Find index of an extra element present in one sorted array

2.3

Given two sorted arrays. There is only 1 difference between the arrays. First array has one element extra added in between. Find the index of the extra element.

Examples:

Input : {2, 4, 6, 8, 9, 10, 12};
        {2, 4, 6, 8, 10, 12};
Output : 4
The first array has an extra element 9.
The extra element is present at index 4.

Input :  {3, 5, 7, 9, 11, 13}
         {3, 5, 7, 11, 13}
Output :  3

Method 1 (Basic)

The basic method is to iterate through the whole second array and check element by element if they are different.

C++

// C++ program to find an extra element present
// in arr1[]
#include <iostream>
using namespace std;

// Returns index of extra element in arr1[].
// n is size of arr2[]. Size of arr1[] is
// n-1.
int findExtra(int arr1[], int arr2[], int n)
{
   for (int i=0; i<n; i++)
     if (arr1[i] != arr2[i])
         return i;

   return n;
}

// Driver code
int main()
{
    int arr1[]  =  {2, 4, 6, 8, 10, 12, 13};
    int arr2[] =  {2, 4, 6, 8, 10, 12};
    int n = sizeof(arr2) / sizeof(arr2[0]);

    // Solve is passed both arrays
    cout << findExtra(arr, arr2, n);
    return 0;
}

Java

// Java program to find an extra 
// element present in arr1[]
class Main
{
    // Returns index of extra element in arr1[].
    // n is size of arr2[]. Size of arr1[] is
    // n-1.
    static int findExtra(int arr1[], int arr2[], int n)
    {
       for (int i=0; i<n; i++)
         if (arr1[i] != arr2[i])
             return i;
     
       return n;
    }
     
    public static void main (String[] args)
    {
        int arr1[]  =  {2, 4, 6, 8, 10, 12, 13};
        int arr2[] =  {2, 4, 6, 8, 10, 12};
        int n = arr2.length;
     
        // Solve is passed both arrays
        System.out.println(findExtra(arr1, arr2, n));
    }
}

/* This code is contributed by Harsh Agarwal */


Output:

 6

Time complexity : O(n)

 

Method 2 (Using Binary search)

We use binary search to check whether the same indices elements are different & reduce our search by a factor of 2 in each step.

C++

// C++ program to find an extra element pre.sent
// in arr1[]
#include <iostream>
using namespace std;

// Returns index of extra element in arr1[].
// n is size of arr2[]. Size of arr1[] is
// n-1.
int findExtra(int arr1[], int arr2[], int n)
{
    int index = 0;  // Initialize result

    // left and right are end points denoting
    // the current range.
    int left = 0, right = n - 1;
    while (left <= right)
    {
        int mid = (left+right) / 2;

        // If middle element is same of both
        // arrays, it means that extra element
        // is after mid so we update left to mid+1
        if (arr2[mid] == arr1[mid])
            left = mid + 1;

        // If middle element is different of the
        // arrays, it means that the index we are
        // searching for is either mid, or before
        // mid. Hence we update right to mid-1.
        else
        {
            index = mid;
            right = mid - 1;
        }
    }

    // when right is greater than left our
    // search is complete.
    return index;
}

// Driver code
int main()
{
    int arr1[]  =  {2, 4, 6, 8, 10, 12, 13};
    int arr2[] =  {2, 4, 6, 8, 10, 12};
    int n = sizeof(arr2) / sizeof(arr2[0]);

    // Solve is passed both arrays
    cout << findExtra(arr1, arr2, n);
    return 0;
}

Java

// Java program to find an extra 
// element present in arr1[]
class Main
{
    // Returns index of extra element in arr1[].
    // n is size of arr2[]. Size of arr1[] is
    // n-1.
    static int findExtra(int arr1[], int arr2[], int n)
    {
        int index = 0;  // Initialize result
     
        // left and right are end points denoting
        // the current range.
        int left = 0, right = n - 1;
        while (left <= right)
        {
            int mid = (left+right) / 2;
     
            // If middle element is same of both
            // arrays, it means that extra element
            // is after mid so we update left to mid+1
            if (arr2[mid] == arr1[mid])
                left = mid + 1;
     
            // If middle element is different of the
            // arrays, it means that the index we are
            // searching for is either mid, or before
            // mid. Hence we update right to mid-1.
            else
            {
                index = mid;
                right = mid - 1;
            }
        }
     
        // when right is greater than left our
        // search is complete.
        return index;
    }
     
    public static void main (String[] args)
    {
        int arr1[]  =  {2, 4, 6, 8, 10, 12,13};
        int arr2[] =  {2, 4, 6, 8, 10, 12};
        int n = arr2.length;
     
        // Solve is passed both arrays
        System.out.println(findExtra(arr1, arr2, n));
    }
}

/* This code is contributed by Harsh Agarwal */


Output:

 6

Time complexity : O(log n)

Asked in: Amazon

Asked in Amazon

This article is contributed by Abhishek Khatri . If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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