Category: C++ Coding Test Tutorials

In this lecture, we will cover the second problem of calculating the range sum. The range sum problem is a very important topic in algorithms and data structures, and it is useful in various situations. We will learn how to efficiently calculate the sum of a given range using an algorithm.

Problem Description

Given an array as follows, there will be several pairs of starting and ending indices provided. Write a program to calculate the range sum for each pair.

Input

• The first line contains the size of the array N (1 ≤ N ≤ 100,000) and the number of queries M (1 ≤ M ≤ 100,000).
• The second line contains N integers that are the elements of the array. Each element’s value is between -10,000 and 10,000 inclusive.
• Subsequently, M lines are given, each containing two integers S and E, which signify calculating the sum of the range [S, E]. (S, E use 1-based indexing.)

Output

Print the range sum for each query on a new line.

Problem Analysis

The problem of calculating the range sum is very inefficient, as directly calculating the sum results in a time complexity of O(N), making it O(N*M) when M is large. Therefore, in such cases, we can improve the calculation to O(1) using the prefix sum technique.

Algorithm

1. Create a prefix sum array.
2. For each query, use the start and end indices to calculate the range sum in O(1) time.

Implementation

Now let’s implement this algorithm in C++.

#include <iostream>
#include <vector>

using namespace std;

int main() {
    int N, M;
    cin >> N >> M;
    
    vector<int> arr(N + 1, 0);
    vector<int> prefixSum(N + 1, 0);

    // Input the array
    for (int i = 1; i <= N; i++) {
        cin >> arr[i];
        prefixSum[i] = prefixSum[i - 1] + arr[i]; // Calculate the prefix sum
    }

    // Processing queries
    for (int i = 0; i < M; i++) {
        int S, E;
        cin >> S >> E;
        // Calculate the range sum
        cout << prefixSum[E] - prefixSum[S - 1] << endl;
    }

    return 0;
}
    

Code Explanation

Each part of the code performs the following tasks:

• Input: Read N and M, then read N numbers and save them in the array arr.
• Create Prefix Sum Array: Generate a prefixSum array to calculate and store the sum up to the i-th element.
• Process Queries: For each query, read S and E, and compute the range sum in O(1) time using the prefixSum.

Complexity Analysis

The time complexity of this algorithm is as follows:

• Creating the prefix sum array: O(N)
• Processing each query: O(M)
• Total time complexity: O(N + M)

Conclusion

In this lecture, we learned an efficient way to calculate the range sum. By using the prefix sum technique, we were able to avoid explicit looping and significantly reduce the time complexity. This technique can be applied to many problems, so it is important to master it.

In the next lecture, we will cover more complex range sum problems using different data structures. Thank you!

Hello! In this session, we will enhance our algorithm problem-solving skills through the ‘Range Sum’ problem, which frequently appears in C++ coding tests. The Range Sum problem aims to find the sum of elements in a specific range of a given array, and we will explore various methods and optimization techniques to solve this problem.

Problem Description

We are given an integer array A and an integer M. For each element in A, we need to find the sum of the range starting from A[i] in the form of A[i] + A[i+1] + ... + A[j]. Below are the specific requirements of the problem.

Problem

Given an integer array A and an integer M, implement a function to find the sum of M consecutive elements starting from A[i].

Input

• The first line contains the length of the array N. (1 ≤ N ≤ 100,000)
• The second line contains N integers forming the array A. (−1,000,000,000 ≤ A[i] ≤ 1,000,000,000)
• The third line contains the integer M. (1 ≤ M ≤ N)

Output

Output the range sum. If the sum of M elements starting from A[i] is possible, print that sum; otherwise, print -1.

Solution Process

There are various approaches to solve this problem, but the most common and efficient method is using the Sliding Window technique. By using the sliding window technique, we can solve the problem with a time complexity of O(N).

Explanation of the Sliding Window Approach

In the sliding window technique, we set the start and end of the range like a window and calculate the value of that range by moving the window. After setting the initial range, we update the sum by adding the next element and subtracting the starting element. This reduces unnecessary calculations and increases efficiency.

Code Implementation

Now let’s implement our approach in C++.

