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: 8Thanks 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.
- 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.
Python3
def findFirstMissing(array, start, end):
if (start > end):
return end + 1
if (start != array[start]):
return start;
mid = int((start + end) / 2)
if (array[mid] == mid):
return findFirstMissing(array,
mid+1, end)
return findFirstMissing(array,
start, mid)
arr = [0, 1, 2, 3, 4, 5, 6, 7, 10]
n = len(arr)
print("Smallest missing element is",
findFirstMissing(arr, 0, n-1))
|
OutputSmallest 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:
- If the first element of the array is not 0, then the smallest missing number is 0.
- If the last elements of the array is N-1, then the smallest missing number is N.
- Otherwise, find the middle element from the first and last index and check if the middle element is equal to the desired element. i.e. first + middle_index.
- If the middle element is the desired element, then the smallest missing element is in the right search space of the middle.
- Otherwise, the smallest missing number is in the left search space of the middle.
Below is the implementation of the above approach:
Python3
def findSmallestMissinginSortedArray(arr):
if (arr[0] != 0):
return 0
if (arr[-1] == len(arr) - 1):
return len(arr)
first = arr[0]
return findFirstMissing(arr, 0,
len(arr) - 1, first)
def findFirstMissing(arr, start, end, first):
if (start < end):
mid = int((start + end) / 2)
if (arr[mid] != mid + first):
return findFirstMissing(arr, start,
mid, first)
else:
return findFirstMissing(arr, mid + 1,
end, first)
return start + first
arr = [ 0, 1, 2, 3, 4, 5, 7 ]
n = len(arr)
print("First Missing element is :",
findSmallestMissinginSortedArray(arr))
|
OutputFirst 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!