LeetCode 3179 - Find the N-th Value After K Seconds

LeetCode Problem 3179

Difficulty: 🟡 Medium
Topics: Array, Math, Simulation, Combinatorics, Prefix Sum

Solution

Problem Understanding

The problem requires computing the value of the last element in an array after a series of sequential cumulative sum operations over a fixed number of seconds. You start with an array a of length n where all elements are initialized to 1. Every second, each element at index i becomes the sum of itself and all elements preceding it in the array. The goal is to determine a[n-1] after k seconds.

The input consists of two integers, n and k, both of which can be up to 1000. This implies that any algorithm must handle potentially large arrays and a significant number of cumulative operations efficiently. A naive implementation that explicitly sums elements repeatedly could lead to timeouts because of repeated O(n) operations per element, potentially resulting in O(n^2 * k) complexity.

Edge cases include the smallest array (n = 1) or the smallest number of seconds (k = 0), both of which should return 1 since no update occurs. The problem also requires taking the result modulo 10^9 + 7 to prevent integer overflow.

Approaches

A brute-force approach involves simulating the array update for k seconds, explicitly computing cumulative sums for each element. While simple, this method is inefficient because each second requires summing up to n elements for every array element, leading to O(n^2 * k) time complexity.

A key observation is that the update rule can be efficiently handled using prefix sums. Instead of recomputing the sum of all previous elements repeatedly, we maintain a running cumulative sum. This allows each update for the next second to be computed in O(n) time. The pattern of the array updates corresponds to combinatorial numbers: a[n-1] after k seconds equals the binomial coefficient C(n + k - 1, k), but for the purpose of this explanation, we will implement it using the prefix sum approach for clarity and simplicity.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n^2 * k)	O(n)	Explicitly compute cumulative sums each second
Optimal	O(n * k)	O(n)	Use prefix sums to accumulate values efficiently

Algorithm Walkthrough

1. Initialize an array a of length n with all elements set to 1. This represents the array at time t=0.
2. Repeat for k seconds:

a. Initialize a variable cumulative to 0 to maintain a running prefix sum.

b. Iterate over the array. For each index i, update cumulative = (cumulative + a[i]) % MOD and set a[i] = cumulative. This ensures each element becomes the sum of itself and all previous elements. 3. Return a[n-1] as the result modulo 10^9 + 7.

Why it works: By maintaining a running cumulative sum, each element is updated to reflect the sum of all preceding elements plus itself. This preserves the invariant of the problem statement and avoids recomputation, producing the correct last element after k seconds.

Python Solution

class Solution:
    def valueAfterKSeconds(self, n: int, k: int) -> int:
        MOD = 10**9 + 7
        a = [1] * n
        
        for _ in range(k):
            cumulative = 0
            for i in range(n):
                cumulative = (cumulative + a[i]) % MOD
                a[i] = cumulative
        
        return a[-1]

The implementation initializes the array a with ones and iterates k times, updating each element using a running cumulative sum. The modulo operation ensures we stay within integer bounds.

Go Solution

func valueAfterKSeconds(n int, k int) int {
    const MOD int = 1e9 + 7
    a := make([]int, n)
    for i := 0; i < n; i++ {
        a[i] = 1
    }

    for t := 0; t < k; t++ {
        cumulative := 0
        for i := 0; i < n; i++ {
            cumulative = (cumulative + a[i]) % MOD
            a[i] = cumulative
        }
    }

    return a[n-1]
}

Go-specific differences include pre-allocating the slice a and using a standard for loop instead of Python's range. The logic mirrors the Python implementation.

Worked Examples

Example 1: n = 4, k = 5

Second	Array State
0	[1, 1, 1, 1]
1	[1, 2, 3, 4]
2	[1, 3, 6, 10]
3	[1, 4, 10, 20]
4	[1, 5, 15, 35]
5	[1, 6, 21, 56]

Return 56.

Example 2: n = 5, k = 3

Second	Array State
0	[1, 1, 1, 1, 1]
1	[1, 2, 3, 4, 5]
2	[1, 3, 6, 10, 15]
3	[1, 4, 10, 20, 35]

Return 35.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * k)	Each of the k seconds requires O(n) to update the array using prefix sums
Space	O(n)	Only the array a of length n is maintained

The algorithm is efficient for n, k ≤ 1000 because 1,000 * 1,000 = 10^6 operations are acceptable.

Test Cases

# test cases
sol = Solution()
assert sol.valueAfterKSeconds(4, 5) == 56  # example 1
assert sol.valueAfterKSeconds(5, 3) == 35  # example 2
assert sol.valueAfterKSeconds(1, 0) == 1   # edge case: single element, no seconds
assert sol.valueAfterKSeconds(1, 10) == 1  # edge case: single element, multiple seconds
assert sol.valueAfterKSeconds(2, 1) == 2   # small array, 1 second
assert sol.valueAfterKSeconds(3, 3) == 10  # small example

Test	Why
n=4, k=5	matches example 1
n=5, k=3	matches example 2
n=1, k=0	tests minimum input, no updates
n=1, k=10	tests single element with updates
n=2, k=1	tests minimal array and small update
n=3, k=3	tests small example with multiple updates

Edge Cases

A key edge case is when n = 1. Since there is only one element, its value never changes, regardless of k. Another edge case is k = 0, which represents no updates; the array remains [1]*n, so the last element is 1. A third edge case involves maximum values n = 1000, k = 1000 where large intermediate sums occur; using the modulo operation prevents integer overflow and ensures correctness. The implementation handles all these cases by design through initialization and modular arithmetic.