What is the time complexity of Search Insert Position?

The optimal solution has a time complexity of O(log n).

What data structures are used to solve Search Insert Position?

This problem uses Array, Binary Search. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Search Insert Position?

The `O(log n)` runtime constraint, coupled with the sorted input array, makes binary search the only viable algorithm. A linear scan would fall short of this performance target. The elegance of this binary search implementation lies in how `l` (the left pointer) naturally converges to the correct insertion index when the target is not found. When the `while l <= r` loop finally breaks, `l` will invariably point to the first element that is greater than or equal to the `target`. If no such element exists (i.e., `target` is greater than all elements), `l` will be `len(nums)`, which is the correct insertion index at the end of the array.

Which companies ask LeetCode 35. Search Insert Position in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Cognizant and 6 more companies.

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Search Insert Position - LeetCode 35 Solution

Q: How to solve Search Insert Position?

The solution employs a classic binary search algorithm, but it's designed to elegantly handle both finding the target and identifying its insertion point. We initialize two pointers, `l` (left) at `0` and `r` (right) at `len(nums) - 1`, defining our search space. Inside a `while l <= r` loop, we calculate the `mid` index. If `nums[mid]` matches our `target`, we've found it and return `mid`. If `nums[mid]` is less than `target`, it means the `target` must lie to the right of `mid` (if it exists at all), so we update `l = mid + 1`. Conversely, if `nums[mid]` is greater than `target`, the `target` must be to the left of `mid` (or `mid` itself is a potential insertion point), so we update `r = mid - 1`. The magic happens when the loop terminates. At this point, `l` will always point to the correct insertion index. If the `target` was found, we would have returned early. If not, `l` has pushed past `r`, and it will be precisely the index where the `target` should be inserted to maintain order. Think of `l` as tracking the first element that is greater than or equal to `target` throughout the search; when the loop ends, this `l` is the ultimate candidate.

Q: Which companies ask LeetCode 35. Search Insert Position in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Cognizant and 6 more companies.

Search Insert Position - Complete Solution Guide

Search Insert Position is LeetCode problem 35, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You must write an algorithm with O(log n) runtime complexity. Example 1: Input: nums = [1,3,5,6], target = 5 Output: 2 Example 2: Input: nums = [1,3,5,6], target = 2 Output: 1 Example 3: Input: nums = [1,3,5,6], target = 7 Output: 4 Constraints: 1 <= nums.length <= 10 4 -10 4 <= nums[i] <= 10 4 nums contains distinct value

Detailed Explanation

This problem asks us to locate a specific `target` value within a `sorted` array of `distinct` integers. The twist isn't just finding it; if the `target` isn't present, we need to determine the index where it *would* be inserted to maintain the array's sorted order. For example, if we have `nums = [1,3,5,6]` and `target = 5`, the output is `2` because `5` is at index `2`. However, if `target = 2`, the output should be `1` since `2` would fit perfectly between `1` (at index `0`) and `3` (at index `1`). Similarly, if `target = 7`, which is larger than all elements, it should be inserted at the very end, resulting in an output of `4` (which is `len(nums)`). The crucial constraint is the `O(log n)` runtime complexity. This immediately signals that a linear scan, while simple, is not sufficient for larger inputs. The sorted nature of the input array combined with the logarithmic time requirement points directly towards a binary search approach, but we need to adapt it slightly to handle the 'insertion point' scenario gracefully.

Solution Approach

The solution employs a classic binary search algorithm, but it's designed to elegantly handle both finding the target and identifying its insertion point. We initialize two pointers, `l` (left) at `0` and `r` (right) at `len(nums) - 1`, defining our search space. Inside a `while l <= r` loop, we calculate the `mid` index. If `nums[mid]` matches our `target`, we've found it and return `mid`. If `nums[mid]` is less than `target`, it means the `target` must lie to the right of `mid` (if it exists at all), so we update `l = mid + 1`. Conversely, if `nums[mid]` is greater than `target`, the `target` must be to the left of `mid` (or `mid` itself is a potential insertion point), so we update `r = mid - 1`. The magic happens when the loop terminates. At this point, `l` will always point to the correct insertion index. If the `target` was found, we would have returned early. If not, `l` has pushed past `r`, and it will be precisely the index where the `target` should be inserted to maintain order. Think of `l` as tracking the first element that is greater than or equal to `target` throughout the search; when the loop ends, this `l` is the ultimate candidate.

