Find index of an extra element present in one sorted array
Last Updated :
12 Jul, 2022
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:
- Traverse through the array from start to end.
- Check if the element at i’th element of the two arrays is similar or not.
- If the elements are not similar then print the index and break
Implementation:
C++
#include <iostream>
using namespace std;
int findExtra(int arr1[],
int arr2[], int n)
{
for (int i = 0; i < n; i++)
if (arr1[i] != arr2[i])
return i;
return n;
}
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]);
cout << findExtra(arr1, arr2, n);
return 0;
}
|
Java
class GFG
{
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;
System.out.println(findExtra(arr1,
arr2, n));
}
}
|
Python3
def findExtra(arr1, arr2, n) :
for i in range(0, n) :
if (arr1[i] != arr2[i]) :
return i
return n
arr1 = [2, 4, 6, 8, 10, 12, 13]
arr2 = [2, 4, 6, 8, 10, 12]
n = len(arr2)
print(findExtra(arr1, arr2, n))
|
C#
using System;
class GfG
{
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 ()
{
int []arr1 = {2, 4, 6, 8,
10, 12, 13};
int []arr2 = {2, 4, 6,
8, 10, 12};
int n = arr2.Length;
Console.Write(findExtra(arr1, arr2, n));
}
}
|
PHP
<?php
function findExtra($arr1,
$arr2, $n)
{
for ($i = 0; $i < $n; $i++)
if ($arr1[$i] != $arr2[$i])
return $i;
return $n;
}
$arr1 = array (2, 4, 6, 8,
10, 12, 13);
$arr2 = array(2, 4, 6,
8, 10, 12);
$n = sizeof($arr2);
echo findExtra($arr1, $arr2, $n);
?>
|
Javascript
<script>
function findExtra(arr1, arr2, n)
{
for (let i = 0; i < n; i++)
if (arr1[i] != arr2[i])
return i;
return n;
}
let arr1 = [2, 4, 6, 8,
10, 12, 13];
let arr2 = [2, 4, 6,
8, 10, 12];
let n = arr2.length;
document.write(findExtra(arr1, arr2, n));
</script>
|
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:
- Create three variables, low = 0, high = n-1, mid, ans = n
- Run a loop until low is less than or equal to high, i.e till our search range is less than zero.
- 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
- Else update the search to the first half of the search range, i.e high = mid – 1, and update the answer to the current index, ans = mid
- Print the index.
Implementation:
C++
#include <iostream>
using namespace std;
int findExtra(int arr1[],
int arr2[], int n)
{
int index = n;
int left = 0, right = n - 1;
while (left <= right)
{
int mid = (left + right) / 2;
if (arr2[mid] == arr1[mid])
left = mid + 1;
else
{
index = mid;
right = mid - 1;
}
}
return index;
}
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]);
cout << findExtra(arr1, arr2, n);
return 0;
}
|
Java
class GFG
{
static int findExtra(int arr1[],
int arr2[], int n)
{
int index = n;
int left = 0, right = n - 1;
while (left <= right)
{
int mid = (left+right) / 2;
if (arr2[mid] == arr1[mid])
left = mid + 1;
else
{
index = mid;
right = mid - 1;
}
}
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;
System.out.println(findExtra(arr1, arr2, n));
}
}
|
Python3
def findExtra(arr1, arr2, n) :
index = n
left = 0
right = n - 1
while (left <= right) :
mid = (int)((left + right) / 2)
if (arr2[mid] == arr1[mid]) :
left = mid + 1
else :
index = mid
right = mid - 1
return index
arr1 = [2, 4, 6, 8, 10, 12, 13]
arr2 = [2, 4, 6, 8, 10, 12]
n = len(arr2)
print(findExtra(arr1, arr2, n))
|
C#
using System;
class GFG {
static int findExtra(int []arr1,
int []arr2,
int n)
{
int index = n;
int left = 0, right = n - 1;
while (left <= right)
{
int mid = (left+right) / 2;
if (arr2[mid] == arr1[mid])
left = mid + 1;
else
{
index = mid;
right = mid - 1;
}
}
return index;
}
