Python Coding Test Course, Finding Range Sum 3

Problem Description

Given an integer array A and several pairs of integers L and R, write a program to find the sum of the interval from index L to index R for each pair.

For example, if A = [1, 2, 3, 4, 5] and the interval pairs are (1, 3), (0, 2), (2, 4),
the results should be 9, 6, and 12 respectively.

Input Format

    - The first line contains an integer N (1 ≤ N ≤ 100,000): size of array A
    - The second line contains N integers A[i] (1 ≤ A[i] ≤ 10,000).
    - The third line contains an integer M (1 ≤ M ≤ 100,000): number of interval pairs.
    - The following M lines contain the interval pairs L, R (0 ≤ L <= R < N).
    

Output Format

    - Output the sum of each interval over M lines.
    

Problem Solving Strategy

While it is possible to solve this problem using simple loops,
the worst-case scenario has N and M up to 100,000, making O(N * M) time complexity impossible.
Therefore, a method with O(N + M) time complexity is needed.

To achieve this, it is useful to create a prefix sum array as a preprocessing step.
Using a prefix sum array allows for quick calculation of each interval’s sum.

Prefix Sum Array Description

First, calculate the prefix sum of array A to create the prefix_sum array.
prefix_sum[i] stores the sum from A[0] to A[i].
Thus, the sum from index L to index R can be calculated as follows:

sum(L, R) = prefix_sum[R] – prefix_sum[L – 1], L > 0

sum(0, R) = prefix_sum[R], L = 0

Code Implementation

    
def compute_prefix_sum(A):
    prefix_sum = [0] * len(A)
    prefix_sum[0] = A[0]
    for i in range(1, len(A)):
        prefix_sum[i] = prefix_sum[i - 1] + A[i]
    return prefix_sum

def range_sum(prefix_sum, L, R):
    if L == 0:
        return prefix_sum[R]
    else:
        return prefix_sum[R] - prefix_sum[L - 1]

def main():
    N = int(input())
    A = list(map(int, input().split()))
    M = int(input())
    
    prefix_sum = compute_prefix_sum(A)

    results = []
    for _ in range(M):
        L, R = map(int, input().split())
        results.append(range_sum(prefix_sum, L, R))
    
    for result in results:
        print(result)

if __name__ == "__main__":
    main()
    
    

Code Explanation

1. The compute_prefix_sum function calculates the prefix sum of the input array A and
returns the prefix_sum array. It initializes the first value and calculates each index’s value by adding the previous value to the current value.

2. The range_sum function quickly calculates the sum of the interval using the prefix sum array
for the given L and R. If L is 0, it returns prefix_sum[R]; otherwise, it calculates the result by subtracting
prefix_sum[L-1] from prefix_sum[R].

3. The main function handles input and calls the range_sum function for each interval pair to display the results.

Time Complexity Analysis

– It takes O(N) time to calculate the prefix sum array.

– Each of the M queries takes O(1) time.
Thus, the overall time complexity is O(N + M).

Conclusion

In this lecture, we covered an efficient approach to finding interval sums.
Utilizing prefix sums allows for reduced time complexity, enabling quick processing even for large inputs.
This technique is useful in various algorithm problems, so it’s important to keep it in mind.

Additional Practice Problems

  • 1. Change the example array A and compute the interval sums.
  • 2. Research methods to calculate interval sums using other algorithms (Segment Tree or Fenwick Tree).
  • 3. Practice problems involving updating the value at a specific index in the array and recalculating the total interval sums.

References

– Competitive Programming Problem Sets

– Materials related to online coding tests

– Interval sum problems from LeetCode and HackerRank

Python Coding Test Course, Finding Interval Sums 2

Hello! Welcome to the Python coding test course. In this course, we will
cover the “Interval Sum Problem 2”. Efficiently calculating the interval sum
is very important in solving algorithm problems.
Below, we will look at the problem description and solution process step by step.

Problem Description

Given N integers A[1], A[2], …, A[N],
Q queries are provided. Each query consists of two integers L and R,
and the goal is to find the sum from A[L] to A[R].
The problem is as follows:

Input Format:
The first line contains the integers N (1 ≤ N ≤ 100,000) and Q (1 ≤ Q ≤ 100,000).
The second line contains N integers separated by spaces. |A[i]| ≤ 1,000,000
The Q queries are given as follows: each query consists of two integers L and R (1 ≤ L ≤ R ≤ N).

