Two-pointer technique in Data Structures Theory - Step-by-Step Execution

Concept Flow - Two-pointer technique

Start: Two pointers at positions

↓

Compare elements at pointers

↓

Move one or both pointers based on condition

↓

Check if pointers have met or crossed

Yes↓

Stop

↩Back to Compare

The two-pointer technique uses two markers moving through data to find or compare elements efficiently until they meet or cross.

Execution Sample

Data Structures Theory

arr = [1, 3, 5, 7, 9]
left = 0
right = 4
while left < right:
    if arr[left] + arr[right] == 10:
        break
    elif arr[left] + arr[right] < 10:
        left += 1
    else:
        right -= 1

This code uses two pointers to find two numbers in a sorted array that add up to 10.

Analysis Table

Step	left	right	arr[left]	arr[right]	Sum	Condition	Action
1	0	4	1	9	10	sum == 10	Break loop - pair found
2	-	-	-	-	-	-	Stop - pointers met condition

💡 Loop stops because sum of arr[left] and arr[right] equals 10 at step 1

State Tracker

Variable	Start	After Step 1	Final
left	0	0	0
right	4	4	4
sum	N/A	10	10

Key Insights - 2 Insights

Why do we move the left pointer when the sum is less than the target?

Because the array is sorted, increasing left moves to a larger number, increasing the sum.

Why does the loop stop when pointers meet or cross?

When pointers meet or cross, all pairs have been checked. The exit_note and execution_table show the loop stops when the condition is met or pointers cross.

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'sum' at step 1?

A 10

B 9

C 8

D 11

💡 Hint

Check the 'Sum' column in execution_table row for step 1

At which step does the loop stop according to the execution_table?

A Step 3

B Step 2

C Step 1

D Step 0

💡 Hint

Look at the 'Action' column where the loop breaks

If the target sum was 12, which pointer would move first based on the technique?

A Right pointer moves left

B Left pointer moves right

C Both pointers move simultaneously

D No pointer moves

💡 Hint

Refer to key_moments about moving left pointer when sum is less than target

Concept Snapshot

Two-pointer technique:
- Use two pointers at start and end of sorted data
- Compare elements at pointers
- Move left pointer right if sum less than target
- Move right pointer left if sum greater than target
- Stop when pointers meet or condition met

Full Transcript

The two-pointer technique involves placing two markers at different positions in a sorted list. We compare the elements at these pointers and decide which pointer to move based on the comparison. For example, if the sum of elements is less than the target, we move the left pointer to the right to increase the sum. If the sum is greater, we move the right pointer to the left to decrease the sum. This continues until the pointers meet or the desired condition is found. This method efficiently finds pairs or solves problems involving sorted data.

Practice

1. What is the main purpose of the two-pointer technique in data structures?

easy

A. To store data in two separate arrays

B. To use two markers to scan data efficiently

C. To sort data using two different algorithms

D. To create two copies of the data for backup

Solution

1. Step 1: Recall the main purpose of the two-pointer technique

   The two-pointer technique uses two markers or pointers to scan through data efficiently, reducing unnecessary checks.

2. Step 2: Compare with other options

   The other options describe unrelated actions like storing data in separate arrays, sorting data, or creating backups, which are not the main purpose of this technique.

3. Final Answer:

   To use two markers to scan data efficiently -> Option B

4. Quick Check:

   Two-pointer technique = efficient scanning [OK]

Hint: Two pointers scan data fast by moving through it together [OK]

Common Mistakes:

• Confusing two-pointer with sorting algorithms
• Thinking it copies data instead of scanning
• Assuming it stores data separately

2. Which of the following is the correct way to initialize two pointers left and right for scanning an array arr from both ends?

easy

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

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

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

D. left = -1; right = len(arr) + 1

Solution

1. Step 1: Identify valid pointer positions

   For scanning from both ends, left starts at 0 (first index) and right at len(arr) - 1 (last index).

2. Step 2: Check other options for errors

   The other options use invalid indices like starting at 1 or beyond array length, which cause errors or miss elements.

3. Final Answer:

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

4. Quick Check:

   Start pointers at array ends = left = 0; right = len(arr) - 1 [OK]

Hint: Start left at 0 and right at last index for two-pointer scan [OK]

Common Mistakes:

• Starting pointers outside array bounds
• Using length instead of length minus one
• Swapping left and right initial values

3. Consider the following code using the two-pointer technique to check if a string is a palindrome:

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

print(is_palindrome('radar'))

What will be the output?

medium

A. True

B. None

C. Error

D. False

Solution

1. Step 1: Understand the palindrome check logic

   The code compares characters from the start and end moving inward. If any pair differs, it returns False; otherwise, True.

2. Step 2: Apply to 'radar'

   Characters match at all pairs: 'r'=='r', 'a'=='a', middle 'd' ignored. So, it returns True.

3. Final Answer:

   True -> Option A

4. Quick Check:

   Palindrome check for 'radar' = True [OK]

Hint: Check characters from ends moving inward for palindrome [OK]

Common Mistakes:

• Confusing return values for palindrome
• Not updating pointers correctly
• Ignoring middle character in odd length strings

4. The following code attempts to find if there exists a pair in a sorted array arr that sums to target using the two-pointer technique:

def has_pair(arr, target):
    left, right = 0, len(arr)
    while left < right:
        current_sum = arr[left] + arr[right]
        if current_sum == target:
            return True
        elif current_sum < target:
            left += 1
        else:
            right -= 1
    return False

What is the error in this code?

medium

A. The function should return True at the end

B. The loop condition should be while left <= right

C. The current_sum calculation is wrong

D. The right pointer is initialized incorrectly

Solution

1. Step 1: Check pointer initialization

   The right pointer is set to len(arr), which is outside valid index range (max index is len(arr)-1).

2. Step 2: Identify correct initialization

   Right pointer should be len(arr) - 1 to point to the last element.

3. Final Answer:

   The right pointer is initialized incorrectly -> Option D

4. Quick Check:

   Right pointer must be last index, not length [OK]

Hint: Right pointer = last index, not array length [OK]

Common Mistakes:

• Setting right pointer to array length
• Using wrong loop condition
• Miscomputing sum inside loop

5. You have two sorted arrays A and B. Using the two-pointer technique, how can you efficiently find all pairs (a, b) such that a + b = target? Choose the best approach.

hard

A. Use nested loops to check every pair from A and B

B. Sort both arrays again and then check pairs

C. Use one pointer on A starting at 0 and one on B starting at end, move pointers based on sum comparison

D. Use a hash map to store sums of all pairs

Solution

1. Step 1: Understand two-pointer approach for two sorted arrays

   Start pointer i at beginning of A and pointer j at end of B. Calculate sum A[i]+B[j].

2. Step 2: Move pointers based on sum

   If sum equals target, record pair and move pointers. If sum less than target, move i forward; if greater, move j backward. This avoids nested loops and is efficient.

3. Final Answer:

   Use one pointer on A starting at 0 and one on B starting at end, move pointers based on sum comparison -> Option C

4. Quick Check:

   Two-pointer on sorted arrays finds pairs efficiently [OK]

Hint: One pointer start, one pointer end; move based on sum [OK]

Common Mistakes:

• Using nested loops causing slow performance
• Resorting already sorted arrays unnecessarily
• Using hash map instead of two-pointer for sorted arrays