Step-by-Step Algorithm

Step 1: Initialize `left` to 0 and `right` to `nums.length - 1`.
Step 2: While `left` is less than or equal to `right`, calculate the middle index `mid = left + (right - left) / 2`. This prevents potential integer overflow.
Step 3: If `nums[mid]` equals the `target`, return `mid`.
Step 4: If `nums[mid]` is less than `target`, update `left` to `mid + 1` (search in the right half).
Step 5: Otherwise, update `right` to `mid - 1` (search in the left half).
Step 6: After the loop, return `left`. This index represents the insertion point if the target is not found.

Key Insights

The `O(log n)` runtime constraint, coupled with the sorted input array, makes binary search the only viable algorithm. A linear scan would fall short of this performance target.
The elegance of this binary search implementation lies in how `l` (the left pointer) naturally converges to the correct insertion index when the target is not found. When the `while l <= r` loop finally breaks, `l` will invariably point to the first element that is greater than or equal to the `target`. If no such element exists (i.e., `target` is greater than all elements), `l` will be `len(nums)`, which is the correct insertion index at the end of the array.
The `mid` calculation `(l + r) // 2` and the update rules `l = mid + 1` and `r = mid - 1` are robust. They ensure that the search space is correctly narrowed down, and crucially, `l` always represents the *lower bound* of where `target` *could* be inserted, eventually settling on the precise spot whether `target` is present or not. This avoids the need for a separate post-loop check to determine the insertion point.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Cognizant, Instacart, Meta, Microsoft, Uber, Yahoo, Zoho.

Search Insert Position | LeetCode 35

Easy

Array Binary Search

Time: O(log n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int searchInsert(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (nums[mid] == target) {
                return mid;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return left;
    }
}

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Problem Info

Description

Problem 35: Search Insert Position - Detailed explanation and solution approaches.

Topics

Array Binary Search

Companies

AdobeAmazonAppleBloombergCognizantInstacartMetaMicrosoft

Stats

Difficulty:Easy

Languages:4

Companies:11

Problem #:35

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize `left` to 0 and `right` to `nums.length - 1`.
2
Step 2: While `left` is less than or equal to `right`, calculate the middle index `mid = left + (right - left) / 2`. This prevents potential integer overflow.
3
Step 3: If `nums[mid]` equals the `target`, return `mid`.
4
Step 4: If `nums[mid]` is less than `target`, update `left` to `mid + 1` (search in the right half).
5
Step 5: Otherwise, update `right` to `mid - 1` (search in the left half).
6
Step 6: After the loop, return `left`. This index represents the insertion point if the target is not found.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

Binary search halves the search space with each iteration, resulting in logarithmic time complexity. The algorithm uses a constant amount of extra space, regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Linear Search

Iterate through the array linearly, comparing each element to the target. If found, return the index. If not found, return the index where it should be inserted.

Time: O(n)Space: O(1)

Trade-offs: Simpler to implement but less efficient for large arrays. Prefer binary search for sorted arrays due to its logarithmic time complexity.

✨

Key Insights

1The `O(log n)` runtime constraint, coupled with the sorted input array, makes binary search the only viable algorithm. A linear scan would fall short of this performance target.
2The elegance of this binary search implementation lies in how `l` (the left pointer) naturally converges to the correct insertion index when the target is not found. When the `while l <= r` loop finally breaks, `l` will invariably point to the first element that is greater than or equal to the `target`. If no such element exists (i.e., `target` is greater than all elements), `l` will be `len(nums)`, which is the correct insertion index at the end of the array.
3The `mid` calculation `(l + r) // 2` and the update rules `l = mid + 1` and `r = mid - 1` are robust. They ensure that the search space is correctly narrowed down, and crucially, `l` always represents the *lower bound* of where `target` *could* be inserted, eventually settling on the precise spot whether `target` is present or not. This avoids the need for a separate post-loop check to determine the insertion point.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: Empty array. The solution correctly handles this by returning 0, the index where the element would be inserted.
Edge case 2: Target smaller than all elements. The solution correctly returns 0, the beginning of the array.
Edge case 3: Target larger than all elements. The solution correctly returns the length of the array, indicating insertion at the end.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect handling of the `left` and `right` pointers in the binary search loop, leading to infinite loops or incorrect results.

✗Common mistake 2: Not considering the case where the target is not found and failing to correctly determine the insertion point.

✗Common mistake 3: Using `(left + right) / 2` instead of `left + (right - left) / 2` for calculating `mid`, which can lead to integer overflow for very large arrays.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Finding a specific book in a library catalog (assuming it's sorted alphabetically).

🏦Real-world scenario 2: Determining the appropriate position of a new data record in a sorted database table to maintain sorted order.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This is one of the first problems I solved when learning Array. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Search Insert Position"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether binary search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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