Output Format:
Print the sum from A[L] to A[R] for each query.

How to Solve the Problem Efficiently

To solve such problems, it is efficient to use the prefix sum
instead of calculating the sum for each query.
The prefix sum allows you to calculate the sum of a specific interval in constant time O(1).

Calculating Prefix Sum

The method to calculate the prefix sum is as follows:

  • Create a prefix sum array S, where S[i] represents the sum from A[1] to A[i].
  • Initialize S[0] = 0. (For convenience, 0 is added so that the calculation can be done by subtracting S[L-1] from S[i].)
  • Then calculate S[i] as follows: S[i] = S[i-1] + A[i]

This allows us to find the interval sum with a single subtraction operation as S[R] – S[L-1].

Solution Code

        
def calculate_prefix_sum(arr):
    n = len(arr)
    prefix_sum = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix_sum[i] = prefix_sum[i - 1] + arr[i - 1]
    return prefix_sum

def range_sum(prefix_sum, L, R):
    return prefix_sum[R] - prefix_sum[L - 1]

N, Q = map(int, input().split())
A = list(map(int, input().split()))
prefix_sum = calculate_prefix_sum(A)

for _ in range(Q):
    L, R = map(int, input().split())
    print(range_sum(prefix_sum, L, R))
        
        

The above code performs the following steps:

  • First, it calls a function to calculate the prefix sum array for the given array A of length N.
  • For each query, it receives L and R, and outputs the sum for that interval.

Time Complexity Analysis

Analyzing the time complexity of this problem yields the following:

  • It takes O(N) time to calculate the prefix sum array.
  • It takes O(1) time to calculate the interval sum for each query.
  • Thus, the overall time complexity is O(N + Q).

Conclusion

The interval sum problem is one of the frequently asked questions in coding tests.
Using prefix sums is a good method to solve the problem efficiently.
Based on what we have covered in this course, it would also be good to try solving various variations of the problem.
If you have any additional questions or want to know more about other algorithm problems, feel free to leave a comment!

Python Coding Test Course, Finding Interval Sum 1

1. Problem Definition

The interval sum problem, which is frequently asked in coding tests, is about efficiently calculating the sum of elements in a specific range within a given array. In this post, we will discuss the ‘Calculating Interval Sum 1’ problem.

Problem Description

An integer N and M are given, along with an array consisting of N integers. Next, M queries are provided, each consisting of two integers S and E. The task is to calculate the sum from S to E. Note that S and E range from 1 to N.

2. Input and Output Format

Input

  • The first line contains N and M separated by a space. (1 ≤ N, M ≤ 100,000)
  • The second line contains N integers separated by spaces. (Each integer is between 1 and 100,000)
  • From the third line onward, M lines contain two queries. (S, E)

Output

  • Output the answer for each query over M lines.

3. Sample Input and Output

Sample Input

    5 3
    1 2 3 4 5
    1 3
    2 4
    3 5
    

Sample Output

    6
    9
    12
    

4. Approach to the Problem

To calculate the interval sum, the simplest method can be used first. However, since the maximum values of N and M are 100,000, it is impossible with a time complexity of O(N * M). Therefore, we need to find an efficient way to calculate the interval sum.

4.1. Simple Approach

The simplest method is to iterate from S to E to calculate the sum. However, this method has a complexity of O(N * M).

4.2. Utilizing Cumulative Sum Array

One of the efficient approaches is to use a Cumulative Sum array. By creating the cumulative sum array first, we quickly calculate the answer for each query using the values of S and E.

When defining the cumulative sum array, the i-th element of the array represents the sum from 1 to i. That is, the i-th value of the cumulative sum array is calculated as:

    sum[i] = sum[i - 1] + array[i - 1]
    

When processing queries, the following formula can be used:

    result = sum[E] - sum[S - 1]
    

5. Algorithm Implementation

Now let’s implement the algorithm using the cumulative sum array described above.

