What is the time complexity of Remove Duplicates from Sorted Array?

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

What data structures are used to solve Remove Duplicates from Sorted Array?

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

What are the key insights for solving Remove Duplicates from Sorted Array?

**Leveraging Sorted Property:** The entire efficiency of this solution hinges on the input array `nums` being sorted. This allows us to detect duplicates by simply comparing `nums[i]` with `nums[i-1]`. If they differ, `nums[i]` *must* be a new unique element because all prior elements up to `nums[i-1]` have already been processed for uniqueness. **Distinct Roles of Two Pointers:** The solution cleverly employs two pointers with different responsibilities: an explicit 'write' pointer (`k`) that points to the next available slot for a unique element and implicitly tracks the count of unique elements, and an implicit 'read' pointer (`i`) that scans through the entire array. The 'write' pointer only advances when a truly new unique element is discovered, effectively compacting the array in-place.

Which companies ask LeetCode 26. Remove Duplicates from Sorted Array in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Bloomberg and 19 more companies.

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Remove Duplicates from Sorted Array - LeetCode 26 Solution

Remove Duplicates from Sorted Array - Complete Solution Guide

Remove Duplicates from Sorted Array is LeetCode problem 26, 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 an integer array nums sorted in non-decreasing order , remove the duplicates in-place such that each unique element appears only once . The relative order of the elements should be kept the same . Then return the number of unique elements in nums . Consider the number of unique elements of nums to be k , to get accepted, you need to do the following things: Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initia

Detailed Explanation

This problem asks us to refine an already sorted array, `nums`, by removing duplicate elements directly within the array itself. The core constraint is 'in-place' modification: we're not allowed to create a new array. Instead, we must rearrange `nums` such that its initial `k` elements contain only the unique values, maintaining their original relative order. What happens to the elements beyond the `k`-th position doesn't matter for the problem's success criteria, though they will typically be duplicates that were overwritten. For example, if `nums = [0,0,1,1,1,2,2,3,3,4]`, the expected outcome is that `nums` is modified to `[0,1,2,3,4,_,_,_,_,_]` (where `_` signifies irrelevant values) and the function returns `k=5`. The challenge lies in efficiently identifying and relocating unique elements without using additional space, leveraging the fact that the array is already sorted in non-decreasing order. This sorted property is a huge hint, simplifying what would otherwise be a much more complex task. The trickiness comes from the simultaneous requirements of in-place modification, preserving relative order, and returning only the count `k`. We're effectively compacting the array, pushing all unique values to the front and reporting the new effective 'length' of the array that contains only unique values. The remaining space at the end of the array is essentially 'dead space' after the operation.

Solution Approach

The most intuitive and efficient approach for this problem, given the sorted input, is to use a two-pointer technique. We can think of these as a 'write' pointer and a 'read' pointer. Let `k` be our 'write' pointer (or more accurately, the index where the *next* unique element should be placed). It also conveniently represents the count of unique elements found so far. We initialize `k` to 1 because the first element `nums[0]` is always unique in any non-empty array. The 'read' pointer, implicitly `i` in the loop `for i in range(1, len(nums))`, iterates through the array starting from the second element. For each element `nums[i]`, we compare it with the element immediately preceding it, `nums[i-1]`. Since the array is sorted, if `nums[i]` is *different* from `nums[i-1]`, it means we've encountered a new unique number that hasn't been added to our unique prefix yet. In this scenario, we take this newly found unique element `nums[i]` and place it at the `nums[k]` position, then increment `k`. If `nums[i]` is the same as `nums[i-1]`, it's a duplicate, and we simply skip it, letting the 'read' pointer `i` advance while the 'write' pointer `k` stays put. This process continues until the 'read' pointer has traversed the entire array. By the end, `k` will hold the total count of unique elements, and the first `k` positions of `nums` will contain these unique elements in their original relative order. This approach works elegantly because the sorted nature guarantees that duplicates are always adjacent, and the two pointers effectively compress the unique elements to the array's front in a single pass.

Step-by-Step Algorithm

Step 1: Handle the empty array case. If the input array is empty, return 0.
Step 2: Initialize `k` to 1. This represents the index of the second unique element (the first unique element is already at index 0).
Step 3: Iterate through the array starting from the second element (index 1).
Step 4: For each element, compare it with the previous element. If they are different, it's a unique element.
Step 5: If it's a unique element, place it at index `k` in the array (`nums[k] = nums[i]`) and increment `k`.
Step 6: After the loop, `k` holds the number of unique elements, which is returned as the result.

