What is the time complexity of Find First and Last Position of Element in Sorted Array?

The optimal solution for Find First and Last Position of Element in Sorted Array runs in O(log n) time and uses O(1) space. This complexity is achieved by halving the search space with each iteration of binary search, making it much faster than the brute force approach.

What is the brute force solution for Find First and Last Position of Element in Sorted Array?

The brute force approach for Find First and Last Position of Element in Sorted Array is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves scanning through the entire array to find the first and last occurrences of the target. This is straightforward but inefficient for large arrays. The optimal Optimal Solution (Binary Search) approach improves this to O(log n).

What algorithm or data structure is used to solve Find First and Last Position of Element in Sorted Array?

Find First and Last Position of Element in Sorted Array uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Array, Two Pointers.

What are the interview tips for Find First and Last Position of Element in Sorted Array?

When you first see this problem, mention that you recognize the need for an efficient search due to the sorted nature of the array. Communicate your thought process clearly, explaining why binary search is suitable and how you will adapt it to find both indices. Mention edge cases such as an empty array or a target that doesn't exist in the array to show your awareness of potential pitfalls.

What are common mistakes when solving Find First and Last Position of Element in Sorted Array?

Students often forget to check if the target exists after finding the first and last indices, leading to incorrect results. Some may not handle edge cases like empty arrays or targets that are less than the smallest or greater than the largest element.

#34

Find First and Last Position of Element in Sorted Array

Medium

ArrayBinary SearchBinary SearchArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Binary Search)★
Time	O(n)	O(log n)
Space	O(1)	O(1)

💡

Intuition

Time O(log n)Space O(1)

Using binary search allows us to efficiently find the first and last positions of the target in O(log n) time. This is possible because the array is sorted, which allows us to eliminate half of the search space at each step.

⚙️

Algorithm

3 steps

1Step 1: Define a helper function to find the first position of the target using binary search.
2Step 2: Define another helper function to find the last position of the target using binary search.
3Step 3: Call both helper functions and return their results as an array.

solution.py31 lines

1# Full working Python code with comments
2
3def searchRange(nums, target):
4    def findFirst(nums, target):
5        left, right = 0, len(nums) - 1
6        while left <= right:
7            mid = left + (right - left) // 2
8            if nums[mid] < target:
9                left = mid + 1
10            else:
11                right = mid - 1
12        return left
13
14    def findLast(nums, target):
15        left, right = 0, len(nums) - 1
16        while left <= right:
17            mid = left + (right - left) // 2
18            if nums[mid] <= target:
19                left = mid + 1
20            else:
21                right = mid - 1
22        return right
23
24    first = findFirst(nums, target)
25    last = findLast(nums, target)
26    if first <= last:
27        return [first, last]
28    return [-1, -1]
29
30# Example usage
31print(searchRange([5,7,7,8,8,10], 8))  # Output: [3, 4]

ℹ

Complexity note: This complexity is achieved by halving the search space with each iteration of binary search, making it much faster than the brute force approach.

1The sorted nature of the array allows for the use of binary search, which is a powerful technique for problems involving sorted data.
2Finding the first and last positions can be efficiently handled by slightly modifying the binary search algorithm.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.