What is the time complexity of Count the Number of Infection Sequences?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Count the Number of Infection Sequences?

This problem uses Array, Math, Combinatorics. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count the Number of Infection Sequences?

Insight 1: The problem is fundamentally a combinatorics problem. The order in which sections of uninfected people between sick people become infected matters. Insight 2: The uninfected people can be viewed as separate groups between the initial infected people. Each group can be infected from either end, leading to 2^(group_size - 1) possibilities within each group (except for the groups at the start and end which have only one way to be infected.)

Which companies ask LeetCode 2954. Count the Number of Infection Sequences in interviews?

This problem is frequently asked at SAP, Tekion.

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Count the Number of Infection Sequences - LeetCode 2954 Solution

Count the Number of Infection Sequences - Complete Solution Guide

Count the Number of Infection Sequences is LeetCode problem 2954, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer n and an array sick sorted in increasing order, representing positions of infected people in a line of n people. At each step, one uninfected person adjacent to an infected person gets infected. This process continues until everyone is infected. An infection sequence is the order in which uninfected people become infected, excluding those initially infected. Return the number of different infection sequences possible, modulo 10 9 +7 . Example 1: Input: n = 5, sick = [0,4

Detailed Explanation

The problem describes a scenario where people in a line of `n` individuals become infected. Initially, some people are infected (their positions are given in the `sick` array, sorted in increasing order). At each step, an uninfected person adjacent to an infected person gets infected. The goal is to count the number of different *infection sequences* possible, modulo 10^9 + 7. An infection sequence is the order in which *uninfected* people become infected. The initial sick individuals are not part of the sequence.

Solution Approach

The solution involves calculating the number of ways to arrange the infection sequence. It breaks down into the following steps: 1. Calculate the number of uninfected people. 2. Calculate the length of each group of consecutive uninfected people between the infected people. 3. Calculate the number of ways to arrange the infection sequence using combinations. This involves calculating factorials and their modular inverses. 4. Account for the multiple ways groups of infected people can be ordered based on which side they are infected from using the power of 2.

Step-by-Step Algorithm

Step 1: Calculate `m`, the number of uninfected people: `m = n - k`, where `k` is the number of initially infected people.
Step 2: Calculate factorials modulo MOD for numbers up to `n` and store them in the `fact` array. This is a precomputation step to avoid repeated calculations.
Step 3: Calculate the number of ways to infect the individuals to the left of the first infected person `sick[0]`. This contributes to the number of combinations.
Step 4: Iterate through the `sick` array from the second element. Calculate the length `l_i` of the uninfected gap between `sick[i]` and `sick[i-1]`: `l_i = sick[i] - sick[i-1] - 1`.
Step 5: If `l_i > 0`, update the combinations term by dividing by `fact[l_i]` and multiply the internal term by `power(2, l_i - 1)` to account for the possible infection order in either direction. Note, the groups at the beginning and end can only be infected in one direction.
Step 6: Calculate the number of ways to infect the individuals to the right of the last infected person `sick[k-1]`. This again contributes to combinations.
Step 7: Return the result of `(combinations_term * internal_term) % MOD`.

Key Insights

Insight 1: The problem is fundamentally a combinatorics problem. The order in which sections of uninfected people between sick people become infected matters.
Insight 2: The uninfected people can be viewed as separate groups between the initial infected people. Each group can be infected from either end, leading to 2^(group_size - 1) possibilities within each group (except for the groups at the start and end which have only one way to be infected.)
Insight 3: We need to calculate combinations (n choose k) where 'n' is the total number of uninfected people, and 'k' represents the sizes of the gaps. Using modular arithmetic to prevent overflow and Fermat's Little Theorem to calculate modular inverses is crucial.
Insight 4: Since the people can be infected in either direction it creates multiple infection sequences, we need to calculate the number of ways each group in between the sick people can be infected from either side. For the groups at the start and end, there is only one way to infect them.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Combinatorics.

Companies

Asked at: SAP, Tekion.

Count the Number of Infection Sequences | LeetCode 2954

Hard

Array Math Combinatorics

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int numberOfSequence(int n, int[] sick) {
        int MOD = 1000000007;

        long[] fact = new long[n + 1];
        fact[0] = 1;
        for (int i = 1; i <= n; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
        }

        
        long k = sick.length;
        long m = n - k;

        long combinations_term = fact[(int)m];

        long l_0 = sick[0];
        combinations_term = (combinations_term * inv(fact[(int)l_0], MOD)) % MOD;

        long internal_term = 1;
        for (int i = 1; i < k; i++) {
            long l_i = sick[i] - sick[i - 1] - 1;
            if (l_i > 0) {
                combinations_term = (combinations_term * inv(fact[(int)l_i], MOD)) % MOD;
                internal_term = (internal_term * power(2, l_i - 1, MOD)) % MOD;
            }
        }

        long l_k = n - 1 - sick[(int)k - 1];
        combinations_term = (combinations_term * inv(fact[(int)l_k], MOD)) % MOD;

        return (int)((combinations_term * internal_term) % MOD);
    }

        private long power(long a, long b, int MOD) {
            long res = 1;
            a %= MOD;
            while (b > 0) {
                if (b % 2 == 1) {
                    res = (res * a) % MOD;
                }
                a = (a * a) % MOD;
                b /= 2;
            }
            return res;
        }

        private long inv(long n, int MOD) {
            return power(n, MOD - 2, MOD);
        }
}