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;
Console.Write(findExtra(arr1, arr2, n));
}
}
|
PHP
<?php
function findExtra($arr1, $arr2, $n)
{
$index = $n;
$left = 0; $right = $n - 1;
while ($left <= $right)
{
$mid = ($left+$right) / 2;
if ($arr2[$mid] == $arr1[$mid])
$left = $mid + 1;
else
{
$index = $mid;
$right = $mid - 1;
}
}
return $index;
}
{
$arr1 = array(2, 4, 6, 8,
10, 12, 13);
$arr2 = array(2, 4, 6,
8, 10, 12);
$n = sizeof($arr2) / sizeof($arr2[0]);
echo findExtra($arr1, $arr2, $n);
return 0;
}
?>
|
Javascript
<script>
function findExtra( arr1, arr2, n)
{
let index = n;
let left = 0, right = n - 1;
while (left <= right)
{
let mid = Math.floor((left + right) / 2);
if (arr2[mid] == arr1[mid])
left = mid + 1;
else
{
index = mid;
right = mid - 1;
}
}
return index;
}
let arr1 = [2, 4, 6, 8, 10, 12, 13];
let arr2 = [2, 4, 6, 8, 10, 12];
let n = arr2.length;
document.write(findExtra(arr1, arr2, n));
</script>
|
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:
- Create a function to calculate the sum of two arrays.
- Find the absolute difference between the sum of two arrays (value).
- Traverse the larger array from start too end
- If the element at any index is equal to value, then print the index and break the loop.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
int sum(int arr[], int n)
{
int summ = 0;
for (int i = 0; i < n; i++)
{
summ += arr[i];
}
return summ;
}
int indexOf(int arr[], int element, int n)
{
for (int i = 0; i < n; i++)
{
if (arr[i] == element)
{
return i;
}
}
return -1;
}
int find_extra_element_index(int arrA[],
int arrB[],
int n, int m)
{
int extra_element = sum(arrA, n) -
sum(arrB, m);
return indexOf(arrA, extra_element, n);
}
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);
}
|
Java
class GFG
{
static int find_extra_element_index(int[] arrA,
int[] arrB)
{
int extra_element = sum(arrA) - sum(arrB);
return indexOf(arrA, extra_element);
}
static int sum(int[] arr)
{
int sum = 0;
for (int i = 0; i < arr.length; i++)
{
sum += arr[i];
}
return sum;
}
static int indexOf(int[] arr, int element)
{
for (int i = 0; i < arr.length; i++)
{
if (arr[i] == element)
{
return i;
}
}
return -1;
}
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));
}
}
|
Python3
def find_extra_element_index(arrA, arrB):
extra_element = sum(arrA) - sum(arrB)
return arrA.index(extra_element)
arrA = [2, 4, 6, 8, 10, 12, 13]
arrB = [2, 4, 6, 8, 10, 12]
print(find_extra_element_index(arrA,arrB))
|
C#
using System;
class GFG
{
static int find_extra_element_index(int[] arrA,
int[] arrB)
{
int extra_element = sum(arrA) - sum(arrB);
return indexOf(arrA, extra_element);
}
static int sum(int[] arr)
{
int sum = 0;
for (int i = 0; i < arr.Length; i++)
{
sum += arr[i];
}
return sum;
}
static int indexOf(int[] arr, int element)
{
for (int i = 0; i < arr.Length; i++)
{
if (arr[i] == element)
{
return i;
}
}
return -1;
}
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));
}
}
|
Javascript
<script>
function find_extra_element_index(arrA, arrB)
{
let extra_element = sum(arrA) - sum(arrB);
return indexOf(arrA, extra_element);
}
function sum(arr)
{
let sum = 0;
for (let i = 0; i < arr.length; i++)
{
sum += arr[i];
}
return sum;
}
function indexOf(arr, element)
{
for (let i = 0; i < arr.length; i++)
{
if (arr[i] == element)
{
return i;
}
}
return -1;
}
let arrA = [2, 4, 6, 8, 10, 12, 13];
let arrB = [2, 4, 6, 8, 10, 12];
document.write(find_extra_element_index(arrA, arrB));
</script>
|
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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