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Huawei Software Written Test Question 2 Analysis on March 10, 2021.

title: Analysis of Huawei Software Written Test Question 2 on March 10, 2021
date: 2021-03-12 16:18:40
tags: ['algorithm']

Recently, my roommate took the Huawei software written test, and after I received the questions, I also tried to solve them. I found the second question to be very clever, so I decided to write a blog post to record my solution.

Problem Description

Driver Xiao Wang is at the warehouse to transport goods. The warehouse manager has already packed the goods into boxes and arranged them in a row. The boxes are numbered starting from 1. The goods stored in each box form a set M={M1, M2, ..., Mn}. The truck can transport a maximum of K goods at a time. In order to maximize efficiency, Xiao Wang wants to select a multiple of K goods at a time and transport them in batches. The warehouse manager requires Xiao Wang to choose boxes with consecutive numbers. For example, he can choose boxes 1, 2, and 3, but not boxes 2, 4, and 6.
Please help Xiao Wang calculate how many ways there are to select a multiple of K goods for transportation.

Input Description

The number of boxes is N, and the maximum number of goods that can be transported by the truck at a time is K.
The number of goods stored in each box: M1, M2, M3, M4, ...
The input consists of two lines in the following format:
The values of N and K are both in the range [1, 100000].
The number of goods stored in each box, Mi, is also in the range [1, 100000].


Since this is a problem of consecutive sequences, I initially considered using a sliding window approach, but upon analysis, I found it to be infeasible.

The general approach of a sliding window is to first expand the window, and when a certain condition is met that makes the window unable to satisfy the condition, then shrink the window. However, in this problem, it is different: it is not possible to determine whether the window needs to be shrunk based on local conditions alone, i.e., it is not possible to guarantee that if the current window does not satisfy the condition of being a multiple of K, then expanding the window further will not satisfy the condition of being a multiple of K.

Then I thought about using dynamic programming to calculate the sum of intervals and then iterate to check if each interval satisfies the condition. Experience tells us that this is correct, but dynamic programming requires a two-dimensional array to store the interval sums, with a space complexity of O(N^2), and it also requires iterating over this two-dimensional array, resulting in a time complexity of O(N^2). In the case where N is in the range [1, 100000], such complexity is unacceptable.

Enumerating through all possibilities? Although it reduces the space complexity, the time complexity is much higher than that of dynamic programming, making it even more unacceptable.

When in doubt, look for a solution. I randomly looked at a few solutions, and this solution provided a solution with a time complexity of O(n+k), which gave me great inspiration.

The basic idea of this solution is based on the following principles:

  1. By subtracting the sum of goods 0 to j from the sum of goods 0 to i, we get the sum of goods i to j. If this sum can be divided by K, i.e., if the sum of goods 0 to j modulo K is equal to the sum of goods 0 to i modulo K, then the interval satisfies the condition.

The mathematical representation of this conclusion is:

(Sum[j]-Sum[i])%k=0 => Sum[j] = n*k + Sum[i] => Sum[j]%k = Sum[i]%k

  1. When multiple positions a, b, c, etc., satisfy Sum[a]%k=Sum[b]%k=Sum[c]%k=..., let the number of positions be t, then any two positions chosen from these positions can form an interval that satisfies the condition. Therefore, the number of intervals that satisfy the condition at this time is C(t,2).

C(t,2) is a mathematical combination, which expands to t*(t-1)/2.

The specific steps are as follows:

  1. Use an array M of length n to store the number of goods in the warehouse. After inputting the values, create another array Sum of length n, where Sum[i] represents the sum of M[0]+M[1]+...+M[i]. At this point, Sum[j]-Sum[i] represents the total number of goods in the interval from goods i to goods j.

  2. Create an array count of length k, and iterate through the Sum array. For any index x between 0 and n, increment count[Sum[x]%k] by 1.

  3. Iterate through the count array, for any index y between 0 and k, if count[y]>1, then result = result + C(count[y],2). Finally, return the result.

In the actual code representation, we can omit the M and Sum arrays and perform the relevant operations on the count array while inputting the values. In this case, we only need to process n inputs, iterate through an array of length k, with a time complexity of O(n+k), and only use a count array of length k, resulting in a space complexity of O(k), which is the optimal solution for this problem.


Here is my Golang code:

package main

import "fmt"

func main() {
        var (
                n, k    int
                tmp     int
                current int
                count   []int
                result  int
        fmt.Scanln(&n, &k)
        count = make([]int, k)
        // Input the list of goods and update count in real-time
        for i := 0; i < n; i++ {
                current = (current + tmp) % k
                count[current] += 1
        for _, v := range count {
                if v > 1 {
                        result += v * (v - 1) / 2
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