5.1. Python Code

def solve(N, M, array, queries):
    # Create cumulative sum array
    sum_array = [0] * (N + 1)
    
    # Calculate cumulative sum
    for i in range(1, N + 1):
        sum_array[i] = sum_array[i - 1] + array[i - 1]
    
    result = []
    
    # Process queries
    for S, E in queries:
        query_sum = sum_array[E] - sum_array[S - 1]
        result.append(query_sum)
    
    return result

# Sample input
N, M = 5, 3
array = [1, 2, 3, 4, 5]
queries = [(1, 3), (2, 4), (3, 5)]

# Output results
output = solve(N, M, array, queries)
for res in output:
    print(res)
    

6. Code Explanation

The above code works in the following order:

  1. First, the cumulative sum array is initialized. The size of the cumulative sum array is set to N + 1 to easily calculate the sum from 1 to N.
  2. Next, each element of the array is iterated to calculate the cumulative sum.
  3. While going through the queries, the sum for the given S and E is quickly calculated using the cumulative sum array.
  4. The calculated results are stored in a list to be output at the end.

7. Time Complexity Analysis

The time complexity of the above algorithm is as follows:

  • O(N) to create the cumulative sum array
  • O(M) to process M queries

Thus, the overall time complexity is O(N + M), allowing us to solve the problem efficiently.

8. Space Complexity Analysis

The space complexity is O(N) to store the cumulative sum array. Additional variables use constant space, so the overall space complexity is O(N).

9. Conclusion

In this post, we discussed the interval sum problem. We learned how to solve the problem efficiently using a cumulative sum array and provided a Python code example to aid understanding. This method can be applied in various problems requiring interval sums and can serve as a basis to tackle more complex issues.

10. Additional Exercises

Now, try to solve some exercises on your own. Below are a few additional practice problems.

  • Write a program to calculate the interval sum from 1 to 100.
  • Implement a program that calculates the sum of the interval each time an element changes.
  • Write a program to calculate the sum of a specific area in a two-dimensional array.

11. Questions and Feedback

If you have any questions or feedback regarding the content discussed in this article, feel free to leave a comment. Let’s solve problems together!

Python coding test course, interval sum

Hello, everyone! Today we will tackle the range sum problem, which frequently appears in coding tests using Python. The range sum problem asks how to quickly calculate the sum of a specific range, and it’s an important concept that forms the basis of many algorithmic problems. We will explore various approaches to solve this problem.

Problem Description

Given an array as follows, write a program to efficiently calculate the sum of specific ranges for multiple query pairs.

Array: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Query: [(1, 3), (2, 5), (3, 7)]

When the above array and queries are provided, the queries are given in the form of (start index, end index). For example, the query (1, 3) means that we need to calculate the sum from index 1 to index 3 of the array (1-based index). In this case, it should be 2 + 3 + 4 = 9.

Approach to the Problem

There are various basic approaches to solve the range sum problem. I will start with the simplest method and explain more efficient ones.

1. Using Simple Iteration

The most intuitive method is to calculate the sum for the required range directly. This method is good when there are few queries, but it is inefficient when there are many. We can implement it as follows.

def simple_range_sum(arr, queries):
    results = []
    for start, end in queries:
        results.append(sum(arr[start - 1:end]))  # 1-based index to 0-based index
    return results


arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
queries = [(1, 3), (2, 5), (3, 7)]
print(simple_range_sum(arr, queries))  # Output: [9, 14, 27]

When we execute the code above, the results for the queries will output [9, 14, 27]. However, the time complexity of this method is O(Q * N), where Q is the number of queries and N is the size of the array. This becomes inefficient for large inputs.

2. Using a Cumulative Sum Array

A more efficient way is to create a cumulative sum array. Once we have the cumulative sum array, we can calculate the sum of each range in O(1) time. The method is as follows.

def prefix_sum(arr):
    n = len(arr)
    prefix = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix[i] = prefix[i - 1] + arr[i - 1]
    return prefix


def efficient_range_sum(arr, queries):
    prefix = prefix_sum(arr)
    results = []
    for start, end in queries:
        results.append(prefix[end] - prefix[start - 1])  # 1-based index adjustment
    return results


arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
queries = [(1, 3), (2, 5), (3, 7)]
print(efficient_range_sum(arr, queries))  # Output: [9, 14, 27]

In the above code, calculating the cumulative sum array takes O(N) time, and processing each query takes O(1) time. Thus, the overall time complexity reduces to O(N + Q). This method allows us to handle multiple queries efficiently.

