Author: root

Hello! In this tutorial, we will explore the “range sum” problem, which is frequently featured in JavaScript coding tests, in depth. The range sum problem involves efficiently calculating the sum of a specific range within a given sequence, and it can be solved using various optimization techniques. The range sum problems we will discuss can be particularly time-consuming when the size of the array is large, so it’s essential to design a more efficient algorithm.

Problem Introduction

Here is a problem related to range sums:

Problem Description

An array of integers arr and several pairs of integers (l, r) are given. Each pair represents the starting point l and endpoint r of a range. Write a program to calculate the sum of arr[l] + arr[l + 1] + ... + arr[r]. The length of the array and the number of pairs are as follows:

• 1 ≤ arr.length ≤ 10⁶
• 1 ≤ l, r ≤ arr.length
• 0 ≤ arr[i] ≤ 10⁹

For example, if the array is arr = [1, 2, 3, 4, 5] and the pair (1, 3) is given, the range sum is arr[1] + arr[2] + arr[3] = 2 + 3 + 4 = 9.

Approach to the Problem

To solve this problem efficiently, simply using loops to calculate the sum for each range is not appropriate. The reason is that in the worst-case scenario, it would have a time complexity of O(N * M), which can lead to exponential time increases if the size of the data is large. Instead, we can use a more effective approach.

Preprocessing Technique: Prefix Sum Array

One way to solve the range sum problem is to create a Prefix Sum Array. Using a prefix sum array allows us to calculate the sum of a range in O(1) time. The definition of the prefix sum array is as follows:

prefix[i] = arr[0] + arr[1] + ... + arr[i-1]

Thus, the sum of the range (l, r) can be calculated as:

sum(l, r) = prefix[r + 1] - prefix[l]

This allows us to compute the range sum for each pair in O(1) time. The overall time complexity of the algorithm is O(N + M), where N is the size of the array and M is the number of pairs.

Implementation Steps

Now, let’s implement the JavaScript code that solves the problem using the prefix sum array.

Step 1: Create the Prefix Sum Array

function createPrefixSum(arr) {
    const prefix = new Array(arr.length + 1).fill(0);
    for (let i = 0; i < arr.length; i++) {
        prefix[i + 1] = prefix[i] + arr[i];
    }
    return prefix;
}

Step 2: Calculate the Range Sum

function rangeSum(prefix, l, r) {
    return prefix[r + 1] - prefix[l];
}

Step 3: Implement the Main Function

function calculateRangeSums(arr, queries) {
    const prefix = createPrefixSum(arr);
    const results = [];
    for (let [l, r] of queries) {
        results.push(rangeSum(prefix, l - 1, r - 1)); // 1-based to 0-based
    }
    return results;
}

// Example usage
const arr = [1, 2, 3, 4, 5];
const queries = [[1, 3], [2, 5], [0, 2]];
const results = calculateRangeSums(arr, queries);
console.log(results); // [9, 14, 6]

Results Analysis

The above code is a program that efficiently calculates and outputs the range sum for each query based on the given array and queries. As a result, it can quickly compute range sums and does not suffer performance degradation depending on the size of the array or the number of queries.

Time Complexity

The time complexity of this algorithm is as follows:

• Creating the prefix sum array: O(N)
• Processing each query: O(1) (If you combine processing all M queries, it becomes O(M))

As a result, the overall time complexity is O(N + M), which is very efficient.

Conclusion

Now we have learned how to efficiently solve the range sum problem using a prefix sum array. Since range sum problems are frequently featured topics in coding tests, understanding and utilizing optimization techniques like this is important. Practice solving various types of problems!

Additional Practice Problems

Try practicing the following variant problems:

• Problems that find the maximum value of the range instead of the sum
• Problems that find the product of the range (note that division operations may need to be considered)
• Consider how to handle different queries (e.g., range increment, decrement, etc.)

Based on the above content, I hope you solve various problems. Thank you!

Problem Description

There is height information for a group of students. You need to specify one student among them
and arrange the line so that this student stands at the front.
In this case, students shorter than the specified student must stand behind, and
the order of students with the same height must remain unchanged.
Write an algorithm that satisfies these conditions for lining up the students.

Input Format

    An array containing the heights of students (e.g., [160, 170, 165, 180, 175])
    Height of the specified student (e.g., 170)
    

Output Format

    An array of the lined-up students' heights (e.g., [170, 160, 165, 180, 175])
    

Problem Solving Process

Step 1: Understand the Problem

The crux of the problem is to place the specified height student at the front of the given array, while
sorting the remaining students by height, maintaining their original order. This problem can primarily be solved using stable sorting.
The algorithm we will implement includes the following steps.

Step 2: Analyze a Simple Example

Step 3: Develop a Solution Strategy

The solution method is as follows.

1. Find the student with the specified height in the given array and add that student as the first element of the result array.
2. Add the remaining students to the result array while maintaining their original order, excluding students with the same height.
3. Finally, return the modified array.

