Find the smallest missing number

Software Development
#1

Hello Everyone,

Given a sorted array of n distinct integers where each integer is in the range from 0 to m-1 and m > n. Find the smallest number that is missing from the array.

Examples

Input: {0, 1, 2, 6, 9}, n = 5, m = 10 Output: 3 Input: {4, 5, 10, 11}, n = 4, m = 12 Output: 0 Input: {0, 1, 2, 3}, n = 4, m = 5 Output: 4 Input: {0, 1, 2, 3, 4, 5, 6, 7, 10}, n = 9, m = 11 Output: 8

Method 1 (Use Binary Search)
For i = 0 to m-1, do binary search for i in the array. If i is not present in the array then return i.
Time Complexity: O(m log n)

Method 2 ( [ Linear Search ])
If arr[0] is not 0, return 0. Otherwise traverse the input array starting from index 0, and for each pair of elements a[i] and a[i+1], find the difference between them. if the difference is greater than 1 then a[i]+1 is the missing number.
Time Complexity: O(n)

Method 3 (Use Modified Binary Search)
In the standard Binary Search process, the element to be searched is compared with the middle element and on the basis of comparison result, we decide whether to search is over or to go to left half or right half.
In this method, we modify the standard Binary Search algorithm to compare the middle element with its index and make decision on the basis of this comparison.

  • If the first element is not same as its index then return first index
  • Else get the middle index say mid
    • If arr[mid] greater than mid then the required element lies in left half.
    • Else the required element lies in right half.

// C++ program to find the smallest elements

// missing in a sorted array.

#include<bits/stdc++.h>

using namespace std;

int findFirstMissing( int array[],

int start, int end)

{

if (start > end)

return end + 1;

if (start != array[start])

return start;

int mid = (start + end) / 2;

// Left half has all elements

// from 0 to mid

if (array[mid] == mid)

return findFirstMissing(array,

mid+1, end);

return findFirstMissing(array, start, mid);

}

// Driver code

int main()

{

int arr[] = {0, 1, 2, 3, 4, 5, 6, 7, 10};

int n = sizeof (arr)/ sizeof (arr[0]);

cout << "Smallest missing element is " <<

findFirstMissing(arr, 0, n-1) << endl;

}

Output

Smallest missing element is 8

Time Complexity: O(Logn)