Find index of an extra element present in one sorted array

Given two sorted arrays. There is only 1 difference between the arrays. The 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
Explanation: 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
Explanation: The first array has an extra element 9.
The extra element is present at index 3.

Method 1: This includes the basic approach to solve this particular problem.

  • Approach: The basic method is to iterate through the whole second array and check element by element if they are different. As the array is sorted, checking the adjacent position of two arrays should be similar until and unless the missing element is found.
  • Algorithm:
    1. Traverse through the array from start to end.
    2. Check if the element at i’th element of the two arrays is similar or not.
    3. If the elements are not similar then print the index and break
  • Implementation:

    C++

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    // 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(arr1, arr2, n);
        return 0;
    }

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    Java

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    // Java program to find an extra 
    // element present in arr1[]
    class GFG 
    {
      
        // 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;
        }
          
        // Driver Code
        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

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    Python3

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    # Python 3 program to find an 
    # extra element present in arr1[]
      
      
    # Returns index of extra .
    # element in arr1[] n is 
    # size of arr2[]. Size of 
    # arr1[] is n-1.
    def findExtra(arr1, arr2, n) :
        for i in range(0, n) :
            if (arr1[i] != arr2[i]) :
                return i
      
        return n
      
      
    # Driver code
    arr1 = [2, 4, 6, 810, 12, 13]
    arr2 = [2, 4, 6, 8, 10, 12]
    n = len(arr2)
      
    # Solve is passed both arrays
    print(findExtra(arr1, arr2, n))
      
    # This code is contributed 
    # by Nikita Tiwari.

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    C#

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    // C# program to find an extra 
    // element present in arr1[]
    using System;
      
    class GfG 
    {
          
        // 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;
        }
          
        // Driver code
        public static void Main ()
        {
            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
            Console.Write(findExtra(arr1, arr2, n));
        }
    }
      
    // This code is contributed by parashar.

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    PHP

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    <?php
    // PHP program to find an extra 
    // element present in arr1[]
      
    // Returns index of extra element 
    // in arr1[]. n is size of arr2[]. 
    // Size of arr1[] is n-1.
    function findExtra($arr1
                       $arr2, $n)
    {
    for ($i = 0; $i < $n; $i++)
        if ($arr1[$i] != $arr2[$i])
            return $i;
      
    return $n;
    }
      
    // Driver code
    $arr1 = array (2, 4, 6, 8, 
                   10, 12, 13);
    $arr2 = array(2, 4, 6, 
                  8, 10, 12);
    $n = sizeof($arr2);
      
    // Solve is passed
    // both arrays
    echo findExtra($arr1, $arr2, $n);
      
    // This code is contributed by ajit
    ?>

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    Output :



     6
    
  • Complexity Analysis:
    • Time complexity : O(n).
      As one traversal through the array is needed, so the time complexity is linear.
    • Space complexity : O(1).
      Since no extra space is required, the time complexity is constant.

 
Method 2: This method is a better way to solve the above problem and uses the concept of binary search.

  • Approach:To find the index of the missing element in less than linear time, binary search can be used, the idea is all the indices greater than or equal to the index of the missing element will have different elements in both the arrays and all the indices less than that index will have the similar elements in both arrays.
  • Algorithm:
    1. Create three variables, low = 0, high = n-1, mid, ans = -1
    2. Run a loop until low is less than or equal to high, i.e till our search range is less than zero.
    3. If the mid element, i.e (low + high)/2, of both arrays is similar then update the search to second half of the search range, i.e low = mid + 1
    4. Else update the search to first half of the search range, i.e high = mid – 1, and update the answer to the current index, ans = i
    5. Print the index.
  • Implementation:

    C++

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    // 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)
    {
        // Initialize result
        int index = n; 
      
        // 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;
    }

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    Java

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    // Java program to find an extra 
    // element present in arr1[]
    class GFG
    {
        // 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)
        {
            // Initialize result
            int index = n; 
          
            // 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
        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 

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    Python3

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    # Python3 program to find an extra
    # element present in arr1[]
      
    # Returns index of extra element 
    # in arr1[]. n is size of arr2[]. 
    # Size of arr1[] is n-1.
    def findExtra(arr1, arr2, n) :
      
        index = n # Initialize result
      
        # left and right are end points 
        # denoting the current range.
        left = 0
        right = n - 1
        while (left <= right) :
            mid = (int)((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
    arr1 = [2, 4, 6, 8, 10, 12, 13]
    arr2 = [2, 4, 6, 8, 10, 12]
    n = len(arr2)
      
    # Solve is passed both arrays
    print(findExtra(arr1, arr2, n))
      
    # This code is contributed by Nikita Tiwari.

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    C#

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    // C# program to find an extra 
    // element present in arr1[]
    using System;
      
    class GFG {
          
        // 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)
        {
              
            // Initialize result
            int index = n; 
          
            // 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
        public static void Main ()
        {
            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
            Console.Write(findExtra(arr1, arr2, n));
        }
    }
      
    // This code is contributed by nitin mittal.

