Welcome to Day 4 of your coding interview journey.

So far, we’ve built a strong foundation:

• ✅ Day 1: HashMaps
• ✅ Day 2: Sliding Window
• ✅ Day 3: Arrays

Today, we unlock another powerful pattern:

👉 Two Pointers

This technique is essential for solving array and string problems efficiently — often reducing complexity from O(n²) to O(n).

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🧠 What is the Two Pointers Technique?

The idea is simple:

👉 Use two indices (pointers) to iterate through the data, usually from different directions.

Instead of nested loops, you:

• Move pointers based on conditions
• Narrow down the search space efficiently

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🔑 When Should You Use It?

Look for these signals in problems:

• Sorted arrays
• Pairs or triplets
• Target sums
• Palindromes
• Problems involving optimization (max/min)

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💡 Problem 1 – Container With Most Water

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🧩 Problem

Given an array height, where each element represents a vertical line:

👉 Find two lines that together form a container that holds the maximum amount of water.

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❌ Brute Force

• Check all pairs
  👉 Time Complexity: O(n²)

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✅ Optimal Approach – Two Pointers

We start with:

• Left pointer at the beginning
• Right pointer at the end

Then:

• Calculate area
• Move the pointer with the smaller height

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🛠️ Solution in Python

def max_area(height):
    left, right = 0, len(height) - 1
    max_water = 0

    while left < right:
        width = right - left
        h = min(height[left], height[right])
        max_water = max(max_water, width * h)

        # Move the smaller pointer
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1

    return max_water

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✅ Example

height = [1,8,6,2,5,4,8,3,7]
print(max_area(height))

Output:

49

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🎯 Key Insight

👉 The limiting factor is always the shorter line

So moving the taller line won’t help — only moving the shorter one might increase area.

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⚡ Complexity

• Time: O(n)
• Space: O(1)

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💡 Problem 2 – 3Sum

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🧩 Problem

Given an array nums, return all unique triplets [a, b, c] such that:

a + b + c = 0

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❌ Brute Force

• Try all triplets
  👉 Time Complexity: O(n³)

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✅ Optimal Approach – Sorting + Two Pointers

Steps:

1. Sort the array
2. Fix one number
3. Use two pointers to find the other two

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🛠️ Solution in Python

def three_sum(nums):
    nums.sort()
    result = []

    for i in range(len(nums)):
        # Skip duplicates
        if i > 0 and nums[i] == nums[i - 1]:
            continue

        left, right = i + 1, len(nums) - 1

        while left < right:
            total = nums[i] + nums[left] + nums[right]

            if total == 0:
                result.append([nums[i], nums[left], nums[right]])

                # Skip duplicates
                while left < right and nums[left] == nums[left + 1]:
                    left += 1
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1

                left += 1
                right -= 1

            elif total < 0:
                left += 1
            else:
                right -= 1

    return result

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✅ Example

nums = [-1,0,1,2,-1,-4]
print(three_sum(nums))

Output:

[[-1, -1, 2], [-1, 0, 1]]

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🎯 Key Insights

• Sorting enables efficient pointer movement
• Avoid duplicates carefully
• Reduce O(n³) → O(n²)

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⚠️ Common Mistakes

• ❌ Forgetting to sort the array
• ❌ Not handling duplicates
• ❌ Moving both pointers incorrectly
• ❌ Using brute force

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🧪 Practice Problems

To master Two Pointers, try:

1. Two Sum II (sorted array)
2. Valid palindrome
3. Remove duplicates from sorted array
4. 4Sum
5. Trapping rain water

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🎯 Key Takeaways

• Two Pointers reduces nested loops
• Works best with:
  • Sorted arrays
  • Pair/triplet problems
• Combine with:
  • Sorting
  • Greedy decisions

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🚀 Your Challenge (Day 4)

Solve this:

Given a sorted array, remove duplicates in-place and return the new length.

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📌 Final Thoughts

Two Pointers is one of the highest ROI patterns for interviews.

Master it, and you’ll solve problems that many candidates struggle with.

You’re already 4 days in — and building real momentum. 💪