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Problem Info

Description

Problem 2954: Count the Number of Infection Sequences - Detailed explanation and solution approaches.

Topics

Array Math Combinatorics

Companies

SAPTekion

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:2954

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Calculate `m`, the number of uninfected people: `m = n - k`, where `k` is the number of initially infected people.
2
Step 2: Calculate factorials modulo MOD for numbers up to `n` and store them in the `fact` array. This is a precomputation step to avoid repeated calculations.
3
Step 3: Calculate the number of ways to infect the individuals to the left of the first infected person `sick[0]`. This contributes to the number of combinations.
4
Step 4: Iterate through the `sick` array from the second element. Calculate the length `l_i` of the uninfected gap between `sick[i]` and `sick[i-1]`: `l_i = sick[i] - sick[i-1] - 1`.
5
Step 5: If `l_i > 0`, update the combinations term by dividing by `fact[l_i]` and multiply the internal term by `power(2, l_i - 1)` to account for the possible infection order in either direction. Note, the groups at the beginning and end can only be infected in one direction.
6
Step 6: Calculate the number of ways to infect the individuals to the right of the last infected person `sick[k-1]`. This again contributes to combinations.
7
Step 7: Return the result of `(combinations_term * internal_term) % MOD`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is the precomputation of factorials which takes O(n) time. The rest of the operations (calculating combinations, powers, and modular inverses) take O(k) time, but k <= n, thus the overall complexity remains O(n). The space complexity is O(n) due to storing the factorial values in the `fact` array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

A dynamic programming approach could potentially be used, but would likely be more complex and less efficient than the current solution due to the combinatorics involved. It would be very difficult to define appropriate states and transitions to achieve the same result without introducing significantly higher complexity.

Time: Likely O(n^2) or higher.Space: Likely O(n^2) or higher.

Trade-offs: Not recommended, significantly more complex and likely less efficient.

✨

Key Insights

1Insight 1: The problem is fundamentally a combinatorics problem. The order in which sections of uninfected people between sick people become infected matters.
2Insight 2: The uninfected people can be viewed as separate groups between the initial infected people. Each group can be infected from either end, leading to 2^(group_size - 1) possibilities within each group (except for the groups at the start and end which have only one way to be infected.)
3Insight 3: We need to calculate combinations (n choose k) where 'n' is the total number of uninfected people, and 'k' represents the sizes of the gaps. Using modular arithmetic to prevent overflow and Fermat's Little Theorem to calculate modular inverses is crucial.
4Insight 4: Since the people can be infected in either direction it creates multiple infection sequences, we need to calculate the number of ways each group in between the sick people can be infected from either side. For the groups at the start and end, there is only one way to infect them.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `sick` array only contains one element. The implementation should still correctly handle the cases with large uninfected regions at the beginning and end of the line.
Edge case 2: All people are sick except one. `n = 2`, `sick=[0]`. The algorithm should correctly compute factorial and modular inverse.
Edge case 3: When there are adjacent sick people (the length of the uninfected region between them `l_i == 0`). The `power` function is correctly called with exponent `-1` when there are adjacent sick people, causing no effect on `internal_term`.
Edge case 4: Sick people at the very start and end `sick[0] == 0` or `sick[k-1] == n -1`

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to apply the modulo operator (%) at each step during factorial and power calculations, leading to integer overflow and incorrect results.

✗Common mistake 2: Incorrectly calculating the lengths of the uninfected regions between the infected people. Ensure to subtract 1 for the infected person themselves.

✗Common mistake 3: Not handling edge cases where `sick[0] == 0` or `sick[len(sick)-1] == n-1` correctly.

✗Common mistake 4: Using integer division instead of floor division when calculating `b //= 2` or `b /= 2` in the power function can lead to wrong results in some languages.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Modeling the spread of a disease in a population where individuals can only be infected by their immediate neighbors.

🏦Real-world scenario 2: Analyzing the order in which tasks can be completed in a system where some tasks depend on the completion of adjacent tasks.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 14, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Count the Number of Infection Sequences"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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