Extra in one of the two sorted arrays

Given two sorted arrays a[] and b[] of distinct elements. The first array has one element extra added in between. Return the index of the extra element.

Note: 0-based indexing is followed.

Examples: 

Input: a[] = [2, 4, 6, 8, 9, 10, 12], b[] = [2, 4, 6, 8, 10, 12]
Output: 4
Explanation: In the first array, 9 is extra added and it's index is 4.

Input: a[] = [3,5,7,8,11,13], b[] = [3,5,7,11,13]
Output: 3
Explanation: In the first array, 8 is extra and it's index is 3.

Input: a[] = [3,5], b[] = [3]
Output: 1
Explanation: In the first array, 5 is extra and it's index is 1.

[Naive Approach] Linear Scan - O(n) Time and O(1) Space

Since both arrays are sorted and identical except for one extra element in a[], compare elements at the same indices. All elements match until the extra element causes a shift, so the first mismatch index gives the answer. If no mismatch is found, the extra element is at the end of a[].

#include <iostream>
#include <vector>
using namespace std;

// Returns index of extra element in a[]
int findExtra(vector<int>& a, vector<int>& b)
{
    for (int i = 0; i < b.size(); i++)
    {
        if (a[i] != b[i])
            return i;
    }
    return b.size();
}

int main()
{
    vector<int> a = {2, 4, 6, 8, 9, 10, 12};
    vector<int> b = {2, 4, 6, 8, 10, 12};

    cout << findExtra(a, b);
    return 0;
}

public class GFG {

    // Returns index of extra element in a
    static int findExtra(int[] a, int[] b) {

        // Compare elements until mismatch is found
        for (int i = 0; i < b.length; i++) {
            if (a[i] != b[i]) {
                return i;
            }
        }

        // If all match, extra element is at the end
        return b.length;
    }

    public static void main(String[] args) {

        int[] a = {2, 4, 6, 8, 9, 10, 12};
        int[] b = {2, 4, 6, 8, 10, 12};

        System.out.println(findExtra(a, b));
    }
}

# Returns index of extra element in a
def findExtra(a, b):

    # Compare elements until mismatch is found
    for i in range(len(b)):
        if a[i] != b[i]:
            return i

    # If all match, extra element is at the end
    return len(b)

if __name__ == "__main__": 
    a = [2, 4, 6, 8, 9, 10, 12]
    b = [2, 4, 6, 8, 10, 12]

print(findExtra(a, b))

using System;

class GFG
{
    // Returns index of extra element in a
    static int FindExtra(int[] a, int[] b)
    {
        // Compare elements until mismatch is found
        for (int i = 0; i < b.Length; i++)
        {
            if (a[i] != b[i])
            {
                return i;
            }
        }

        // If all match, extra element is at the end
        return b.Length;
    }

    static void Main()
    {
        int[] a = { 2, 4, 6, 8, 9, 10, 12 };
        int[] b = { 2, 4, 6, 8, 10, 12 };

        Console.WriteLine(FindExtra(a, b));
    }
}

// Returns index of extra element in a
function findExtra(a, b) {

    // Compare elements until mismatch is found
    for (let i = 0; i < b.length; i++) {
        if (a[i] !== b[i]) {
            return i;
        }
    }

    // If all match, extra element is at the end
    return b.length;
}

// Driver code
let a = [2, 4, 6, 8, 9, 10, 12];
let b = [2, 4, 6, 8, 10, 12];

console.log(findExtra(a, b));

4

[Expected Approach] Using Binary Search - O(log(n)) Time and O(1) Space

Since both arrays are sorted and identical except for one extra element, use binary search to find the first index where elements differ. If elements at mid are equal, the extra element lies on the right, otherwise, it lies on the left (including mid). keep narrowing the range to find the earliest mismatch.

a[] = [2, 4, 6, 8, 9, 10, 12], b[] = [2, 4, 6, 8, 10, 12]

left = 0, right = b.size() - 1 = 5
mid = 2

Compare a[2] and b[2], since they are same, we move to the right half
left = 3, right = 5
mid = 4

Compare a[4] and b[4], since they are not same, we move to the left half and update index = 3
left = 3, right = 4
mid = 3

Compare a[3] and b[3], since they are same, we move to the right half
left = 4 right = 4
mid = 4

Compare a[4] and b[4], since they are not same, we move to the right half and update index = 4
left = 4 right = 3

Since left is now more than right, stop.

