Search Insert Position Leetcode Problem 35 [Python Solution]

In the world of programming, there are problems that seem simple at first glance but reveal hidden complexity when examined more closely.

The Search Insert Position problem is one such challenge.

In this blog post, we’ll explore the problem, break it down step by step, and present an efficient Python solution.

Whether you’re a seasoned coder or a beginner, this guide has something for everyone.

So, let’s dive into the world of binary search and algorithm optimization.

Problem Overview

The Search Insert Position problem can be summarized as follows: Given a sorted array of distinct integers and a target value, you need to return the index if the target is found in the array.

If the target is not present, return the index where it would be if it were inserted in order.

Let’s illustrate this with an example:

Example 1:

Input: nums = [1,3,5,6], target = 5
Output: 2

In this example, the target value, which is 5, is found at index 2 in the given sorted array.

If it were not present, it would be inserted at index 2 to maintain the sorted order.

But why should we care about this problem?

Well, it’s a classic binary search problem, and it’s not only about solving it but also about solving it efficiently.

The problem statement specifies that the algorithm must have a runtime complexity of O(log n).

Understanding Constraints

Before we delve into the solution, let’s understand the constraints of the problem:

• The length of the nums array can be as large as 10^4.
• The integers in the nums array can range from -10^4 to 10^4.
• The nums array contains distinct values sorted in ascending order.
• The target value also falls within the same range as the values in nums.

These constraints are crucial as they influence our approach to solving the problem efficiently.

Search Insert Position LeetCode Problem Solution

Now, let’s get into the heart of the problem and discuss the solution.

The most efficient way to solve this problem is by applying binary search.

Binary search is a divide-and-conquer algorithm that works exceptionally well when the input is sorted.

Efficient Python Code Solution:

def searchInsert(self, nums, target):
    # `O(log n)` time complexity and `O(1)` space complexity

    left, right = 0, len(nums)
    while left < right:
        mid = left + (right - left) // 2
        if target > nums[mid]:
            left = mid + 1
        else:
            right = mid
    return left

This elegant Python code snippet showcases a binary search approach that meets the O(log n) runtime complexity requirement.

Let’s break it down step by step.

1. Initialize left and right pointers to 0 and the length of the nums array, respectively.

These pointers will define the search range.

1. Enter a while loop that continues as long as the left pointer is less than the right pointer.

This loop will iterate until we’ve narrowed down our search to a single element.

1. Compute the mid value, which represents the middle index of the current search range.

We use the left + (right - left) // 2 formula to ensure integer division and prevent potential integer overflow issues.

1. Check if the target value is greater than the element at index mid in the nums array.

If it is, we update the left pointer to mid + 1, effectively shifting the search range to the right half of the current range.

1. If the target is not greater than the element at index mid, we update the right pointer to mid.
2. This means we shift the search range to the left half.
3. Repeat steps 3-5 until the left and right pointers converge, signifying that we’ve found the insertion position or the target element itself.
4. Finally, return the left pointer as the result.

This left pointer represents the index where the target value would be inserted while maintaining the sorted order.

Time and Space Complexity

It’s essential to analyze the time and space complexity of our solution to understand its efficiency.

Time Complexity: The binary search algorithm has a time complexity of O(log n), where n is the number of elements in the input array nums.

This is significantly faster than a linear search (O(n)) and makes the algorithm highly efficient, especially when dealing with large arrays.

Space Complexity: Our solution uses O(1) additional space, meaning the memory required for the algorithm remains constant, regardless of the input size.

This is optimal in terms of space efficiency.

Reasoning Behind Our Approach

Let’s take a moment to understand why the binary search approach works so well for this problem.

The key insight is that binary search allows us to eliminate half of the remaining possibilities with each iteration, leading to a significant reduction in the number of checks required.

Binary search operates by continuously dividing the search range in half, which means it has a logarithmic time complexity.

When we compare the target to the middle element, we determine whether to search in the left or right half of the current range, effectively halving the search space.

This process repeats until the search range contains only one element.

In this way, binary search takes advantage of the sorted nature of the input array.

It’s like flipping through a dictionary to find a word “Auditorical” — which you probably not find, — instead of reading page by page, you open the book in the middle and decide whether to continue left or right, reducing your search time drastically.

Edge Cases a Valid Solution Must Consider

While our efficient solution handles the majority of cases, it’s essential to consider edge cases to ensure the algorithm’s robustness.

Here are some edge cases to keep in mind:

1. Empty Input Array: What if the input array nums is empty?

Our binary search approach still works correctly in this case.

Since left and right initially point to the same index, the algorithm will terminate, and left will be returned as 0, indicating that the target would be inserted at the beginning of the array.

1. Target Outside Array Bounds: If the target is less than the minimum value in the array or greater than the maximum value in the array, the algorithm will correctly determine the insertion position.

It will handle these scenarios by effectively shifting the left and right pointers to the appropriate boundaries.

1. Duplicate Values in the Array: The problem statement specifies that the input array nums contains distinct values.

If this condition is not met and there are duplicate values, the algorithm can still determine the correct insertion position.

It will place the left pointer at the first occurrence of the target value.

By considering these edge cases, we ensure that our solution is robust and handles various scenarios effectively.

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Conclusion

In this blog post, we explored the Search Insert Position problem, which is a classic binary search challenge on LeetCode.

We discussed the problem statement, constraints, and an efficient Python solution with a time complexity of O(log n).

By understanding the binary search approach, we gained insights into how to optimize the algorithm for sorted arrays.

Binary search is a powerful technique

in algorithm design, and it showcases how thinking in terms of algorithms and data structures can lead to efficient solutions.

It’s a fundamental skill for any programmer, and mastering it can help you tackle a wide range of problems efficiently.

As you continue your coding journey, remember that practice is key to becoming a better problem solver.

Feel free to explore additional LeetCode problems, experiment with different data structures and algorithms, and always seek to optimize your code.

If you have any questions, suggestions, or insights, please don’t hesitate to share them in the comments.

Coding is a collaborative endeavor, and we can all learn from each other’s experiences and perspectives.

Happy coding!

Question Link: Search Insert Position LeetCode Problem

Now, it’s your turn to practice and explore more coding challenges.

Keep coding, keep learning, and keep pushing your problem-solving skills to new heights.