LeetCode #2954 — HARD

Count the Number of Infection Sequences

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 10⁹+7.

Example 1:

Input: n = 5, sick = [0,4]

Output: 4

Explanation:

There is a total of 6 different sequences overall.

Valid infection sequences are [1,2,3], [1,3,2], [3,2,1] and [3,1,2].
[2,3,1] and [2,1,3] are not valid infection sequences because the person at index 2 cannot be infected at the first step.

Example 2:

Input: n = 4, sick = [1]

Output: 3

Explanation:

There is a total of 6 different sequences overall.

Valid infection sequences are [0,2,3], [2,0,3] and [2,3,0].
[3,2,0], [3,0,2], and [0,3,2] are not valid infection sequences because the infection starts at the person at index 1, then the order of infection is 2, then 3, and hence 3 cannot be infected earlier than 2.

Constraints:

2 <= n <= 10⁵
1 <= sick.length <= n - 1
0 <= sick[i] <= n - 1
sick is sorted in increasing order.

Step 01

Brute Force Baseline

Problem summary: 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 109+7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

5
[0,4]

Example 2

4
[1]

Core Insight

What unlocks the optimal approach

Consider infected children as <code>0</code> and non-infected as <code>1</code>, then divide the array into segments with the same value.
For each segment of non-infected children whose indices are <code>[i, j]</code> and indices <code>(i - 1)</code> and <code>(j + 1)</code>, if they exist, are already infected. Then if <code>i == 0</code> or <code>j == n - 1</code>, each second there is only one kid that can be infected (which is at the other endpoint).
If <code>i > 0</code> and <code>j < n - 1</code>, we have two choices per second since the children at the two endpoints can both be the infect candidates. So there are <code>2j - i</code> orders to infect all children in the segment.
Each second we can select a segment and select one endpoint from it.
The answer is: <code>S! / (len[1]! * len[2]! * ... * len[m]! * lenstart! * lenend!) * 2k</code> where <code>len[1], len[2], ..., len[m]</code> are the lengths of each segment of non-infected children that have an infected child at both endpoints, <code>lenstart</code> and <code>lenend</code> denote the number of non-infected children with infected child at one endpoint, <code>S</code> is the total length of all segments of non-infected children, and <code>k = (len[1] - 1) + (len[2] - 1) + ... + (len[m] - 1)</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2954: Count the Number of Infection Sequences
class Solution {
    private static final int MOD = (int) (1e9 + 7);
    private static final int MX = 100000;
    private static final int[] FAC = new int[MX + 1];

    static {
        FAC[0] = 1;
        for (int i = 1; i <= MX; i++) {
            FAC[i] = (int) ((long) FAC[i - 1] * i % MOD);
        }
    }

    public int numberOfSequence(int n, int[] sick) {
        int m = sick.length;
        int[] nums = new int[m + 1];
        nums[0] = sick[0];
        nums[m] = n - sick[m - 1] - 1;
        for (int i = 1; i < m; i++) {
            nums[i] = sick[i] - sick[i - 1] - 1;
        }
        int s = 0;
        for (int x : nums) {
            s += x;
        }
        int ans = FAC[s];
        for (int x : nums) {
            if (x > 0) {
                ans = (int) ((long) ans * qpow(FAC[x], MOD - 2) % MOD);
            }
        }
        for (int i = 1; i < nums.length - 1; ++i) {
            if (nums[i] > 1) {
                ans = (int) ((long) ans * qpow(2, nums[i] - 1) % MOD);
            }
        }
        return ans;
    }

    private int qpow(long a, long n) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % MOD;
            }
            a = a * a % MOD;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #2954: Count the Number of Infection Sequences
const MX = 1e5
const MOD = 1e9 + 7

var fac [MX + 1]int

func init() {
	fac[0] = 1
	for i := 1; i <= MX; i++ {
		fac[i] = fac[i-1] * i % MOD
	}
}

func qpow(a, n int) int {
	ans := 1
	for n > 0 {
		if n&1 == 1 {
			ans = (ans * a) % MOD
		}
		a = (a * a) % MOD
		n >>= 1
	}
	return ans
}

func numberOfSequence(n int, sick []int) int {
	m := len(sick)
	nums := make([]int, m+1)

	nums[0] = sick[0]
	nums[m] = n - sick[m-1] - 1
	for i := 1; i < m; i++ {
		nums[i] = sick[i] - sick[i-1] - 1
	}

	s := 0
	for _, x := range nums {
		s += x
	}
	ans := fac[s]
	for _, x := range nums {
		if x > 0 {
			ans = ans * qpow(fac[x], MOD-2) % MOD
		}
	}
	for i := 1; i < len(nums)-1; i++ {
		if nums[i] > 1 {
			ans = ans * qpow(2, nums[i]-1) % MOD
		}
	}
	return ans
}

