Search algorithms always returns -1

Question

I have the following implementation of binary and sequential search. I'm not very experienced with Java, and somehow implementation of both methods always returns -1 when searching an element:

import java.util.Random;
import java.util.Scanner;

public class searchArrayDemo {

    private static int binarySearch(int[] inputArr, int key) {
        int start = 0;
        int end = inputArr.length - 1;

        while (start <= end) {

            int middle = (start + end) / 2;

            if (key < inputArr[middle]) {
                end = middle - 1;
            }

            if (start > inputArr[middle]) {
                start = middle + 1;
            }

            if (start == inputArr[middle]) {
                return middle;
            }
        }
        return -1;
        }

    // This function returns index of element x in arr[]
    private static int search(int arr[],int x)
    {
        for (int i = 0; i < arr.length; i++) {
            // Return the index of the element if the element
            // is found
            if (arr[i] == x)
                return i;
        }

        // return -1 if the element is not found
        return -1;
    }


    // Driver method to test above
    public static void main(String args[])
    {
        int c, n, array[];
        Scanner in = new Scanner(System.in);
        System.out.println("Enter number of elements: ");
        n = in.nextInt();
        array = new int[n];
        Random rand = new Random();
        for (c = 0; c < n; c++)
        {
            array[c] = rand.nextInt(100);
        }
        System.out.println("Elements: ");
        for (int i = 0; i < array.length; i++)
        {
            System.out.print(array[i] + " ");
        }

        Scanner sc = new Scanner(System.in);
        System.out.println("Enter the element to be searched: ");
        int k = sc.nextInt();

            System.out.println("\nSelect type of search\n");
            System.out.println("1. Binary search");
            System.out.println("2. Sequential search");

        int choice = in.nextInt();
        int binarySearchResult = binarySearch(array, choice);
        int sequentialSearchResult = search(array, choice);
        switch(choice) {
            case 1:
                if (binarySearchResult == -1) {
                    System.out.println("\n" + k + " element not found");
                }
                else {
                    System.out.println("\n"+ k +" element found at position "+ binarySearchResult);
                }
                break;
            case 2:
                if (sequentialSearchResult == -1) {

                    System.out.println(sequentialSearchResult + "");
                    System.out.println("\n" + k + " element not found");
                }
                else {
                    System.out.println("\n" + k + " element found at position " + sequentialSearchResult);
                }
                break;
        }
    }
}

I'm suposing that this is some silly mistake, but can you please help me.

Possible duplicate of [What is a debugger and how can it help me diagnose problems?](https://stackoverflow.com/questions/25385173/what-is-a-debugger-and-how-can-it-help-me-diagnose-problems) — UnholySheep, Dec 14 '18 at 00:40

score 0 · Accepted Answer · answered Dec 14 '18 at 00:39

0

Yes. This

int binarySearchResult = binarySearch(array, choice);
int sequentialSearchResult = search(array, choice);

should be

int binarySearchResult = binarySearch(array, k);
int sequentialSearchResult = search(array, k);

Because k is the element you want to search for and choice is supposed to be the type of search. It's also worth noting that your binarySearch will not function correctly until you sort the array.

answered Dec 14 '18 at 00:39

Elliott Frisch

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1

You've missed `if (start > inputArr[middle])`... also all look like "typograpical error" and don't need full answer. – Alexei Levenkov Dec 14 '18 at 00:41
@AlexeiLevenkov Made the answer community wiki. Add it in... it's a little cruel (imo) to just close it as a typographical error (especially since OP is clearly just starting out). – Elliott Frisch Dec 14 '18 at 00:43
Thanks, that's the thing. – enoent Dec 14 '18 at 09:34

score 0 · Answer 2 · answered Dec 14 '18 at 00:52

I believe you want to replace "start" with "key" in the if statements of the first function. Also (side note), how does your function handle the scenario when start + end is odd and, also, is that how you want your function to handle that scenario?

Also, check your start = middle + 1;

Search algorithms always returns -1

2 Answers2