Two Pointer Technique on Arrays in DSA Javascript - Time & Space Complexity

Time Complexity: Two Pointer Technique on Arrays

O(n)

Understanding Time Complexity

We want to understand how fast the two pointer technique works on arrays.

How does the number of steps grow as the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function twoPointerSum(arr, target) {
  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;
}
    

This code checks if two numbers in a sorted array add up to a target using two pointers.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop moves two pointers through the array.
• How many times: Each pointer moves at most the length of the array once.

How Execution Grows With Input

As the array size grows, the total steps increase roughly in a straight line.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The steps grow directly with the size of the array.

Final Time Complexity

Time Complexity: O(n)

This means the time to finish grows in a straight line with the array size.

Common Mistake

[X] Wrong: "The two pointer method checks every pair, so it must be O(n²)."

[OK] Correct: The pointers move inward without repeating pairs, so total steps are just about n, not n squared.

Interview Connect

Knowing this technique helps you solve many array problems efficiently and shows you understand smart ways to reduce work.

Self-Check

"What if the array was not sorted? How would the time complexity change?"

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