Two Pointer Technique on Arrays in DSA Typescript - Build from Scratch

Two Pointer Technique on Arrays

📖 Scenario: You are helping a friend organize a list of numbers. Your friend wants to find pairs of numbers that add up to a specific target number. To do this efficiently, you will use the two pointer technique on a sorted array.

🎯 Goal: Build a program that uses the two pointer technique to find if there is a pair of numbers in a sorted array that adds up to a given target number.

📋 What You'll Learn

Create a sorted array called numbers with exact values [1, 3, 4, 5, 7, 10, 12]

Create a variable called target and set it to 9

Use two pointers called left and right to scan the array

Implement the two pointer technique to check if any two numbers add up to target

Print "Pair found" if such a pair exists, otherwise print "No pair found"

💡 Why This Matters

🌍 Real World

Finding pairs of numbers that add up to a target is common in budgeting, shopping discounts, and matching problems.

💼 Career

The two pointer technique is a fundamental algorithm skill useful in coding interviews and efficient data processing.

1

Create the sorted array

Create a sorted array called numbers with these exact values: [1, 3, 4, 5, 7, 10, 12]

DSA Typescript

// Create the sorted array called numbers
// Your code here

Hint

Use const numbers = [1, 3, 4, 5, 7, 10, 12]; to create the array.

2

Set the target sum

Create a variable called target and set it to 9

DSA Typescript

const numbers = [1, 3, 4, 5, 7, 10, 12];
// Create the target variable
// Your code here

Hint

Use const target = 9; to set the target sum.

3

Implement the two pointer technique

Create two variables called left and right. Set left to 0 and right to numbers.length - 1. Use a while loop with the condition left < right. Inside the loop, calculate the sum of numbers[left] and numbers[right]. If the sum equals target, set a variable found to true and break the loop. If the sum is less than target, increase left by 1. Otherwise, decrease right by 1. Initialize found as false before the loop.

DSA Typescript

const numbers = [1, 3, 4, 5, 7, 10, 12];
const target = 9;
// Implement the two pointer technique
// Your code here

Hint

Use two pointers starting at the ends of the array and move them based on the sum compared to the target.

4

Print the result

Use an if statement to check if found is true. If yes, print "Pair found". Otherwise, print "No pair found".

DSA Typescript

const numbers = [1, 3, 4, 5, 7, 10, 12];
const target = 9;

let left = 0;
let right = numbers.length - 1;
let found = false;

while (left < right) {
    const sum = numbers[left] + numbers[right];
    if (sum === target) {
        found = true;
        break;
    } else if (sum < target) {
        left++;
    } else {
        right--;
    }
}

// Print the result
// Your code here

Hint

Use console.log("Pair found") or console.log("No pair found") based on the found variable.

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