Two Pointer Technique on Arrays in DSA Python

Mental Model

Use two markers moving through the array to find pairs or conditions without checking every pair.

Analogy: Imagine two friends starting at opposite ends of a hallway, walking towards each other to find a meeting point quickly.

Array: [1] -> [2] -> [3] -> [4] -> [5]
Left pointer ↑                      Right pointer ↑

Dry Run Walkthrough

Input: array: [1, 3, 5, 7, 9], target sum: 12

Goal: Find two numbers in the array that add up to 12 using two pointers

Step 1: Initialize left pointer at index 0 (value 1) and right pointer at index 4 (value 9)

[1] ↑ -> [3] -> [5] -> [7] -> [9] ↑

Why: Start checking pairs from both ends to find the target sum efficiently

Step 2: Calculate sum: 1 + 9 = 10, which is less than 12, move left pointer right

[1]    [3] ↑ -> [5] -> [7] -> [9] ↑

Why: Sum too small, increase sum by moving left pointer to higher value

Step 3: Calculate sum: 3 + 9 = 12, found target sum

[1]    [3] -> [5] -> [7] -> [9] ↑

Why: Sum matches target, pair found

Result:

Final pointers at indices 1 and 4 with values 3 and 9
Output: Pair found (3, 9)

Annotated Code

DSA Python

from typing import List, Optional

class TwoPointerArray:
    def __init__(self, arr: List[int]):
        self.arr = arr

    def find_pair_with_sum(self, target: int) -> Optional[tuple]:
        left, right = 0, len(self.arr) - 1
        while left < right:
            current_sum = self.arr[left] + self.arr[right]
            if current_sum == target:
                return (self.arr[left], self.arr[right])
            elif current_sum < target:
                left += 1
            else:
                right -= 1
        return None

if __name__ == '__main__':
    arr = [1, 3, 5, 7, 9]
    target = 12
    tp = TwoPointerArray(arr)
    result = tp.find_pair_with_sum(target)
    if result:
        print(f"Pair found {result}")
    else:
        print("No pair found")

left, right = 0, len(self.arr) - 1

initialize two pointers at array ends

while left < right:

loop until pointers meet or cross

current_sum = self.arr[left] + self.arr[right]

calculate sum of values at pointers

if current_sum == target:

check if sum matches target

return (self.arr[left], self.arr[right])

return the pair if found

elif current_sum < target:

if sum too small, move left pointer right to increase sum

left += 1

advance left pointer

else:

if sum too large, move right pointer left to decrease sum

right -= 1

move right pointer left

Output

Pair found (3, 9)

Complexity Analysis

Time: O(n) because each pointer moves at most n steps through the array once

Space: O(1) because only two pointers and a few variables are used

vs Alternative: Compared to checking all pairs (O(n^2)), two pointer technique is much faster for sorted arrays

Edge Cases

empty array

returns None immediately as no pairs exist

DSA Python

while left < right:

array with one element

returns None because no pair can be formed

DSA Python

while left < right:

no pair sums to target

returns None after pointers cross

DSA Python

return None

When to Use This Pattern

When you need to find pairs or conditions involving two elements in a sorted array, use two pointers moving inward to reduce time from quadratic to linear.

Common Mistakes

Mistake: Not moving the correct pointer based on sum comparison
Fix: If sum is less than target, move left pointer right; if sum is greater, move right pointer left

Mistake: Using two pointers on unsorted array without sorting first
Fix: Ensure array is sorted before applying two pointer technique

Summary

Find pairs or conditions in sorted arrays efficiently using two pointers from both ends.

Use when searching for pairs or sums in sorted arrays to reduce time complexity.

The key is moving pointers inward based on comparison to target sum to avoid checking all pairs.

Practice

1. What is the main idea behind the Two Pointer Technique when used on arrays?

easy

A. Use two arrays to store temporary results

B. Sort the array twice before processing

C. Use two indices to scan the array from different ends or positions simultaneously

D. Use recursion to divide the array into two parts

Solution

1. Step 1: Understand the two pointer concept

   The technique uses two pointers (indices) to traverse the array, often starting from opposite ends or different positions.

2. Step 2: Recognize the purpose

   This helps solve problems efficiently by reducing the number of iterations compared to nested loops.

