Two Pointer Technique on Arrays in DSA Javascript - Execution Trace

Concept Flow - Two Pointer Technique on Arrays

Initialize left pointer at start

↓

Initialize right pointer at end

↓

Check condition: left < right?

No→Stop

Yes↓

Compare or process elements at left and right

↓

Move left pointer right or right pointer left based on condition

↩Back to condition check

Start with two pointers at opposite ends of the array, move them towards each other based on conditions until they meet or cross.

Execution Sample

DSA Javascript

const arr = [1, 3, 5, 7, 9];
let left = 0;
let right = arr.length - 1;
while (left < right) {
  if (arr[left] + arr[right] === 10) {
    console.log(left, right);
    break;
  } else if (arr[left] + arr[right] < 10) {
    left++;
  } else {
    right--;
  }
}

Find two numbers in a sorted array that add up to 10 using two pointers moving inward.

Execution Table

Step	Operation	Left Pointer	Right Pointer	Sum arr[left]+arr[right]	Action	Array State
1	Initialize pointers	0 (value=1)	4 (value=9)	1+9=10	Sum equals 10, print indices and break	[1, 3, 5, 7, 9]
2	Exit loop	-	-	-	Break triggered, stop	[1, 3, 5, 7, 9]

💡 Sum equals target 10 at step 1, loop breaks.

Variable Tracker

Variable	Start	After Step 1	After Step 2 (End)
left	0	0	0
right	4	4	4

Key Moments - 3 Insights

Why do we stop when left pointer is no longer less than right pointer?

Because when left >= right, pointers have met or crossed, meaning all pairs have been checked (see execution_table step 1 condition).

Why do we move the left pointer right when sum is less than target?

Because the array is sorted, increasing left pointer moves to a bigger number, increasing the sum (see execution_table step 1 action).

What happens if no pair sums to the target?

The pointers move inward until left >= right, then loop ends without finding a pair (not shown here but implied by exit condition).

Visual Quiz - 3 Questions

Look at the execution_table at step 1, what is the sum of arr[left] + arr[right]?

A 9

B 8

C 10

D 11

💡 Hint

Check the 'Sum arr[left]+arr[right]' column in execution_table row 1.

At which step does the loop stop because the sum equals the target?

A Step 3

B Step 1

C Step 2

D Step 4

💡 Hint

Look at the 'Action' column in execution_table to find when break happens.

If the array was [1, 2, 3, 4, 5] and target was 10, how would the pointers move?

A Left moves right until it meets right, no pair found

B Right moves left until it meets left, pair found

C Both pointers stay fixed

D Left moves left and right moves right

💡 Hint

Think about the sum compared to target and pointer movement rules from key_moments.

Concept Snapshot

Two Pointer Technique on Arrays:
- Use two pointers at start and end of sorted array
- Move pointers inward based on sum comparison
- Stop when pointers meet or sum matches target
- Efficient for pair sum and similar problems

Full Transcript

The two pointer technique uses two indexes starting at opposite ends of a sorted array. We check the sum of elements at these pointers. If the sum matches the target, we stop. If the sum is less, we move the left pointer right to increase sum. If the sum is more, we move the right pointer left to decrease sum. This continues until pointers meet or cross, ensuring all pairs are checked efficiently.

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