Two Pointer Technique on Arrays in DSA C++ - Build from Scratch

Two Pointer Technique on Arrays

📖 Scenario: You are working on a simple program to find pairs of numbers in a sorted list that add up to a specific target. This technique is useful in many real-world tasks like finding matching expenses or pairs of items that fit a budget.

🎯 Goal: Build a program that uses the two pointer technique on a sorted array to find all pairs of numbers that sum up to a given target value.

📋 What You'll Learn

Create a sorted array of integers with exact values

Create a target sum variable

Use the two pointer technique with variables left and right to find pairs

Print all pairs that add up to the target sum in the format (num1, num2)

💡 Why This Matters

🌍 Real World

Finding pairs of expenses that match a budget or pairs of items that fit a price limit in shopping apps.

💼 Career

Two pointer technique is a common pattern in coding interviews and helps solve problems efficiently in software development.

1

Create the sorted array

Create a sorted array called numbers with these exact values: 1, 2, 3, 4, 5, 6, 7, 8, 9

DSA C++

// Your code here

Hint

Use an array with the exact values and calculate its size using sizeof.

2

Set the target sum

Create an integer variable called target and set it to 10

DSA C++

int numbers[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int size = sizeof(numbers) / sizeof(numbers[0]);
// Your code here

Hint

Just create a variable named target and assign it the value 10.

3

Implement the two pointer technique

Create two integer variables called left and right. Set left to 0 and right to size - 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, print the pair and move both pointers inward. If the sum is less than target, move left forward. If the sum is greater than target, move right backward.

DSA C++

int numbers[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int size = sizeof(numbers) / sizeof(numbers[0]);
int target = 10;
// 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 pairs found

Make sure your program prints each pair found in the format (num1, num2) on a new line. The output should be exactly as shown.

DSA C++

int numbers[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int size = sizeof(numbers) / sizeof(numbers[0]);
int target = 10;
int left = 0;
int right = size - 1;
while (left < right) {
    int sum = numbers[left] + numbers[right];
    if (sum == target) {
        // Your code here
    } else if (sum < target) {
        left++;
    } else {
        right--;
    }
}

Hint

Printing is already done inside the loop when a pair is found. Make sure the format matches exactly.

Practice

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

easy

A. Use two indices to scan the array from different ends or positions simultaneously

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

C. Sort the array first and then use binary search

D. Use a single pointer to iterate through the array

Solution

1. Step 1: Understand the technique

   The Two Pointer Technique uses two indices to traverse the array, often from opposite ends or different positions, to solve problems efficiently.

2. Step 2: Compare with other options

   Options B and C describe different methods, and D uses only one pointer, so they do not match the Two Pointer Technique.

3. Final Answer:

   Use two indices to scan the array from different ends or positions simultaneously -> Option A

4. Quick Check:

   Two pointers = two indices scanning array [OK]

Hint: Two pointers move independently to cover array efficiently [OK]

Common Mistakes:

• Thinking it always requires sorting first
• Confusing with nested loops approach
• Using only one pointer instead of two

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

easy

A. int left = n - 1, right = 0;

B. int left = 1, right = n;

C. int left = n, right = 0;

D. int left = 0, right = n - 1;

Solution

1. Step 1: Understand pointer initialization

   For two pointers scanning from start and end, left starts at 0 and right at n-1 (last index).

2. Step 2: Check each option

   int left = 0, right = n - 1; correctly sets left=0 and right=n-1. Options A, B, and C have incorrect indices that may cause out-of-bound errors.

3. Final Answer:

   int left = 0, right = n - 1; -> Option D

4. Quick Check:

   Start = 0, End = n-1 [OK]

Hint: Start left at 0 and right at n-1 for full array scan [OK]

Common Mistakes:

• Setting right pointer to n (out of bounds)
• Swapping left and right initial values
• Starting left pointer at 1 instead of 0

3. What will be the output of the following C++ code snippet?

int arr[] = {1, 2, 3, 4, 5, 6};
int left = 0, right = 5;
int sum = 0;
while (left < right) {
    sum += arr[left] + arr[right];
    left++;
    right--;
}
std::cout << sum << std::endl;

medium

A. 27

B. 18

C. 21

D. 15

Solution

1. Step 1: Trace the loop iterations

   Initial pointers: left=0, right=5
   Iteration 1: sum=0+arr[0]+arr[5]=0+1+6=7, left=1, right=4
   Iteration 2: sum=7+arr[1]+arr[4]=7+2+5=14, left=2, right=3
   Iteration 3: sum=14+arr[2]+arr[3]=14+3+4=21, left=3, right=2 (loop ends)

2. Step 2: Final sum value

   Sum after loop is 21.

3. Final Answer:

   21 -> Option C

4. Quick Check:

   Sum pairs (1+6)+(2+5)+(3+4) = 21 [OK]

Hint: Add pairs from ends moving inward until pointers cross [OK]

Common Mistakes:

• Including middle element when pointers cross
• Adding elements twice
• Incorrect loop condition (<= instead of <)

4. Identify the error in the following code that uses the two pointer technique to find if a sorted array contains two numbers that sum to a target value.

bool hasPairSum(int arr[], int n, int target) {
    int left = 0, right = n - 1;
    while (left <= right) {
        int sum = arr[left] + arr[right];
        if (sum == target) return true;
        else if (sum < target) left++;
        else right--;
    }
    return false;
}

medium

A. The loop condition should be left < right instead of left <= right

B. The pointers should be initialized as left = 1 and right = n

C. The sum calculation should be arr[left] - arr[right]

D. The function should return false inside the loop

Solution

1. Step 1: Analyze loop condition

   The loop runs while left <= right, which allows pointers to be equal, causing sum of same element twice.

2. Step 2: Correct loop condition

   Changing condition to left < right ensures pointers do not overlap, avoiding incorrect sum calculation.

3. Final Answer:

   The loop condition should be left < right instead of left <= right -> Option A

4. Quick Check:

   Pointers must not cross or overlap [OK]

Hint: Use left < right to avoid using same element twice [OK]

Common Mistakes:

• Allowing pointers to be equal causing wrong sums
• Incorrect pointer initialization
• Returning false too early inside loop

5. Given a sorted array arr of size n, how can you use the two pointer technique to find all unique pairs that sum up to a target value t? Which approach is correct?

hard

A. Use nested loops to check every pair and store unique pairs in a set

B. Initialize left = 0 and right = n - 1, move pointers inward, skip duplicates after finding a pair

C. Sort the array, then use binary search for each element to find complement

D. Initialize left = 1 and right = n, move pointers outward

Solution

1. Step 1: Set pointers and move inward

   Start with left at 0 and right at n-1. Move pointers inward based on sum compared to target.

2. Step 2: Skip duplicates to find unique pairs

   After finding a pair, increment left and decrement right while skipping duplicate values to avoid repeated pairs.

3. Final Answer:

   Initialize left = 0 and right = n - 1, move pointers inward, skip duplicates after finding a pair -> Option B

4. Quick Check:

   Two pointers inward + skip duplicates = unique pairs [OK]

Hint: Skip duplicates after finding pairs to avoid repeats [OK]

Common Mistakes:

• Not skipping duplicates causing repeated pairs
• Using nested loops causing higher time complexity
• Incorrect pointer initialization or movement