#include <iostream>
#include <vector>
using namespace std;

// Returns index of extra element in a
int findExtra(vector<int>& a, vector<int>& b) {
    int left = 0;
    int right = b.size() - 1;
    int index = b.size();

    // Binary search to find first mismatch
    while (left <= right) {
        int mid = (left + right) / 2;

        // If elements match, extra element is on right side
        if (a[mid] == b[mid]) 
            left = mid + 1;
            
        else {
            // Possible answer, move left to find earlier mismatch
            index = mid;
            right = mid - 1;
        }
    }
    return index;
}

int main() {
    vector<int> a = {2, 4, 6, 8, 9, 10, 12};
    vector<int> b = {2, 4, 6, 8, 10, 12};
    cout << findExtra(a, b);
    return 0;
}

public class GFG {

    // Returns index of extra element in a
    static int findExtra(int[] a, int[] b) {
        int left = 0;
        int right = b.length - 1;
        int index = b.length;

        // Binary search to find first mismatch
        while (left <= right) {
            int mid = (left + right) / 2;

            // If elements match, extra element is on right side
            if (a[mid] == b[mid]) {
                left = mid + 1;
            } else {
                // Possible answer, move left to find earlier mismatch
                index = mid;
                right = mid - 1;
            }
        }
        return index;
    }

    public static void main(String[] args) {
        int[] a = {2, 4, 6, 8, 9, 10, 12};
        int[] b = {2, 4, 6, 8, 10, 12};

        System.out.println(findExtra(a, b));
    }
}

# Returns index of extra element in a
def findExtra(a, b):
    left = 0
    right = len(b) - 1
    index = len(b)

    # Binary search to find first mismatch
    while left <= right:
        mid = (left + right) // 2

        # If elements match, extra element is on right side
        if a[mid] == b[mid]:
            left = mid + 1
        else:
            # Possible answer, move left to find earlier mismatch
            index = mid
            right = mid - 1

    return index
    
if __name__ == "__main__": 

    a = [2, 4, 6, 8, 9, 10, 12]
    b = [2, 4, 6, 8, 10, 12]

print(findExtra(a, b))

using System;

class GFG {
    // Returns index of extra element in a
    static int FindExtra(int[] a, int[] b) {
        int left = 0;
        int right = b.Length - 1;
        int index = b.Length;

        // Binary search to find first mismatch
        while (left <= right) {
            int mid = (left + right) / 2;

            // If elements match, extra element is on right side
            if (a[mid] == b[mid]) {
                left = mid + 1;
            }
            else {
                // Possible answer, move left to find earlier mismatch
                index = mid;
                right = mid - 1;
            }
        }
        return index;
    }

    static void Main() {
        int[] a = { 2, 4, 6, 8, 9, 10, 12 };
        int[] b = { 2, 4, 6, 8, 10, 12 };

        Console.WriteLine(FindExtra(a, b));
    }
}

// Returns index of extra element in a
function findExtra(a, b) {
    let left = 0;
    let right = b.length - 1;
    let index = b.length;

    // Binary search to find first mismatch
    while (left <= right) {
        let mid = Math.floor((left + right) / 2);

        // If elements match, extra element is on right side
        if (a[mid] === b[mid]) {
            left = mid + 1;
        } else {
            // Possible answer, move left to find earlier mismatch
            index = mid;
            right = mid - 1;
        }
    }

    return index;
}
// Driver code
let a = [2, 4, 6, 8, 9, 10, 12];
let b = [2, 4, 6, 8, 10, 12];

console.log(findExtra(a, b));

4