterative version not quite easy to read, recursion makes it nice and easy :-)

public class BinarySearch {

    private static int binarySearchMain(int key, int[] arr, int start, int end) {
        int middle = (end-start+1)/2 + start; //get index of the middle element of a particular array portion

        if (arr[middle] == key) {
            return middle;
        }

        if (key < arr[middle] && middle > 0) {
            return binarySearchMain(key, arr, start, middle-1); //recurse lower half
        }

        if (key > arr[middle] && middle < arr.length-1) {
            return binarySearchMain(key, arr, middle+1, end); //recurse higher half
        }

        return Integer.MAX_VALUE; 
    }

    public static int binarySearch(int key, int[] arr) { //entry point here
        return binarySearchMain(key, arr, 0, arr.length-1);
    }

}

user2200660 · Accepted Answer · 2014-03-13 04:46:06Z

Here is a solution without heap. The same thing can be done in an array. If we need to find 'k' largest numbers, we take an array of size 'k' populated with first k items from the main data source. Now, keep on reading an item, and place it in the result array, if it has a place.

public static void largestkNumbers() {
    int k = 4;    // find 4 largest numbers
    int[] arr = {4,90,7,10,-5,34,98,1,2};
    int[] result = new int[k];

    //initial formation of elems
    for (int i = 0; i < k; ++i) {
      result[i] = arr[i];
    }
    Arrays.sort(result);

    for ( int i = k; i < arr.length; ++i ) {
      int index = binarySearch(result, arr[i]);
      if (index > 0) {
        // insert arr[i] at result[index] and remove result[0]
        insertInBetweenArray(result, index, arr[i]);
      }
    }
  }

  public static void insertInBetweenArray(int[] arr, int index, int num) {
    // insert num at arr[index] and remove arr[0]
    for ( int i = 0 ; i < index; ++i ) {
      arr[i] = arr[i+1];
    }
    arr[index-1] = num;
  }

  public static int binarySearch(int[] arr, int num) {

    int lo = 0;
    int hi = arr.length - 1;
    int mid = -1;

    while( lo <= hi ) {
      mid = (lo+hi)/2;
      if ( arr[mid] > num ) {
        hi = mid-1;
      } else if ( arr[mid] < num ) {
        lo = mid+1;
      } else {
        return mid;
      }
    }
    return mid;
  }

Community · Accepted Answer · 2017-05-23 12:32:26Z

0

int BinSearch(int[] array, int size, int value)
{
    if(size == 0) return -1;
    if(array[size-1] == value) return size-1;
    if(array[0] == value) return 0;
    if(size % 2 == 0) {
        if(array[size-1] == value) return size-1;
        BinSearch(array,size-1,value);
    }
    else
    {
        if(array[size/2] == value) return (size/2);
        else if(array[size/2] > value) return BinSearch(array, (size/2)+1, value);
    else if(array[size/2] < value) return (size/2)+BinSearch(array+size/2, size/2, value);
    }
}

or

Binary Search in Array

edited May 23, 2017 at 12:32

CommunityBot

11 silver badge

answered Jun 26, 2015 at 18:34

Mark Kazmenko

1

1 Comment

Mark Kazmenko Over a year ago

Change T* to int[] and T value to int

Shiran Gabriel · Accepted Answer · 2016-07-05 23:12:09Z

0

   /**
     * Find whether 67 is a prime no
     * Domain consists 25 of prime numbers
     * Binary Search
     */

    int primes[] = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97};

    int min = 0,
            mid,
            max = primes.length,
            key = 67,
            count= 0;

    boolean isFound = false;


    while (!isFound) {
        if (count < 6) {
          mid = (min + max) / 2;
            if (primes[mid] == key) {
                isFound = true;
                System.out.println("Found prime at: " + mid);
            } else if (primes[mid] < key) {
                min = mid + 1; 
                isFound = false;
            } else if (primes[mid] > key) { 
                max = mid - 1; 
                isFound = false;
            }
            count++;

        } else {
            System.out.println("No such number");
            isFound = true;
        }

    }

edited Jul 5, 2016 at 23:12

answered Jul 5, 2016 at 22:59

Shiran Gabriel

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1 Comment

coatless Over a year ago

Please explain the code that you are submitting as an answer.

Sᴀᴍ Onᴇᴌᴀ · Accepted Answer · 2017-10-24 18:06:31Z

/**
HOPE YOU LIKE IT
A.K.A Binary Search
Take number array of 10 elements, input a number a check whether the number 
is present:
**/
package array;
import java.io.InputStreamReader;
import java.io.BufferedReader;
import java.io.IOException;
class BinaryS
{
    public static void main(String args[]) throws IOException
    {       
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));      
    System.out.print("Enter a number: ");
    int n=Integer.parseInt(br.readLine());
    int a[]={10,20,30,40,50,60,70,80,90,100};
    int upper=a.length-1,lower=0,mid;
    boolean found=false;
    int pos=0;
    while(lower<=upper)
    {
        mid=(upper+lower)/2;
        if(n<a[mid])upper=mid-1;
        else if(n>a[mid])lower=mid+1;
        else
        {
            found=true;
            pos=mid;
            break;
        }
    }
    if(found)System.out.println(n+" found at index "+pos);
    else System.out.println(n+" not found in array");
}
}

Nirav Chhatrola · Accepted Answer · 2019-10-02 08:09:45Z

Well I know I am posting this answer much later.
But according to me its always better to check boundary condition at first.
That will make your algorithm more efficient.

    public static int binarySearch(int[] array, int element){
    if(array == null || array.length == 0){  // validate array
        return -1;
    }else if(element<array[0] || element > array[array.length-1]){ // validate value our of range that to be search 
        return -1;
    }else if(element == array[0]){  // if element present at very first element of array
        return 0;
    }else if(element == array[array.length-1]){ // if element present at very last element of array
        return array.length-1;
    }

    int start = 0;
    int end = array.length-1;

    while (start<=end){
        int midIndex = start + ((end-start)/2);   // calculate midIndex
        if(element < array[midIndex]){  // focus on left side of midIndex
            end = midIndex-1;
        }else if(element > array[midIndex]){// focus on right side of midIndex
            start = midIndex+1;
        }else {
            return midIndex;   // You are in luck :)
        }
    }
    return -1; // better luck next time :(
}

Kanagalingam · Accepted Answer · 2021-03-25 15:23:50Z

static int binarySearchAlgorithm() {
        // Array should be in sorted order. Mandatory requirement
        int[] a = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
        int lowIndex = 0;
        int valueToFind = 8;
        int highIndex = a.length - 1;

        while (lowIndex <= highIndex) {
            //Finding the midIndex;
            int midIndex = (highIndex + lowIndex) / 2;
            // Checking if midIndex value of array contains the value to be find.
            if (a[midIndex] == valueToFind) {
                return midIndex;
            } 
            // Checking the mid Index value is less than the value to be find.
            else if (a[midIndex] < valueToFind) {
                // If Yes, changing the lowIndex value to midIndex value + 1;
                lowIndex = midIndex + 1;
            } else if (a[midIndex] > valueToFind) {
                // If Yes, changing the highIndex value to midIndex value - 1;
                highIndex = midIndex - 1;
            } else {
                return -1;
            }
        }
        return -1;
    }

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