Java Coding Test Course, Calculating Prefix Sums 2

1. Problem Description

In this session, we will discuss the interval sum problem. This is one of the problems you often encounter in algorithms and coding tests, where we will learn how to efficiently calculate the sum of a specific range in a given array.

The problem is as follows:

Given an array A consisting of integers and various queries, find the sum from A[l] to A[r] for each range l and r specified in the queries.

Input format:

1. Size N of the array A (1 ≤ N ≤ 100,000)
2. Elements of the array A (1 ≤ A[i] ≤ 100,000)
3. Number of queries Q (1 ≤ Q ≤ 100,000)
4. Each query contains two integers l and r (1 ≤ l ≤ r ≤ N).

Output format:

Print the interval sum for each query.

2. Approach

The approach to the problem can be divided into two methods:

1. A straightforward method using loops to calculate the sum (inefficient)
2. A method using a prefix sum array to calculate the interval sum in O(1) time (efficient)

In the first approach, if we directly calculate the sum for each query, the time complexity becomes O(N * Q), as it is the product of the number of queries Q and the size of the array N. In this case, in the worst case, it requires 10¹⁰ operations, which is not efficient.

The second approach involves creating a prefix sum array to calculate the interval sum in O(1) time. The overall approach of this method is as follows:

1. First, create a prefix sum array of size N.
2. The i-th index of the prefix sum array stores the sum from A[0] to A[i].
3. The sum of each query (l, r) can be calculated using the prefix sum array as S[r] – S[l-1].

3. Code Implementation

      public class IntervalSum {
        public static void main(String[] args) {
            int N = 5; // Size of array
            int[] A = {1, 2, 3, 4, 5}; // Input values
            int Q = 3; // Number of queries
            int[][] queries = {{1, 3}, {2, 5}, {1, 5}}; // Sample queries
            
            // Create prefix sum array
            long[] prefixSum = new long[N + 1];
            for (int i = 1; i <= N; i++) {
                prefixSum[i] = prefixSum[i - 1] + A[i - 1];
            }
            
            // Process each query
            for (int i = 0; i < Q; i++) {
                int l = queries[i][0];
                int r = queries[i][1];
                long sum = prefixSum[r] - prefixSum[l - 1];
                System.out.println("Sum from " + l + " to " + r + ": " + sum);
            }
        }
      }
    

4. Code Explanation

The above code operates as follows:

1. It receives the array A, the number of queries Q, and l and r for each query.
2. First, it creates the prefix sum array. The i-th index of this array stores the sum from the 0-th index to the (i-1)-th index of array A.
3. When processing each query, the corresponding interval sum is calculated and output as prefixSum[r] – prefixSum[l-1].

5. Time Complexity Analysis

The time complexity of this algorithm is as follows:

1. It takes O(N) time to create the prefix sum array.
2. It takes O(1) time to process each query.

Therefore, the total time complexity is O(N + Q). This is a very efficient method that performs well even when both the size of the array and the number of queries are at their maximum values.

6. Final Summary

The interval sum problem is an important concept in understanding algorithms and data structures. In this session, we learned an efficient way to calculate interval sums using the prefix sum array. This method can significantly maximize performance when processing large amounts of data.

Such types of problems may also appear in various modified forms, making it important to practice further. I hope you continue to solve various algorithm problems to deepen and broaden your understanding.