Why Two Pointer Technique on Arrays in DSA Javascript?

The Big Idea

What if you could find pairs in a list without checking every single combination?

The Scenario

Imagine you have a long list of numbers and you want to find pairs that add up to a certain value. Doing this by checking every possible pair one by one feels like searching for a needle in a haystack.

The Problem

Checking every pair means lots of repeated work and takes a long time, especially if the list is very long. It's easy to make mistakes or miss pairs because you have to remember which pairs you already checked.

The Solution

The two pointer technique uses two markers that move through the list from different ends. This way, you can find pairs quickly without checking every combination, saving time and effort.

Before vs After

✗ Before

for (let i = 0; i < arr.length; i++) {
  for (let j = i + 1; j < arr.length; j++) {
    if (arr[i] + arr[j] === target) {
      console.log(arr[i], arr[j]);
    }
  }
}

✓ After

let sorted = arr.slice().sort((a, b) => a - b);
let left = 0;
let right = sorted.length - 1;
while (left < right) {
  let sum = sorted[left] + sorted[right];
  if (sum === target) {
    console.log(sorted[left], sorted[right]);
    left++;
    right--;
  } else if (sum < target) {
    left++;
  } else {
    right--;
  }
}

What It Enables

This technique makes it easy and fast to find pairs or solve problems involving two elements moving from opposite ends.

Real Life Example

Think about two people starting at opposite ends of a hallway trying to find the perfect spot to meet in the middle based on their walking speeds.

Key Takeaways

Manual pair checking is slow and error-prone.

Two pointers move from both ends to find pairs efficiently.

This method reduces time and simplifies the search process.

Practice

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

easy

A. Use two indices moving towards each other or in the same direction to solve problems efficiently

B. Use two nested loops to compare every pair of elements

C. Sort the array first and then use binary search

D. Use recursion to divide the array into two parts

Solution

1. Step 1: Understand the technique

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

2. Step 2: Purpose of two pointers

   This helps solve problems like finding pairs, removing duplicates, or partitioning without extra space and in linear time.

3. Final Answer:

   Use two indices moving towards each other or in the same direction to solve problems efficiently -> Option A

4. Quick Check:

   Two pointers = two indices moving efficiently [OK]

Hint: Think of two fingers scanning array ends or start [OK]

Common Mistakes:

• Confusing with nested loops approach
• Assuming sorting is always required
• Thinking recursion is part of two pointers

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

easy

A. var left = 1, right = arr.length;

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

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

D. const left = 0, right = arr.length - 1;

Solution

1. Step 1: Understand pointer initialization

   For two pointers starting at the array ends, left starts at 0 (start), right at arr.length - 1 (end).

2. Step 2: Check each option

   const left = 0, right = arr.length - 1; correctly sets left = 0 and right = arr.length - 1. Others have incorrect indices.

3. Final Answer:

   const left = 0, right = arr.length - 1; -> Option D

4. Quick Check:

   Start pointer at 0, end pointer at length-1 [OK]

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

Common Mistakes:

• Setting right pointer to arr.length (out of bounds)
• Swapping left and right initialization
• Using 1-based indexing instead of 0-based

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

const arr = [1, 2, 3, 4, 5];
let left = 0;
let right = arr.length - 1;
const result = [];
while (left <= right) {
  if (arr[left] + arr[right] === 6) {
    result.push([arr[left], arr[right]]);
    left++;
    right--;
  } else if (arr[left] + arr[right] < 6) {
    left++;
  } else {
    right--;
  }
}
console.log(result);

medium

A. [[2,4],[3,3]]

B. [[1,5],[3,3]]

C. [[1,5],[2,4],[3,3]]

D. [[1,5],[2,4]]

Solution

1. Step 1: Trace the two pointers and sums

   Initially left=0 (1), right=4 (5), sum=6 -> push [1,5], move left=1, right=3.
   Now left=1 (2), right=3 (4), sum=6 -> push [2,4], move left=2, right=2.
   Now left=2 (3), right=2 (3), sum=6 -> push [3,3], move left=3, right=1 but loop ends since left > right.

2. Step 2: Check the result array

   Result contains [[1,5],[2,4],[3,3]]. But note the condition left <= right allows the middle element pairing only if left equals right.
   Here, at left=2 and right=2, sum=6, so [3,3] is added.

3. Final Answer:

   [[1,5],[2,4],[3,3]] -> Option C

4. Quick Check:

   Pairs summing to 6 = [[1,5],[2,4],[3,3]] [OK]

Hint: Check sums at left and right, move pointers accordingly [OK]

Common Mistakes:

• Ignoring the case when left equals right
• Not updating pointers correctly after finding a pair
• Assuming only pairs with different indices count

4. Find the bug in the following code that uses two pointers to check if an array is sorted in non-decreasing order.

function isSorted(arr) {
  let left = 0;
  let right = arr.length - 1;
  while (left < right) {
    if (arr[left] > arr[left + 1]) {
      return false;
    }
    left++;
  }
  return true;
}

medium

A. The loop condition should be left <= right

B. The right pointer is unused and unnecessary

C. The comparison should be between arr[left] and arr[right]

D. The function does not handle empty arrays

Solution

1. Step 1: Analyze pointer usage

   The variable right is set but never used inside the loop or condition, making it unnecessary.

2. Step 2: Check loop and comparison logic

   The loop correctly checks adjacent elements using left pointer only, so right can be removed without affecting logic.

3. Final Answer:

   The right pointer is unused and unnecessary -> Option B

4. Quick Check:

   Unused variable right pointer = bug [OK]

Hint: Check if all declared variables are used [OK]

Common Mistakes:

• Thinking right pointer must be used in two pointer technique always
• Changing loop condition unnecessarily
• Confusing adjacent element checks with two pointer sum checks

5. Given a sorted array arr and a target sum target, which code snippet correctly uses the two pointer technique to find if there exists a pair that sums to target?

const arr = [1, 3, 4, 5, 7, 10, 12];
const target = 9;
// Which snippet is correct?

hard

A. let left = 0, 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;

B. let left = 0, 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;

C. for (let left = 0, right = arr.length - 1; left < right; left++, right--) {
if (arr[left] + arr[right] === target) return true;
}
return false;

D. let left = 0, right = 0;
while (left < arr.length && right < arr.length) {
if (arr[left] + arr[right] === target) return true;
else if (arr[left] + arr[right] < target) right++;
else left++;
}
return false;

Solution

1. Step 1: Understand two pointer approach for pair sum

   Start with pointers at array ends: left=0, right=length-1. Move pointers based on sum compared to target.

2. Step 2: Check each option logic

   let left = 0, 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; correctly moves left pointer forward if sum < target, right pointer backward if sum > target, and returns true if sum equals target.
   let left = 0, 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; incorrectly moves left pointer forward when sum > target.
   for (let left = 0, right = arr.length - 1; left < right; left++, right--) { if (arr[left] + arr[right] === target) return true; } return false; moves both pointers every iteration, missing some pairs.
   let left = 0, right = 0; while (left < arr.length && right < arr.length) { if (arr[left] + arr[right] === target) return true; else if (arr[left] + arr[right] < target) right++; else left++; } return false; starts both pointers at 0, which is incorrect for this problem.

3. Final Answer:

   Option A code snippet -> Option A

4. Quick Check:

   Correct pointer moves based on sum vs target [OK]

Hint: Move left if sum less, right if sum more [OK]

Common Mistakes:

• Moving wrong pointer when sum is greater or smaller
• Using for loop that moves both pointers simultaneously
• Starting both pointers at same index