Two Pointer Technique on Arrays in DSA Typescript

Mental Model

Use two markers moving through the array to find or compare elements efficiently without extra space.

Analogy: Imagine two friends starting at opposite ends of a hallway, walking towards each other to find a meeting point or compare items along the way.

Array: [1] -> [2] -> [3] -> [4] -> [5] -> null
Pointers: ↑left                      ↑right

Dry Run Walkthrough

Input: array: [1, 3, 5, 7, 9], find if two numbers sum to 10

Goal: Find two numbers in the array that add up to 10 using two pointers

Step 1: Set left pointer at index 0 (value 1), right pointer at index 4 (value 9)

[1]↑left -> [3] -> [5] -> [7] -> [9]↑right -> null

Why: Start with pointers at both ends to check sums from smallest and largest values

Step 2: Calculate sum 1 + 9 = 10, found target sum

[1]↑left -> [3] -> [5] -> [7] -> [9]↑right -> null

Why: Sum matches target, so we found the pair

Result:

[1]↑left -> [3] -> [5] -> [7] -> [9]↑right -> null
Answer: True (1 + 9 = 10)

Annotated Code

DSA Typescript

class TwoPointer {
  static hasPairWithSum(arr: number[], target: number): boolean {
    let left = 0
    let right = arr.length - 1
    while (left < right) {
      const sum = arr[left] + arr[right]
      if (sum === target) {
        return true
      } else if (sum < target) {
        left++
      } else {
        right--
      }
    }
    return false
  }
}

const arr = [1, 3, 5, 7, 9]
const target = 10
const result = TwoPointer.hasPairWithSum(arr, target)
console.log(result ? 'True (pair found)' : 'False (no pair)')

let left = 0 let right = arr.length - 1

Initialize two pointers at start and end of array

while (left < right) {

Loop until pointers meet or cross

const sum = arr[left] + arr[right]

Calculate sum of elements at pointers

if (sum === target) { return true }

If sum matches target, return success

else if (sum < target) { left++ }

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

else { right-- }

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

Output

True (pair found)

Complexity Analysis

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

Space: O(1) because no extra space is used besides pointers

vs Alternative: Better than nested loops O(n^2) which check all pairs; two pointers reduce checks by moving inward

Edge Cases

empty array

returns false immediately because no pairs exist

DSA Typescript

while (left < right) {

array with one element

returns false because at least two elements are needed

DSA Typescript

while (left < right) {

no pair sums to target

returns false after pointers cross without finding sum

DSA Typescript

return false

When to Use This Pattern

When asked to find pairs or compare elements in a sorted array efficiently, use two pointers moving inward from both ends to reduce time.

Common Mistakes

Mistake: Not moving pointers correctly when sum is less or greater than target
Fix: Move left pointer right if sum is less than target, move right pointer left if sum is greater

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

Summary

Find pairs or compare elements in a sorted array using two pointers moving inward.

Use when you need to find pairs with a target sum or similar conditions efficiently.

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

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