Why Two Pointer Technique Beats Brute Force in DSA C - See It Work

Why Two Pointer Technique Beats Brute Force

📖 Scenario: Imagine you have a list of numbers representing the prices of items in a store. You want to find two items that together cost exactly a certain amount of money. This is a common problem in shopping apps and budgeting tools.

🎯 Goal: You will build a simple program that finds two numbers in a sorted list that add up to a target value. First, you will try the slow way (brute force), then you will use the faster two pointer technique. You will see why the two pointer method is better.

📋 What You'll Learn

Create an array of integers called prices with the exact values: 2, 7, 11, 15

Create an integer variable called target and set it to 9

Write a brute force nested loop to find two numbers in prices that add up to target

Write a two pointer technique to find two numbers in prices that add up to target

Print the pairs found by both methods

💡 Why This Matters

🌍 Real World

Finding pairs of numbers that add up to a target is common in shopping apps, budgeting tools, and data analysis.

💼 Career

Understanding efficient search techniques like two pointers helps in software development roles that require optimization and handling large data sets.

1

Create the data array

Create an integer array called prices with these exact values: 2, 7, 11, 15

DSA C

# Your code here

Hint

Use curly braces to list the numbers inside the array.

2

Set the target sum

Create an integer variable called target and set it to 9

DSA C

int prices[] = {2, 7, 11, 15};
# Your code here

Hint

Use simple assignment to set the target value.

3

Write brute force nested loops

Write nested for loops with variables i and j to check every pair in prices and print the pair if their sum equals target

DSA C

int prices[] = {2, 7, 11, 15};
int target = 9;
// Your code here

Hint

Use two loops to check all pairs. Print when sum matches target.

4

Write two pointer technique and print result

Use two integer variables left and right starting at the beginning and end of prices. Move pointers inward to find and print the pair that sums to target

DSA C

int prices[] = {2, 7, 11, 15};
int target = 9;
int n = 4;
for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
        if (prices[i] + prices[j] == target) {
            printf("Brute force found: %d + %d = %d\n", prices[i], prices[j], target);
        }
    }
}
// Your code here

Hint

Use left and right pointers. Move left forward if sum is less than target, else move right backward.