#include <iostream>
#include <vector>
using namespace std;

// Function: Calculate range sum
long long rangeSum(const vector<int> &A, int M) {
    long long sum = 0; // Initial sum variable
    int N = A.size(); // Array size

    // Calculate the sum of the first range
    for (int i = 0; i < M; i++) {
        sum += A[i];
    }
    
    // Print the sum of the first range
    cout << "Range Sum: " << sum << endl;

    // Update the range sum using the sliding window
    for (int i = M; i < N; i++) {
        sum += A[i] - A[i - M]; // Subtract the old element and add the new element
        cout << "Range Sum: " << sum << endl; // Print the updated range sum
    }
    
    return sum; // Final return value
}

int main() {
    int N, M;
    cout << "Length of the array N: ";
    cin >> N; // Input array size
    vector<int> A(N);

    cout << "Integer array A: ";
    for (int i = 0; i < N; i++) {
        cin >> A[i]; // Input array elements
    }

    cout << "Length of the range M: ";
    cin >> M; // Input length of range sum

    long long totalSum = rangeSum(A, M); // Call the range sum function
    if (N < M) {
        cout << "-1" << endl; // Print -1 if range sum is not possible
    } else {
        cout << "Total of range sums: " << totalSum << endl; // Print the total sum
    }

    return 0;
}
    

Code Explanation

The above code shows the process of calculating the range sum as follows:

1. First, we input the array A of size N and the range length M.
2. To calculate the initial range sum, we sum the first M elements.
3. By applying the sliding window technique, we get the next range sum by subtracting the existing element and adding the new element.
4. Output all range sums.

Conclusion

In this session, we solved the problem of calculating the range sum using C++. The sliding window technique is a very simple yet efficient method, making it useful for solving various problems. Familiarizing yourself with techniques like this will greatly help you prepare for coding tests.

In the next lecture, we will cover more diverse problems and help deepen your understanding of algorithms. Until then, practice and best of luck!

The range sum problem is an algorithm problem that efficiently calculates the sum of a specific range within a given array. This problem frequently appears in various programming languages, particularly with important implementations in C++. In this tutorial, we will set a specific problem for the range sum problem and explain the process to solve it in detail.

1. Problem Description

Let’s assume we are given an array composed of integers. The size of the integer array A is N. We need to calculate the sum of specific ranges in the array through M queries. The starting index of the range is provided as L, and the ending index is provided as R. We need to calculate A[L] + A[L+1] + ... + A[R].

For example, suppose the array A is as follows:

A = [10, 20, 30, 40, 50]

And let’s assume the following queries are given:

1. L = 1, R = 3
2. L = 0, R = 4
3. L = 2, R = 2

Calculating the range sum for the given queries yields the following results:

1. A[1] + A[2] + A[3] = 20 + 30 + 40 = 90
2. A[0] + A[1] + A[2] + A[3] + A[4] = 10 + 20 + 30 + 40 + 50 = 150
3. A[2] = 30

2. Problem Solving Approach

The traditional method to solve the range sum problem is to iterate through the array directly. However, this can be inefficient. For example, if each query among M queries spends O(R - L + 1) time, in the worst case, it will take O(N * M) time. To improve this, we can use the following method.

2.1. Preprocessing Method – Cumulative Sum Array

To quickly calculate the sum of all queries, we can utilize a cumulative sum array. First, we define the cumulative sum array S for the given array A. The cumulative sum array is calculated as follows:

S[i] = S[i - 1] + A[i]

Here, S[0] is initialized to 0. Using the cumulative sum array, the sum of a range can be calculated easily as follows:

sum(L, R) = S[R] - S[L - 1]

Now we can calculate the range sum with a time complexity of O(1). Below is a C++ solution that uses the cumulative sum array.

3. C++ Code Example

#include 
#include 
using namespace std;

int main() {
    int N, M;
    cin >> N >> M;

    vector A(N);
    vector S(N + 1, 0); // S[0] is initialized to 0

    for (int i = 0; i < N; i++) {
        cin >> A[i];
        S[i + 1] = S[i] + A[i]; // Calculate cumulative sum
    }

    for (int i = 0; i < M; i++) {
        int L, R;
        cin >> L >> R;
        // Output the range sum
        cout << S[R] - S[L - 1] << endl;
    }
    return 0;
}

4. Code Explanation

Let’s explain the above code step by step:

1. Input Gathering: We receive the size of the array N and the number of queries M, and we input the values of the array A.
2. Cumulative Sum Array Creation: We calculate the cumulative sum for each index and store it in the S array. This array helps us easily find the sums of the values of the original array A.
3. Query Processing: We input the starting and ending indices L and R for each query, and we calculate and output the sum of the range using the cumulative sum array S.

