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Python3 Program to Find the smallest missing number

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

Thanks to Ravichandra for suggesting following two methods.

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) 
Thanks to yasein and Jams for suggesting this method. 
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.




# Python3 program to find the smallest
# elements missing in a sorted array.
 
def findFirstMissing(array, start, end):
 
    if (start > end):
        return end + 1
 
    if (start != array[start]):
        return start;
 
    mid = int((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 program to test above function
arr = [0, 1, 2, 3, 4, 5, 6, 7, 10]
n = len(arr)
print("Smallest missing element is",
      findFirstMissing(arr, 0, n-1))
 
# This code is contributed by Smitha Dinesh Semwal

Output
Smallest missing element is 8

Note: This method doesn’t work if there are duplicate elements in the array.
Time Complexity: O(Logn)

Space Complexity: O(1)
The space complexity of this algorithm is O(1) because it is using constant space to search the smallest missing element.

Another Method: The idea is to use Recursive Binary Search to find the smallest missing number. Below is the illustration with the help of steps:

Below is the implementation of the above approach:




# Python3 program for above approach
 
# Function to find Smallest
# Missing in Sorted Array
def findSmallestMissinginSortedArray(arr):
     
    # Check if 0 is missing
    # in the array
    if (arr[0] != 0):
        return 0
     
    # Check is all numbers 0 to n - 1
    # are present in array
    if (arr[-1] == len(arr) - 1):
        return len(arr)
     
    first = arr[0]
     
    return findFirstMissing(arr, 0,
            len(arr) - 1, first)
 
# Function to find missing element
def findFirstMissing(arr, start, end, first):
     
    if (start < end):
        mid = int((start + end) / 2)
         
        # Index matches with value
        # at that index, means missing
        # element cannot be upto that point
        if (arr[mid] != mid + first):
            return findFirstMissing(arr, start,
                                    mid, first)
        else:
            return findFirstMissing(arr, mid + 1,
                                    end, first)
     
    return start + first
 
# Driver code
arr = [ 0, 1, 2, 3, 4, 5, 7 ]
n = len(arr)
 
# Function Call
print("First Missing element is :",
    findSmallestMissinginSortedArray(arr))
 
# This code is contributed by rag2127

Output
First Missing element is : 6

Time Complexity: O(Log n) 
Auxiliary Space: O(log n) where log(n) is the size of the recursive call stack

Please refer complete article on Find the smallest missing number for more details!


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