Problem Optimization and Selection

The range sum problem can be utilized in various situations, and the algorithm chosen may vary depending on the conditions of the problem. It is essential to choose the optimal method considering the size of the input array, the number of queries, and the length of each query’s range.

Besides the methods described above, more complex data structures such as segment trees can also be used, but this tutorial focused on basic methods for solving practical problems. In the next lesson, we will address more complicated dynamic query problems.

Conclusion

In this lesson, we covered the range sum problem using Python. We found that while simple iteration can be inefficient, using a cumulative sum array allows us to solve the problem much more efficiently. The range sum problem is a very useful topic for building basic algorithm skills, so I encourage you to practice it multiple times and try various variations!

Python coding test course, finding the interval product

Python Coding Test Course: Calculating the Product of Intervals

When solving programming problems, there are often cases where calculations are performed based on specific ranges of given values. In this article, we will address the problem of ‘calculating the product of intervals.’ This problem involves implementing an algorithm that multiplies the values belonging to a specific range within a given array. During this process, we will also discuss strategies to solve the problem efficiently.

Problem Description

There is a given array of integers and several queries. Each query consists of two indices (l, r), and the task is to output the product of all elements in the array within that range, including the specified indices. It is important to note that the result of the product of the interval can be a very large number, so it needs to be optimized for calculation.

Example:
Array: [1, 2, 3, 4, 5]
Query: (1, 3)
Output: 2 * 3 * 4 = 24

Problem-Solving Strategy

1. Basic Approach

The most intuitive approach is to iterate over the elements at the specified indices for each query in the given array and multiply them. However, this method has a worst-case time complexity of O(Q * N), which is inefficient. Here, N represents the size of the array and Q is the number of queries. This method can become significantly slower as the number of queries increases.

2. Approach Using Cumulative Product

One efficient method is to use the cumulative product (Prefix Product). The way to calculate the cumulative product is to compute and store the product of all previous elements for each element in the array.

  • For example, if the array is A = [1, 2, 3, 4, 5]:
    Cumulative product array P:
    P[0] = 1
    P[1] = 1 * 1 = 1
    P[2] = 1 * 1 * 2 = 2
    P[3] = 1 * 1 * 2 * 3 = 6
    P[4] = 1 * 1 * 2 * 3 * 4 = 24
    P[5] = 1 * 1 * 2 * 3 * 4 * 5 = 120
    

After constructing the cumulative product array, the product of the interval can be calculated using the following formula:

Interval Product = P[r+1] / P[l]

Here, P[i] represents the product up to the i-th element. By using this method, we can calculate the product of the interval with a time complexity of O(1).

Python Code Implementation

Now, let’s implement the algorithm for calculating the product of intervals in Python, referring to the approach above.


def calculate_prefix_product(arr):
    n = len(arr)
    prefix_product = [1] * (n + 1)
    
    for i in range(n):
        prefix_product[i + 1] = prefix_product[i] * arr[i]
    
    return prefix_product

def range_product(arr, queries):
    prefix_product = calculate_prefix_product(arr)
    results = []
    
    for l, r in queries:
        product = prefix_product[r + 1] // prefix_product[l]
        results.append(product)
    
    return results

# Example array and queries
arr = [1, 2, 3, 4, 5]
queries = [(1, 3), (0, 2), (2, 4)]

# Function call
result = range_product(arr, queries)
print(result)  # Output: [24, 6, 60]

Final Check

Now it’s time to verify if the code works properly. We will check if the correct results are obtained through various test cases. It is important to ensure that the range of queries is valid and that there are no issues at boundary values.

  • Test Case 1: arr = [1,2,3,4,5], queries = [(0, 4)] => Result: 120
  • Test Case 2: arr = [10,20,30], queries = [(0, 2), (1, 1)] => Result: [6000, 20]
  • Test Case 3: arr = [0, 1, 2, 3], queries = [(0, 3), (1, 1)] => Result: [0, 1]

Conclusion

The problem of calculating the product of intervals discussed in this tutorial demonstrates how to derive an efficient solution by appropriately utilizing cumulative products. This principle can also be applied to solve other similar problems. Practicing various algorithmic problems will enhance your preparation for coding tests. Make good use of this technique in your future coding tests!

Additional Questions

If you have any unresolved issues or additional questions, feel free to reach out to me at any time. Happy Coding!