Key Insights

**Leveraging Sorted Property:** The entire efficiency of this solution hinges on the input array `nums` being sorted. This allows us to detect duplicates by simply comparing `nums[i]` with `nums[i-1]`. If they differ, `nums[i]` *must* be a new unique element because all prior elements up to `nums[i-1]` have already been processed for uniqueness.
**Distinct Roles of Two Pointers:** The solution cleverly employs two pointers with different responsibilities: an explicit 'write' pointer (`k`) that points to the next available slot for a unique element and implicitly tracks the count of unique elements, and an implicit 'read' pointer (`i`) that scans through the entire array. The 'write' pointer only advances when a truly new unique element is discovered, effectively compacting the array in-place.
**Problem's 'In-place' Definition:** Understanding that 'in-place' removal doesn't mean truly shrinking the array or invalidating memory. It specifically means constructing a unique-element prefix within the *original* `nums` array and returning the length of this prefix (`k`). Elements beyond index `k-1` are inconsequential to the problem's success criteria, making the modification straightforward without complex memory management.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Two Pointers.

Companies

Asked at: Accenture, Adobe, Amazon, Apple, Bloomberg, Capgemini, Cisco, Cognizant, Flipkart, Infosys, Meta, Microsoft, Morgan Stanley, Nagarro, Oracle, Qualcomm, SAP, Siemens, Uber, Wipro, Yahoo, ZScaler, Zoho, tcs.

Remove Duplicates from Sorted Array | LeetCode 26

Easy

Array Two Pointers

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

Solution Code

Available in:

class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }
        int k = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[i - 1]) {
                nums[k] = nums[i];
                k++;
            }
        }
        return k;
    }
}

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Problem Info

Description

Problem 26: Remove Duplicates from Sorted Array - Detailed explanation and solution approaches.

Topics

Array Two Pointers

Companies

AccentureAdobeAmazonAppleBloombergCapgeminiCiscoCognizant

Stats

Difficulty:Easy

Languages:4

Companies:24

Problem #:26

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: Handle the empty array case. If the input array is empty, return 0.
2
Step 2: Initialize `k` to 1. This represents the index of the second unique element (the first unique element is already at index 0).
3
Step 3: Iterate through the array starting from the second element (index 1).
4
Step 4: For each element, compare it with the previous element. If they are different, it's a unique element.
5
Step 5: If it's a unique element, place it at index `k` in the array (`nums[k] = nums[i]`) and increment `k`.
6
Step 6: After the loop, `k` holds the number of unique elements, which is returned as the result.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The solution iterates through the input array once, resulting in a linear time complexity. It uses only a constant amount of extra space to store the pointers `i` and `k`, making the space complexity constant.

📖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

HashSet Approach

Use a HashSet to store unique elements encountered so far. Iterate through the array, adding elements to the HashSet only if they are not already present. Then, create a new array containing only the elements from the HashSet.

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

Trade-offs: While this approach also has O(n) time complexity, it uses O(n) extra space for the HashSet, making it less space-efficient than the two-pointer approach. The two-pointer method is preferred for its in-place nature and constant space usage.

✨

Key Insights

1**Leveraging Sorted Property:** The entire efficiency of this solution hinges on the input array `nums` being sorted. This allows us to detect duplicates by simply comparing `nums[i]` with `nums[i-1]`. If they differ, `nums[i]` *must* be a new unique element because all prior elements up to `nums[i-1]` have already been processed for uniqueness.
2**Distinct Roles of Two Pointers:** The solution cleverly employs two pointers with different responsibilities: an explicit 'write' pointer (`k`) that points to the next available slot for a unique element and implicitly tracks the count of unique elements, and an implicit 'read' pointer (`i`) that scans through the entire array. The 'write' pointer only advances when a truly new unique element is discovered, effectively compacting the array in-place.
3**Problem's 'In-place' Definition:** Understanding that 'in-place' removal doesn't mean truly shrinking the array or invalidating memory. It specifically means constructing a unique-element prefix within the *original* `nums` array and returning the length of this prefix (`k`). Elements beyond index `k-1` are inconsequential to the problem's success criteria, making the modification straightforward without complex memory management.

📚

Key Terms Explained

For Students

Array

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

Two Pointers

Using two indices to traverse data from different positions simultaneously.

Edge Cases

Edge case 1: Empty array. The code correctly handles this by returning 0.
Edge case 2: Array with all duplicates. The code correctly identifies and returns 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of an empty input array.

✗Common mistake 2: Incorrectly managing the indices during in-place modification, potentially leading to incorrect results or array out-of-bounds errors.

✗Common mistake 3: Not understanding the 'in-place' requirement and creating a new array instead of modifying the original.

✅

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: Data deduplication in databases. Removing duplicate entries from a sorted dataset to improve database efficiency and reduce storage space.

🏦Real-world scenario 2: Log file processing. Removing duplicate log entries to reduce the size of log files and simplify analysis.

💡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.

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: June 27, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

When is the two-pointer technique most effective?

Two pointers work best on sorted arrays or when searching for pairs/triplets that satisfy a condition. Common patterns include: moving pointers inward from both ends (like in two-sum variants), fast/slow pointers for cycle detection, or left/right pointers for partitioning.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Remove Duplicates from Sorted Array"?

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 two pointers techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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