Swift Coding Test Course, Calculate the Sum of Remainders

Recently, the importance of algorithm coding tests in the field of software development has been growing. This is because many companies evaluate candidates’ problem-solving abilities through coding tests during the interview process. In this article, we will solve the ‘Subarray Sum Remainder’ problem using the Swift language, covering everything from understanding the problem to the overall resolution process.

Problem Description

Given an integer array nums and an integer k, write a function to calculate the remainder of the sum of all subarrays of nums when divided by k, and return the total sum of these remainders.

Example Problem

Input: nums = [1, 2, 3, 4, 5], k = 3
Output: 10
Description: The sums of the subarrays are 1, 1+2=3, 1+2+3=6, 1+2+3+4=10, ..., and we need to find the remainders of each when divided by 3.

Approach to the Problem

To solve this problem, we first need to generate all subarrays, calculate their sums, and then find the remainder when divided by k. However, generating all subarrays directly can lead to a time complexity of more than O(n^3), so a more efficient approach is needed.

We can accumulate the sums of the subarrays and use this to solve the problem. In this process, we can think about how to store the remainders.

Swift Implementation


func subarraySumRemainder(nums: [Int], k: Int) -> Int {
    var totalRemainderSum = 0
    let n = nums.count
    
    for start in 0..


Code Explanation
The above function subarraySumRemainder takes nums and k as inputs and calculates the sum of remainders.

totalRemainderSum: A variable that stores the sum of all subarray remainders.
n: Stores the size of the array.
for start in 0..: Sets the starting index for each subarray.

currentSum: A variable that stores the sum of the current subarray.
for end in start..: Sets the ending index for the subarray and updates the current sum.

totalRemainderSum += currentSum % k: Calculates the remainder and adds it to the total sum.

Time Complexity Analysis
The time complexity of this algorithm is O(n^2). This is because, in the worst case, we need to traverse all subarrays of the array. Since calculating the remainder can be done in constant time, the total complexity is determined by the time spent generating the subarrays.
Other Approaches
This problem can be solved in various ways. For example, using the prefix sum technique can provide better time complexity. By using prefix sums, we can quickly find the sum over a specific range, thus reducing the time required to calculate the remainder.
Method Using Prefix Sum

func subarraySumRemainderUsingPrefixSum(nums: [Int], k: Int) -> Int {
    var prefixSums = [Int](repeating: 0, count: nums.count + 1)
    for i in 1...nums.count {
        prefixSums[i] = prefixSums[i - 1] + nums[i - 1]
    }
    
    var totalRemainderSum = 0
    for start in 0..

Conclusion
In this article, we explored the method of solving the 'Subarray Sum Remainder' problem using the Swift language. We provided a comprehensive explanation from understanding the problem and approach to implementation and time complexity analysis. There are multiple approaches to algorithm problems, and they require creative thinking to solve. I hope you continue to develop your coding skills through consistent practice.
References

LeetCode (https://leetcode.com)
GeeksforGeeks (https://www.geeksforgeeks.org)
Algorithm Visualizer (https://algorithm-visualizer.org)




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