How do you solve Search Insert Position on LeetCode?

Search Insert Position (LeetCode #35) can be solved using Brute Force or Optimal Solution (Binary Search). The optimal approach is Optimal Solution (Binary Search) with O(log n) time complexity and O(1) space complexity. Key insight: The sorted nature of the array allows for the use of binary search, which is a powerful technique for searching in sorted data.

What is the brute force solution for Search Insert Position?

The brute force approach for Search Insert Position is Brute Force, with O(n) time complexity and O(1) space complexity. The simplest way to solve this problem is to iterate through the array and check each element to see if it matches the target. If we don't find it, we can determine where it should be inserted by finding the first element that is greater than the target. The optimal Optimal Solution (Binary Search) approach improves this to O(log n).

What algorithm or data structure is used to solve Search Insert Position?

Search Insert Position uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Array.

What are the interview tips for Search Insert Position?

When you first see this problem, clarify whether the array is sorted and if the integers are distinct. Communicate your thought process clearly, explaining why you would choose binary search over a linear search. Mention edge cases such as an empty array or the target being less than the smallest element or greater than the largest element.

What are common mistakes when solving Search Insert Position?

Students often forget to handle the case where the target is greater than all elements, leading to incorrect index returns. Some students may confuse the conditions for moving left and right in binary search, which can lead to infinite loops or incorrect results.

#35

Search Insert Position

Easy

ArrayBinary SearchBinary SearchArray

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)

Since the array is sorted, we can use binary search to efficiently find the target or the insertion point. This approach reduces the search space by half with each iteration, leading to a logarithmic time complexity.

⚙️

Algorithm

6 steps

1Step 1: Initialize two pointers, left at 0 and right at the length of the array.
2Step 2: While left is less than right, calculate the mid index.
3Step 3: If the mid element equals the target, return mid.
4Step 4: If the mid element is less than the target, move the left pointer to mid + 1.
5Step 5: If the mid element is greater than the target, move the right pointer to mid.
6Step 6: After the loop, left will be the insertion point.

solution.py18 lines

1# Full working Python code with comments
2
3def search_insert(nums, target):
4    left, right = 0, len(nums)
5    while left < right:
6        mid = (left + right) // 2
7        if nums[mid] == target:
8            return mid  # Target found
9        elif nums[mid] < target:
10            left = mid + 1  # Move right
11        else:
12            right = mid  # Move left
13    return left  # Insertion point
14
15# Example usage
16print(search_insert([1, 3, 5, 6], 5))  # Output: 2
17print(search_insert([1, 3, 5, 6], 2))  # Output: 1
18print(search_insert([1, 3, 5, 6], 7))  # Output: 4

ℹ

Complexity note: This complexity is achieved because we are halving the search space with each iteration, 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 searching in sorted data.
2Understanding the difference between searching for an element and finding an insertion point is crucial in array problems.

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