5. Performance Analysis

The time complexity of the above C++ code is O(N) for calculating the cumulative sum of the array A and O(1) for processing each query, making the overall time complexity O(N + M). This is a very efficient method, allowing for quick results even with thousands of queries.

6. Various Modified Problems

The range sum problem can have various modifications. For example:

• Range Minimum/Maximum: It can be modified to find the minimum or maximum value for a given parameter.
• Range Updates: In cases where a specific range of values needs to be changed and the range sum must be recalculated. In this case, segment trees can be utilized.
• 2D Range Sum: A problem of calculating the sum of a specific range in a 2D array, where double cumulative sums can be used.

7. Conclusion

The range sum problem is one of the very useful techniques in programming and provides significant efficiency, especially when large-scale data processing is required. In this tutorial, we learned how to solve the range sum problem using a cumulative sum array. We can further enhance this technique through various modified problems. Practice solving various problems to build experience and achieve high scores on C++ coding tests.

As we conclude the tutorial, please leave any additional questions or requests in the comments.

Understanding data structures and algorithms is an essential element for growth as a developer. C++ is a high-performance programming language that is widely used to solve algorithm problems, especially in coding tests. In this course, we will address the problem of ‘Calculating the Product of an Interval’ and explain various approaches to solve it.

Problem Definition

Let’s consider the problem of quickly calculating the product of a specified interval for a given array and several queries.

Problem:

There is a given integer array A and the number of queries Q.
A[i] represents the i-th element of the array, and 1 ≤ i ≤ N. Each query contains two numbers L and R, and the goal is to calculate the value of A[L] × A[L+1] × ... × A[R] when L ≤ R.

Input:

• Size of the array N (1 ≤ N ≤ 10⁵)
• Integer array A[1...N] (1 ≤ A[i] ≤ 10⁹)
• Number of queries Q (1 ≤ Q ≤ 10⁵)
• Each query contains two integers L and R (1 ≤ L ≤ R ≤ N)

Output:

Print the product of the interval for each query, one per line.

Problem Approach

Several approaches can be considered to solve this problem.

The simplest method is to sequentially calculate the product of the given interval for each query.
However, this method is inefficient as it takes O(N * Q) time in the worst-case scenario.

The second approach is to use a Segment Tree.
The Segment Tree is effective for calculating sums, products, or maximum/minimum values for intervals in a short amount of time.

Each node of the Segment Tree stores the product of its segment, allowing us to calculate the interval product in O(log N) time.

Structure of the Segment Tree

The Segment Tree is structured like a binary tree, with each node storing information about a particular interval of the array.
Typically, a Segment Tree can perform the following main operations:

1. Interval product query: query() function
2. Element update: update() function

Final Code Implementation

Below is the C++ code for calculating the interval product using a Segment Tree:

#include <iostream>
#include <vector>
using namespace std;

// Definition of the Segment Tree class
class SegmentTree {
private:
    vector tree; // Storage for the tree
    int size; // Size of the original array
    
    // Function to calculate the product of a sub-array
    long long buildTree(const vector& arr, int node, int start, int end) {
        if(start == end) {
            return tree[node] = arr[start];
        }
        int mid = (start + end) / 2;
        return tree[node] = buildTree(arr, node*2, start, mid) * buildTree(arr, node*2+1, mid+1, end);
    }

public:
    // Constructor
    SegmentTree(const vector& arr) {
        size = arr.size();
        tree.resize(4 * size);
        buildTree(arr, 1, 0, size - 1);
    }

    // Function to get the product of a specific interval
    long long query(int node, int start, int end, int L, int R) {
        if(R < start || end < L) return 1; // Multiplication identity
        if(L <= start && end <= R) return tree[node]; // Range of the whole node
        int mid = (start + end) / 2;
        return query(node*2, start, mid, L, R) * query(node*2 + 1, mid + 1, end, L, R);
    }

    // Function callable from outside
    long long query(int L, int R) {
        return query(1, 0, size - 1, L, R);
    }
};

int main() {
    int N, Q;
    cin >> N; // Size of the array
    vector A(N);
    for(int i = 0; i < N; i++) {
        cin >> A[i]; // Input array elements
    }
    SegmentTree segtree(A); // Creating the Segment Tree

    cin >> Q; // Number of queries
    for(int i = 0; i < Q; i++) {
        int L, R;
        cin >> L >> R; // Input for query range
        cout << segtree.query(L - 1, R - 1) << endl; // Print result (converting to 0-based index)
    }
    return 0;
}
    

Code Explanation

• buildTree() function is used to construct the Segment Tree.
• query() function calculates the product for a given interval.
• main() function receives inputs and outputs the processed interval products.

