Two Pointer Technique on Arrays in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the Two Pointer Technique on Arrays?

It is a method where two pointers move through the array to solve problems efficiently, often from different ends or at different speeds.

beginner

How do the two pointers usually move in the Two Pointer Technique?

They often start at opposite ends and move towards each other, or one moves faster than the other, depending on the problem.

intermediate

Why is the Two Pointer Technique useful?

It reduces time complexity by avoiding nested loops, making some problems solvable in linear time.

intermediate

Example problem: Find if there is a pair in a sorted array that sums to a target. How does Two Pointer Technique help?

Start one pointer at the beginning and one at the end. Move pointers inward based on sum comparison to target, finding the pair in one pass.

intermediate

What is a common mistake when using Two Pointer Technique?

Not moving pointers correctly or forgetting to check all conditions can cause infinite loops or missed solutions.

In the Two Pointer Technique, if you want to find two numbers that add up to a target in a sorted array, where do you start the pointers?

A One at the start, one at the end

B Both at the start

C Both at the end

D Random positions

Starting one pointer at the beginning and one at the end allows checking sums efficiently by moving pointers inward.

What is the main advantage of using the Two Pointer Technique over nested loops?

A Works on unsorted arrays only

B Uses less memory

C Simpler code

D Faster time complexity

Two Pointer Technique often reduces time complexity from quadratic to linear by avoiding nested loops.

Which type of array is best suited for the Two Pointer Technique when searching for pairs with a target sum?

A Array with duplicates only

B Sorted array

C Unsorted array

D Empty array

Sorted arrays allow moving pointers based on sum comparisons, which is key to the technique.

If the sum of elements at two pointers is less than the target, what should you do?

A Stop the search

B Move the right pointer left

C Move the left pointer right

D Move both pointers

Moving the left pointer right increases the sum because the array is sorted.

Can the Two Pointer Technique be used on unsorted arrays without sorting first?

A Usually no, sorting is needed

B No, never

C Only if pointers move at different speeds

D Yes, always

The technique relies on sorted order to decide pointer movement, so sorting is usually required first.

Explain the Two Pointer Technique and how it helps solve array problems efficiently.

Think about how two pointers can scan the array without nested loops.

Describe a step-by-step approach to find two numbers in a sorted array that add up to a target using the Two Pointer Technique.

Focus on pointer movement rules depending on sum compared to target.

Practice

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

easy

A. Use a single pointer to iterate through the array once

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

C. Sort the array before processing it with one pointer

D. Use recursion to divide the array into two parts

Solution

1. Step 1: Understand the technique

   The Two Pointer Technique involves using two indices or pointers to traverse the array, often from opposite ends or different positions.

2. Step 2: Compare with other options

   Options A, C, and D describe different methods like single pointer iteration, sorting, or recursion, which are not the core idea of two pointers.

3. Final Answer:

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

4. Quick Check:

   Two pointers = two indices scanning array [OK]

Hint: Two pointers move independently to cover array efficiently [OK]

Common Mistakes:

• Thinking it uses only one pointer
• Confusing with sorting or recursion
• Assuming pointers move only forward

2. Which of the following is the correct TypeScript syntax to initialize two pointers at the start and end of an array named arr?

easy

A. let left = arr[0], right = arr[arr.length];

B. let left = 1, right = arr.length;

C. let left = 0, right = arr.length - 1;

D. let left = arr.length, right = 0;

Solution

1. Step 1: Identify correct pointer initialization

   For two pointers at array ends, left starts at 0 (first index), right at arr.length - 1 (last index).

2. Step 2: Check other options for errors

   let left = 1, right = arr.length; uses invalid indices (arr.length is out of bounds). let left = arr[0], right = arr[arr.length]; assigns values from array elements, not indices. let left = arr.length, right = 0; reverses pointers incorrectly.

3. Final Answer:

   let left = 0, right = arr.length - 1; -> Option C

4. Quick Check:

   Start = 0, End = length-1 [OK]

Hint: Array indices start at 0, last index is length minus one [OK]

Common Mistakes:

• Using arr.length as an index (out of bounds)
• Assigning pointer to array values instead of indices
• Swapping left and right initialization

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

const arr = [1, 3, 5, 7, 9];
let left = 0;
let right = arr.length - 1;
let sumPairs = [];
while (left < right) {
  sumPairs.push(arr[left] + arr[right]);
  left++;
  right--;
}
console.log(sumPairs);

medium

A. [10, 16]

B. [10, 12]

C. [10, 8]

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, sumPairs=[10]
   left=1, right=3
   Iteration 2: arr[1]+arr[3] = 3+7=10, sumPairs=[10,10]
   left=2, right=2 (loop ends as left is not less than right)

2. Step 2: Final output array

   The sumPairs array contains [10, 10] after the loop finishes.

3. Final Answer:

   [10, 10] -> Option D

4. Quick Check:

   Pairs sums = [10, 10] [OK]

Hint: Add elements at left and right, move pointers inward [OK]

Common Mistakes:

• Including middle element sum when left equals right
• Incorrect pointer updates causing infinite loop
• Wrong sum calculations or indices

4. Identify the bug in the following TypeScript code that uses two pointers to find if a sorted array contains two numbers summing to target.

function hasPairSum(arr: number[], target: number): boolean {
  let left = 0;
  let right = arr.length;
  while (left < right) {
    const sum = arr[left] + arr[right];
    if (sum === target) return true;
    else if (sum < target) left++;
    else right--;
  }
  return false;
}

medium

A. Right pointer initialized to arr.length instead of arr.length - 1

B. Left pointer should start at 1, not 0

C. The loop condition should be left <= right

D. The sum calculation should use arr[left] - arr[right]

Solution

1. Step 1: Check pointer initialization

   The right pointer is set to arr.length, which is out of array bounds. It should be arr.length - 1 to point to the last element.

2. Step 2: Validate other parts

   Left pointer at 0 is correct. Loop condition left < right is standard. Sum calculation uses addition correctly.

3. Final Answer:

   Right pointer initialized to arr.length instead of arr.length - 1 -> Option A

4. Quick Check:

   Right pointer must be last index (length-1) [OK]

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

Common Mistakes:

• Using arr.length as index causing runtime error
• Changing loop condition incorrectly
• Mistaking sum operation for subtraction

5. Given a sorted array arr of integers, how can you use the Two Pointer Technique to find all unique pairs that sum up to a target value target? Choose the best approach.

hard

A. Initialize two pointers at start and end; move inward adjusting pointers based on sum; skip duplicates to collect unique pairs

B. Use nested loops to check all pairs and add those matching target

C. Sort the array first, then use a hash map to find complements

D. Initialize pointers both at start and move forward checking sums

Solution

1. Step 1: Use two pointers at array ends

   Start with left at 0 and right at arr.length - 1. Calculate sum of elements at these pointers.

2. Step 2: Move pointers and skip duplicates

   If sum equals target, record pair and move both pointers skipping duplicates. If sum less than target, move left pointer right. If sum greater, move right pointer left.

3. Step 3: Collect unique pairs efficiently

   This approach avoids nested loops and handles duplicates by skipping repeated values, ensuring unique pairs.

4. Final Answer:

   Initialize two pointers at start and end; move inward adjusting pointers based on sum; skip duplicates to collect unique pairs -> Option A

5. Quick Check:

   Two pointers + skip duplicates = unique pairs [OK]

Hint: Move pointers inward and skip duplicates for unique pairs [OK]

Common Mistakes:

• Using nested loops causing higher time complexity
• Not skipping duplicates leading to repeated pairs
• Incorrect pointer movement causing missed pairs