Finding an element in an array such that the index is equal to its value using binary search

Question

I have an array in ascending order, consisting of positive integers and with condition duplicates are not allowed.In the array i must find m such that array[m]=m using binary search I have used below code:

public class Main {

public static void main(String[] args) {
    Scanner a = new Scanner(System.in);
    int n = a.nextInt();
    int arr[] = new int[n];
    for (int i = 0; i < n; i++) {
        arr[i] = a.nextInt();
    }
    int result = binarysearch(arr,n);
    if(result!=-1)
        System.out.println(result);
    else
        System.out.println("NOT_FOUND");
}

public static  int binarysearch(int[] array,int n){
int start=1,end=n;
while(start<=end){
    int mid = (start+end)/2;
    if(mid==array[mid-1])
        return mid;
    else  if(mid<array[mid-1])
        end=mid-1;
    else
        start=mid+1;
}
return -1;
}

Please note that the array is 1-indexed and not 0-indexed The problem I am facing is when i try the following case : array={1,2,3,4,5} the expected output is 1 but due to binary search the method returns 3 as Output. Is there any workaround to bypass this type of cases?

This problem trivial if the array only has positive numbers without duplicates in ascending order. You only need to check the first number, if it is 1 then return it. Otherwise, the rest of the array would never be equal to its index. — MT756
– MT756, Commented Mar 11, 2020 at 19:25
@ArvindKumarAvinash Apparently, the OP is viewing the array as 1-indexed — MT756
– MT756, Commented Mar 11, 2020 at 19:32

MT756 · Accepted Answer · 2020-03-11 19:42:30Z

Looking at the conditions: "I have an array in ascending order, consisting of positive integers and with condition duplicates are not allowed"

ascending order
positive integers
no duplicates

This problem has a trivial solution.

public static int binarysearch(int[] array,int n){
    If(a[0] == 1) return 1;
    else return -1;
}

Because if the first number in the array is not 1, the rest of the array would not have any value that is equal to its index.

Explanation:

Suppose the array has terms a1, a2, a3 ... an, where non of the terms are equal and aj < ak if j < k

The problem is essentially finding a term ai where ai = i

If a1 is not 1, it has to be greater than 1 according to condition #2.

By condition 1 and 3, a2 has to be greater than 2. Then, a3 has to be greater than 3, and so on. Term an has to be greater than n. Yielding no solution.

On the other hand, if a1 is 1, you have the term and you can return it.

Isaiah · Accepted Answer · 2020-03-11 19:31:20Z

Your code looks mostly correct, but you should not assume the array is already sorted when the user inputs it.

I think the real problem lies in understanding how binary search works. Binary search always starts in the middle of an array. Thus, if you provide the input {1, 2, 3, 4, 5}, you will always get 3 because the algorithm starts by searching in the middle and notices that the number 3 equals its index.

If your goal is to find the smallest integer m such that a[m] is equal to m, then you should not use binary search. Instead, you should start from the beginning of the array (since we assume it is sorted in ascending order) and use linear search:

for (int i = 0; i < n; ++i) {
  if (a[i] == i) {
    return i;
  }
}
return -1;

Something like the above should give the intended result. (Note that the above uses 0-based index logic, but you should be able to tweak it as needed. Also, arrays in Java are always 0-indexed, so you may want to double-check that you understand the problem correctly.)

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