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    PHP

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    <?php
    // PHP program to find an extra 
    // element present in arr1[]
      
    // Returns index of extra element 
    // in arr1[]. n is size of arr2[]. 
    // Size of arr1[] is n-1.
    function findExtra($arr1, $arr2, $n)
    {
        // Initialize result
        $index = $n
      
        // left and right are 
        // end points denoting
        // the current range.
        $left = 0; $right = $n - 1;
        while ($left <= $right)
        {
            $mid = ($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
    {
        $arr1 = array(2, 4, 6, 8, 
                      10, 12, 13);
        $arr2 = array(2, 4, 6, 
                      8, 10, 12);
        $n = sizeof($arr2) / sizeof($arr2[0]);
      
        // Solve is passed both arrays
        echo findExtra($arr1, $arr2, $n);
        return 0;
      
    // This code is contributed by nitin mittal
    ?>

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    Output :

     6
    
  • Complexity Analysis:
    • Time complexity : O(log n).
      The time complexity of binary search is O(log n)
    • Space complexity : O(1).
      As no extra space is required, so the time complexity is constant.

Method 3: This method solves the given problem using the predefined function.

  • Approach: To find the element which is different, find the sum of each array and subtract the sums and find the absolute value. Search the larger array and check if the absolute is equal to an index and return that index. If an element is missing and all the other elements are the same, then the difference of sums will be equal to missing element.
  • Algorithm:
    1. Create a function to calculate the sum of two ararys.
    2. Find the absolute differnce between the sum of two arrays (value).
    3. Traverse the larger array from start too end
    4. If the element at any index is equal to value, then print the index and break the loop.
  • Implementation:

    C++

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    // C++ code for above approach
    #include<bits/stdc++.h>
    using namespace std;
      
    // function return sum of array elements
    int sum(int arr[], int n)
    {
        int summ = 0;
        for (int i = 0; i < n; i++)
        {
            summ += arr[i];
        }
        return summ;
    }
      
    // function return index of given element
    int indexOf(int arr[], int element, int n)
    {
        for (int i = 0; i < n; i++)
        {
            if (arr[i] == element)
            {
                return i;
            }
        }
        return -1;
    }
      
    // Function to find Index
    int find_extra_element_index(int arrA[], 
                                 int arrB[], 
                                 int n, int m)
    {
      
        // Calculating extra element
        int extra_element = sum(arrA, n) - 
                            sum(arrB, m);
          
        // returns index of extra element
        return indexOf(arrA, extra_element, n);
    }
      
    // Driver Code
    int main()
    {
        int arrA[] = {2, 4, 6, 8, 10, 12, 13};
        int arrB[] = {2, 4, 6, 8, 10, 12};
        int n = sizeof(arrA) / sizeof(arrA[0]);
        int m = sizeof(arrB) / sizeof(arrB[0]);
        cout << find_extra_element_index(arrA, arrB, n, m);
    }
      
    // This code is contributed by mohit kumar

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    Java

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    // Java code for above approach
    class GFG 
    {
      
        // Function to find Index 
        static int find_extra_element_index(int[] arrA, 
                                            int[] arrB) 
        {
      
            // Calculating extra element
            int extra_element = sum(arrA) - sum(arrB);
              
            // returns index of extra element
            return indexOf(arrA, extra_element);
        }
          
        // function return sum of array elements
        static int sum(int[] arr)
        {
            int sum = 0;
            for (int i = 0; i < arr.length; i++)
            {
                sum += arr[i];
            }
            return sum;
        }
          
        // function return index of given element
        static int indexOf(int[] arr, int element) 
        {
            for (int i = 0; i < arr.length; i++)
            {
                if (arr[i] == element)
                {
                    return i;
                }
            }
            return -1;
        }
          
        // Driver Code
        public static void main(String[] args) 
        {
            int[] arrA = {2, 4, 6, 8, 10, 12, 13};
            int[] arrB = {2, 4, 6, 8, 10, 12};
            System.out.println(find_extra_element_index(arrA, arrB));
        }
    }
      
    /* This code contributed by PrinciRaj1992 */

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    Python3

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    # Python3 code for above approach
      
    # Function to find Index 
    def find_extra_element_index(arrA, arrB):
          
        # Calculating extra element
        extra_element = sum(arrA) - sum(arrB)
          
        # returns index of extra element
        return arrA.index(extra_element)
      
    # Driver Code
    arrA = [2, 4, 6, 8, 10, 12, 13]
    arrB = [2, 4, 6, 8, 10, 12]
    print(find_extra_element_index(arrA,arrB))
      
    # This code is contributed by Dravid

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    C#

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    // C# code for above approach
    using System;
      
    class GFG 
    {
      
        // Function to find Index 
        static int find_extra_element_index(int[] arrA, 
                                            int[] arrB) 
        {
      
            // Calculating extra element
            int extra_element = sum(arrA) - sum(arrB);
              
            // returns index of extra element
            return indexOf(arrA, extra_element);
        }
          
        // function return sum of array elements
        static int sum(int[] arr)
        {
            int sum = 0;
            for (int i = 0; i < arr.Length; i++)
            {
                sum += arr[i];
            }
            return sum;
        }
          
        // function return index of given element
        static int indexOf(int[] arr, int element) 
        {
            for (int i = 0; i < arr.Length; i++)
            {
                if (arr[i] == element)
                {
                    return i;
                }
            }
            return -1;
        }
          
        // Driver Code
        public static void Main(String[] args) 
        {
            int[] arrA = {2, 4, 6, 8, 10, 12, 13};
            int[] arrB = {2, 4, 6, 8, 10, 12};
            Console.WriteLine(find_extra_element_index(arrA, arrB));
        }
    }
      
    // This code has been contributed by 29AjayKumar

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    Output :

     6
    
  • Complexity Analysis:

    • Time Complexity: O(n).
      Since only three traversals through the array is needed, So the time complexity is linear.
    • Space Complexity: O(1).
      As no extra space is required, so the time complexity is constant.

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