Step 4: Implement JavaScript Code

Based on the above strategy, I will write a JavaScript function. This function will take two parameters and
serve to move the specified height student to the front.

function lineUpStudents(students, targetHeight) {
    // Declare an array to store the result
    let result = [];

    // First, add the student corresponding to targetHeight
    const targetStudents = students.filter(height => height === targetHeight);
    result.push(...targetStudents);

    // Add the remaining students (maintaining original order)
    const otherStudents = students.filter(height => height !== targetHeight);
    result.push(...otherStudents);

    return result;
}
        

Step 5: Test and Validate the Code

I will run a few test cases to confirm that the function works correctly.

console.log(lineUpStudents([160, 170, 165, 180, 175], 170)); // [170, 160, 165, 180, 175]
console.log(lineUpStudents([160, 160, 165, 170, 180], 160)); // [160, 160, 165, 170, 180]
console.log(lineUpStudents([180, 170, 160, 150], 160)); // [160, 180, 170, 150]
        

Step 6: Complexity Analysis

The time complexity of this algorithm is O(n). This is because we iterate through the given array once.
The space complexity is also O(n) since a separate result array is created.

Conclusion

In this tutorial, we learned how to solve the problem of lining up students based on their height information
using JavaScript.
Maintaining a stable sort by height was the key to this problem.
Such problems are very important as they frequently appear in coding tests.
Practicing various variations of this problem can also be good preparation for developing algorithmic thinking.

Coding tests are becoming mandatory for more and more companies, and JavaScript is one of the most popular languages in web development.
In this course, we will solve commonly occurring algorithm problems found in coding tests using JavaScript.
Today’s topic is the problem ‘Fast Travel with a Time Machine’.

Problem Description

You are a scientist who can operate a time machine. The time machine can move to a specific time, and
given two integers a and b, you need to calculate the minimum time (number of moves) it takes to move from a to b.
The time machine follows these rules:

• You can add 1 or subtract 1 from your current position.
• You can double your current position.

Write a function that calculates the minimum number of moves required to travel from a to b for given a and b.

Input Format

– Two integers a (0 ≤ a ≤ 10⁵), b (0 ≤ b ≤ 10⁵) are given.

Output Format

– Output the minimum number of moves required to travel from a to b as an integer.

Examples

        Input: a = 2, b = 9
        Output: 4
    

        Input: a = 5, b = 22
        Output: 7
    

Approach to Solve the Problem

To solve this problem, we will use BFS (Breadth-First Search). BFS is an algorithm suitable for finding the shortest path,
exploring all possible states from the given state to find the shortest move sequence to reach the target state.
In this case, each state represents a time the time machine could occupy.

Step 1: Prepare to Use BFS Algorithm

To implement BFS, we will use a queue. First, we add the starting position a to the queue.
Then, we will repeat adding all possible moves to the queue until we reach the target position b.
The possible moves from each position are as follows:

• Add 1 to the current position
• Subtract 1 from the current position
• Double the current position

We will keep track of the number of moves taken for each move until we reach the target position, and output the count when we reach the target.

Step 2: Implement the Code

function minimumMoves(a, b) {
    if (a >= b) return a - b; // If a is greater than or equal to b, determine moves by simply subtracting the difference
    const queue = [[a, 0]]; // [current position, number of moves]
    const visited = new Set(); // Record visited positions
    visited.add(a); // Mark the starting position as visited

    while (queue.length > 0) {
        const [current, moves] = queue.shift(); 

        // Possible moves
        const nextMoves = [current - 1, current + 1, current * 2]; 

        for (let next of nextMoves) {
            if (next === b) return moves + 1; // Return move count when the target position is reached
            if (next < 0 || next > 100000 || visited.has(next)) continue; // Only for valid range and unvisited positions
            visited.add(next); // Mark the next position as visited
            queue.push([next, moves + 1]); // Add the next position and move count to the queue
        }
    }
}
    

Step 3: Analyze Algorithm Complexity

The time complexity of the algorithm is O(n).
In the worst case, we need to explore all possible positions, which corresponds to O(n),
and the space complexity is also O(n). This complexity arises from the space needed for the queue and visited records.

Step 4: Optimization Possibilities

This algorithm is already based on BFS, which explores the shortest path
and does not require additional optimization. However, depending on the situation,
DFS (Depth-First Search) could also be applied, but BFS is more effective for this problem.

Step 5: Conclusion

Through the ‘Fast Travel with a Time Machine’ problem, we have briefly understood the principle of BFS and learned how to solve a problem using JavaScript.
In this way, various problems can be solved, and it is crucial to master basic algorithms to achieve good results in coding tests.

Additional Resources

– To further study the BFS algorithm and practice various problem-solving, we recommend the following resources.

• BFS Algorithm on GeeksforGeeks
• LeetCode Problem Set
• HackerRank – 10 Days of Java Tutorial