Why Remove Duplicates from Sorted Array Two Pointer in DSA Go?

The Big Idea

Discover how a simple two-pointer trick can save you hours of tedious duplicate checking!

The Scenario

Imagine you have a long list of names sorted alphabetically, but some names appear more than once. You want to write down only unique names on a new list without duplicates.

The Problem

If you try to check each name manually and compare it with all others, it will take a lot of time and you might miss some duplicates or write the same name twice by mistake.

The Solution

The two-pointer method helps you quickly scan the sorted list once, keeping track of unique names and skipping duplicates automatically, making the process fast and error-free.

Before vs After

✗ Before

for i := 0; i < len(names); i++ {
    for j := i + 1; j < len(names); j++ {
        if names[i] == names[j] {
            // remove duplicate
        }
    }
}

✓ After

uniqueIndex := 0
for currentIndex := 1; currentIndex < len(names); currentIndex++ {
    if names[currentIndex] != names[uniqueIndex] {
        uniqueIndex++
        names[uniqueIndex] = names[currentIndex]
    }
}

What It Enables

This method enables you to efficiently remove duplicates from sorted lists in just one pass, saving time and effort.

Real Life Example

When cleaning up a sorted contact list on your phone to keep only one entry per person, this method quickly removes repeated contacts.

Key Takeaways

Manual duplicate removal is slow and error-prone.

Two-pointer technique scans once and keeps unique items.

It works best on sorted lists for fast, simple cleanup.

Practice

1. What is the main purpose of using two pointers in the Remove Duplicates from Sorted Array problem?

easy

A. To efficiently track unique elements and overwrite duplicates in-place

B. To sort the array before removing duplicates

C. To create a new array with unique elements

D. To count the total number of duplicates in the array

Solution

1. Step 1: Understand the two-pointer approach

   One pointer tracks the position to place the next unique element, the other scans the array. This method avoids extra space by overwriting duplicates directly in the original array.

2. Final Answer:

   To efficiently track unique elements and overwrite duplicates in-place -> Option A

3. Quick Check:

   Two pointers help remove duplicates in-place [OK]

Hint: Two pointers move to overwrite duplicates without extra space [OK]

Common Mistakes:

• Thinking two pointers create a new array
• Assuming sorting is needed inside the method
• Confusing counting duplicates with removing them

2. Which of the following Go code snippets correctly initializes the two pointers for removing duplicates from a sorted array named nums?

easy

A. i, j := 0, 1

B. i, j := 1, 0

C. i, j := 0, 0

D. i, j := 1, 1

Solution

1. Step 1: Identify pointer roles and correct initialization

   Pointer i tracks last unique index starting at 0 (first element), pointer j scans from the next element at 1 to compare adjacent elements.

2. Final Answer:

   i, j := 0, 1 -> Option A

3. Quick Check:

   i=0, j=1 starts correct two-pointer scan [OK]

Hint: Start i at 0 and j at 1 to compare neighbors [OK]

Common Mistakes:

• Swapping i and j initialization
• Starting both pointers at 0
• Starting both pointers at 1

3. What will be the content of the array nums and the returned length after running the two-pointer duplicate removal on nums := []int{1, 1, 2, 3, 3, 4}?

medium

A. nums = [1, 2, 3, 3, 4, 4], length = 3

B. nums = [1, 1, 2, 3, 4, 4], length = 5

C. nums = [1, 2, 3, 4, 4, 4], length = 4

D. nums = [1, 2, 3, 4, 3, 4], length = 4

Solution

1. Step 1: Trace two-pointer process

   Start i=0, j=1; compare nums[j] with nums[i]. When different, increment i and set nums[i]=nums[j]. Unique elements placed at start: 1, 2, 3, 4; length = 4; leftover elements remain unchanged. Final array: [1, 2, 3, 4, 3, 4].

2. Final Answer:

   nums = [1, 2, 3, 4, 3, 4], length = 4 -> Option D

3. Quick Check:

   Unique elements count = 4, array start updated [OK]

Hint: Overwrite duplicates, keep unique at front, count unique [OK]

Common Mistakes:

• Assuming leftover elements are removed
• Counting duplicates as unique
• Misplacing unique elements

4. Identify the error in the following Go code snippet for removing duplicates from a sorted array nums:

func removeDuplicates(nums []int) int {
    i := 0
    for j := 1; j < len(nums); j++ {
        if nums[j] != nums[i] {
            i = j
            nums[i] = nums[j]
        }
    }
    return i + 1
}

medium

A. Loop should start from j = 0 instead of 1

B. Pointer i is incorrectly updated to j instead of incrementing

C. Return statement should be i instead of i + 1

D. Comparison should be nums[j] == nums[i] instead of !=

Solution

1. Step 1: Analyze pointer update logic

   Pointer i should move forward by one (i++) to place unique element, not jump to j (i = j). Correct code uses i++ then nums[i] = nums[j] to overwrite duplicates properly.

2. Final Answer:

   Pointer i is incorrectly updated to j instead of incrementing -> Option B

3. Quick Check:

   i must increment, not jump to j [OK]

Hint: Increment i to place unique, don't assign i = j [OK]

Common Mistakes:

• Setting i = j causes skipping elements
• Starting loop at wrong index
• Wrong return value calculation

5. Given a sorted array nums := []int{0,0,1,1,1,2,2,3,3,4}, what will be the final array state and length after applying the two-pointer duplicate removal method?

hard

A. nums = [0,1,1,2,3,4,4,3,3,4], length = 5

B. nums = [0,0,1,2,3,4,3,3,4,4], length = 6

C. nums = [0,1,2,3,4,2,2,3,3,4], length = 5

D. nums = [0,1,2,3,4,0,1,1,2,2], length = 4

Solution

1. Step 1: Apply two-pointer logic step-by-step

   Start with i=0, j=1; move j forward, when nums[j] != nums[i], increment i and set nums[i]=nums[j]. Unique elements are 0,1,2,3,4 placed at start; length = 5; rest of array remains unchanged: [0,1,2,3,4,2,2,3,3,4].

2. Final Answer:

   nums = [0,1,2,3,4,2,2,3,3,4], length = 5 -> Option C

3. Quick Check:

   Unique count 5, array start updated correctly [OK]

Hint: Place unique elements at front, count unique length [OK]

Common Mistakes:

• Assuming leftover elements are removed
• Miscounting unique elements
• Overwriting wrong indices