3. Final Answer:

   Use two indices to scan the array from different ends or positions simultaneously -> Option C

4. Quick Check:

   Two pointers scan array ends or positions [OK]

Hint: Think of two fingers moving inward or outward on the array [OK]

Common Mistakes:

• Confusing two pointers with using two separate arrays
• Assuming sorting is always required
• Thinking recursion is part of two pointer technique

2. Which of the following is the correct syntax to initialize two pointers left and right for an array arr in Python?

easy

A. left = 0; right = len(arr) - 1

B. left = 1; right = len(arr)

C. left = -1; right = len(arr)

D. left = 0; right = len(arr)

Solution

1. Step 1: Identify valid pointer positions

   In Python, array indices start at 0, so the left pointer should start at 0.

2. Step 2: Set right pointer correctly

   The right pointer should be at the last valid index, which is length of array minus 1.

3. Final Answer:

   left = 0; right = len(arr) - 1 -> Option A

4. Quick Check:

   Start left at 0 and right at last index [OK]

Hint: Remember Python arrays start at 0, last index is length-1 [OK]

Common Mistakes:

• Setting right pointer to len(arr) causes index error
• Starting left pointer at 1 misses first element
• Using negative indices incorrectly

3. What will be the output of the following Python code?

arr = [1, 3, 5, 7, 9]
left, right = 0, 4
result = []
while left < right:
    result.append(arr[left] + arr[right])
    left += 1
    right -= 1
print(result)

medium

A. [10, 14]

B. [10, 12]

C. [10, 12, 14]

D. [10, 10]

Solution

1. Step 1: Trace the loop iterations

   Initial pointers: left=0, right=4 - Iteration 1: arr[0]+arr[4] = 1+9=10, result=[10] - left=1, right=3 - Iteration 2: arr[1]+arr[3] = 3+7=10, result=[10,10] - left=2, right=2 (loop ends because left is not less than right)

2. Step 2: Check loop condition and final result

   Loop stops when left == right (2), so only two sums added: [10, 10]

3. Final Answer:

   [10, 10] -> Option D

4. Quick Check:

   Sum pairs from ends until pointers meet [OK]

Hint: Add pairs from ends until pointers cross [OK]

Common Mistakes:

• Including middle element sum when pointers meet
• Not updating pointers correctly
• Confusing indices causing wrong sums

4. Find the error in the following code that uses two pointers to check if an array is a palindrome:

def is_palindrome(arr):
    left, right = 0, len(arr)
    while left < right:
        if arr[left] != arr[right]:
            return False
        left += 1
        right -= 1
    return True

print(is_palindrome([1, 2, 3, 2, 1]))

medium

A. left should start at 1

B. right should be initialized as len(arr) - 1

C. Use 'while left <= right' instead of 'while left < right'

D. Return True should be inside the loop

Solution

1. Step 1: Check pointer initialization

   In Python, last index is len(arr) - 1, but code sets right = len(arr), which is out of range.

2. Step 2: Correct pointer initialization

   Change right to len(arr) - 1 to avoid index error and correctly compare elements.

3. Final Answer:

   right should be initialized as len(arr) - 1 -> Option B

4. Quick Check:

   Last index = length - 1 [OK]

Hint: Last index is length minus one, not length [OK]

Common Mistakes:

• Setting right pointer to length causes index error
• Changing loop condition incorrectly
• Returning True too early inside loop

5. Given a sorted array arr and a target sum target, how can you use the two pointer technique to find if there exists two numbers that add up to target? Choose the correct approach.

hard

A. Initialize left at 0 and right at len(arr)-1; move pointers inward based on sum comparison

B. Initialize both pointers at 0 and move right pointer only

C. Sort the array again and use binary search for each element

D. Use two nested loops to check all pairs

Solution

1. Step 1: Set pointers at array ends

   Start left pointer at 0 (start) and right pointer at last index (len(arr)-1).

2. Step 2: Move pointers based on sum

   Calculate sum of arr[left] + arr[right]. If sum equals target, return True. If sum is less, move left pointer right to increase sum. If sum is more, move right pointer left to decrease sum.

3. Final Answer:

   Initialize left at 0 and right at len(arr)-1; move pointers inward based on sum comparison -> Option A

4. Quick Check:

   Two pointers from ends adjust to find target sum [OK]

Hint: Move left/right pointers based on sum compared to target [OK]

Common Mistakes:

• Moving only one pointer without condition
• Using nested loops instead of two pointers
• Not starting pointers at array ends