Find the N-th Value After K Seconds

Find the N-th Value After K Seconds - Problem

Array Math Simulation Combinatorics Prefix Sum Medium

You are given two integers n and k. Initially, you start with an array a of n integers where a[i] = 1 for all 0 <= i <= n - 1.

After each second, you simultaneously update each element to be the sum of all its preceding elements plus the element itself. For example, after one second, a[0] remains the same, a[1] becomes a[0] + a[1], a[2] becomes a[0] + a[1] + a[2], and so on.

Return the value of a[n - 1] after k seconds. Since the answer may be very large, return it modulo 10^9 + 7.

Input & Output

Example 1 — Basic Case

$ Input: n = 4, k = 2

› Output: 10

💡 Note: Initial: [1,1,1,1]. After 1s: [1,2,3,4]. After 2s: [1,3,6,10]. Return a[3] = 10.

Example 2 — Single Element

$ Input: n = 1, k = 3

› Output: 1

💡 Note: Array has only one element [1], which never changes regardless of k.

Example 3 — No Time Elapsed

$ Input: n = 3, k = 0

› Output: 1

💡 Note: No seconds pass, so array remains [1,1,1]. Return a[2] = 1.

Constraints

1 ≤ n ≤ 1000
0 ≤ k ≤ 1000

Visualization

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Asked in

G Google 25 M Microsoft 20 a Amazon 15

The key insight is recognizing this follows Pascal's triangle pattern. Each element after k seconds equals the binomial coefficient C(k+i, i). Best approach uses the mathematical formula to directly calculate C(k+n-1, n-1). Time: O(n), Space: O(1)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force Simulation

⏱️ Time: O(n × k) Space: O(n)

For each of the k seconds, create a new array where each element is the sum of all previous elements plus itself from the current array. This directly follows the problem description.

Mathematical Formula

⏱️ Time: O(n) Space: O(1)

The problem follows Pascal's triangle pattern. After k seconds, a[i] equals C(k+i, i) where C(n,r) is the binomial coefficient. We can calculate this directly using the mathematical formula.

Optimized Prefix Sum

⏱️ Time: O(n × k) Space: O(1)

Instead of creating a new array each time, we can update the array in-place by calculating prefix sums from left to right. This saves space and is slightly more efficient.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Smart Actions

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