# Accepted solution for LeetCode #2954: Count the Number of Infection Sequences
mod = 10**9 + 7
mx = 10**5
fac = [1] * (mx + 1)
for i in range(2, mx + 1):
    fac[i] = fac[i - 1] * i % mod


class Solution:
    def numberOfSequence(self, n: int, sick: List[int]) -> int:
        nums = [b - a - 1 for a, b in pairwise([-1] + sick + [n])]
        ans = 1
        s = sum(nums)
        ans = fac[s]
        for x in nums:
            if x:
                ans = ans * pow(fac[x], mod - 2, mod) % mod
        for x in nums[1:-1]:
            if x > 1:
                ans = ans * pow(2, x - 1, mod) % mod
        return ans

// Accepted solution for LeetCode #2954: Count the Number of Infection Sequences
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2954: Count the Number of Infection Sequences
// class Solution {
//     private static final int MOD = (int) (1e9 + 7);
//     private static final int MX = 100000;
//     private static final int[] FAC = new int[MX + 1];
// 
//     static {
//         FAC[0] = 1;
//         for (int i = 1; i <= MX; i++) {
//             FAC[i] = (int) ((long) FAC[i - 1] * i % MOD);
//         }
//     }
// 
//     public int numberOfSequence(int n, int[] sick) {
//         int m = sick.length;
//         int[] nums = new int[m + 1];
//         nums[0] = sick[0];
//         nums[m] = n - sick[m - 1] - 1;
//         for (int i = 1; i < m; i++) {
//             nums[i] = sick[i] - sick[i - 1] - 1;
//         }
//         int s = 0;
//         for (int x : nums) {
//             s += x;
//         }
//         int ans = FAC[s];
//         for (int x : nums) {
//             if (x > 0) {
//                 ans = (int) ((long) ans * qpow(FAC[x], MOD - 2) % MOD);
//             }
//         }
//         for (int i = 1; i < nums.length - 1; ++i) {
//             if (nums[i] > 1) {
//                 ans = (int) ((long) ans * qpow(2, nums[i] - 1) % MOD);
//             }
//         }
//         return ans;
//     }
// 
//     private int qpow(long a, long n) {
//         long ans = 1;
//         for (; n > 0; n >>= 1) {
//             if ((n & 1) == 1) {
//                 ans = ans * a % MOD;
//             }
//             a = a * a % MOD;
//         }
//         return (int) ans;
//     }
// }

// Accepted solution for LeetCode #2954: Count the Number of Infection Sequences
const MX = 1e5;
const MOD: bigint = BigInt(1e9 + 7);
const fac: bigint[] = Array(MX + 1);

const init = (() => {
    fac[0] = 1n;
    for (let i = 1; i <= MX; ++i) {
        fac[i] = (fac[i - 1] * BigInt(i)) % MOD;
    }
    return 0;
})();

function qpow(a: bigint, n: number): bigint {
    let ans = 1n;
    for (; n > 0; n >>= 1) {
        if (n & 1) {
            ans = (ans * a) % MOD;
        }
        a = (a * a) % MOD;
    }
    return ans;
}

function numberOfSequence(n: number, sick: number[]): number {
    const m = sick.length;
    const nums: number[] = Array(m + 1);
    nums[0] = sick[0];
    nums[m] = n - sick[m - 1] - 1;
    for (let i = 1; i < m; i++) {
        nums[i] = sick[i] - sick[i - 1] - 1;
    }

    const s = nums.reduce((acc, x) => acc + x, 0);
    let ans = fac[s];
    for (let x of nums) {
        if (x > 0) {
            ans = (ans * qpow(fac[x], Number(MOD) - 2)) % MOD;
        }
    }
    for (let i = 1; i < nums.length - 1; ++i) {
        if (nums[i] > 1) {
            ans = (ans * qpow(2n, nums[i] - 1)) % MOD;
        }
    }
    return Number(ans);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.