Time Complexity

The time to construct the Segment Tree is O(N), and the response time for each query is O(log N).
Therefore, the worst-case time complexity of the entire algorithm is O(N + Q log N).
This allows for efficient processing of large input data.

Conclusion

Choosing the appropriate data structure and algorithm for problems presented in coding tests is very important.
In this course, we addressed the problem of calculating the interval product using a Segment Tree, and explored what an efficient method entails.
I hope to enhance coding skills by solving algorithm problems and will continue to cover various topics in the future.

Author: [Your Name]

Date: [Date]

1. Problem Description

The number of stairs refers to the combinations of ways to climb stairs by choosing either to go up or down. The given problem is as follows.

Given an integer N, write a program to calculate the number of ways to climb N stairs.

However, you can climb either one or two stairs at a time, and the last stair number must end with a digit from 1 to 9.

For example, the ways to climb 4 stairs are as follows:

• 1111 (1+1+1+1)
• 112 (1+1+2)
• 121 (1+2+1)
• 211 (2+1+1)
• 22 (2+2)

Considering the above cases, you need to write code to correctly calculate the number of stairs.

2. Problem Analysis

Let us analyze some important points to solve the problem.

• State Definition: The Nth stair is composed differently depending on the state of the previous stairs. For example, look at the counts for N=1 and N=2.
• Recurrence Relation: The way to climb N stairs is the sum of the number of ways to climb 1 stair from N-1 stairs and the number of ways to climb 2 stairs from N-2 stairs. The last stair number must be one of the digits from 1 to 9.
• Result: The number of ways to climb N stairs is the sum of all the counts of each case.

3. Deriving the Recurrence Relation

The general recurrence relation can be expressed as follows:

dp[i][j] = dp[i-1][j-1] + dp[i-2][j]
            

Here, dp[i][j] represents the number of ways ending with j at the i-th stair. The initial conditions can be set as dp[1][1] = 1, dp[1][2] = 1 … dp[1][9] = 1. Based on this, we can write the following code.

4. Code Design

Now let’s write code in C++. This code takes the given N as an input and prints the corresponding number of ways to climb stairs.

#include <iostream>

using namespace std;

const int MOD = 1000000000; // MOD setting for large number processing
int dp[100][10]; // dp table

void init() {
    for (int i = 1; i <= 9; i++) {
        dp[1][i] = 1; // Number of ways to climb one stair
    }
  
    for (int i = 2; i <= 100; i++) {
        for (int j = 0; j <= 9; j++) {
            // Case ending with 0 (j == 0)
            if (j == 0) {
                dp[i][0] = dp[i - 1][1] % MOD; 
            }
            // Case ending with 9 (j == 9)
            else if (j == 9) {
                dp[i][9] = dp[i - 1][8] % MOD; 
            }
            // Other cases
            else {
                dp[i][j] = (dp[i - 1][j - 1] + dp[i - 1][j + 1]) % MOD;
            }
        }
    }
}

int main() {
    int N;
    cin >> N; // Input number of stairs
    init(); // Initialize dp table

    long long total = 0;
    for (int i = 0; i <= 9; i++) {
        total = (total + dp[N][i]) % MOD; // Calculate total number of cases
    }

    cout << total << endl; // Output result
    return 0;
}
            

5. Code Explanation

The above code is written in a dynamic programming style to calculate the number of stairs.

• Input Handling: It takes N as input from the user to calculate the number of stairs.
• Table Initialization: The first row of the dp table is initialized to set each ending digit to 1.
• Applying the Recurrence Relation: It calculates the possible counts for each number of stairs and saves the values divided by MOD.
• Result Calculation: It sums all counts from 1 to 9 and outputs the final result.

6. Complexity Analysis

The time complexity of this algorithm is O(N), which takes linear time for the given N. The space complexity also has a size of O(N). This ensures efficiency in both memory and time.

7. Conclusion

The stair number problem is suitable for learning the basic concepts of dynamic programming. Performing this problem greatly aids in understanding recursive thinking and the concept of memoization. By solving this problem through the C++ Programming Language, you can further strengthen your algorithm problem-solving skills.

Hope this course helps you in preparing for the C++ coding test! Learn it, and try solving more problems!