What is the best way to approach 'Find the N-th Value After K Seconds'?

Start by simulating the array updates, then optimize using prefix sums to avoid redundant recalculations and improve performance.

How can I optimize my solution for large n and k values?

Use the prefix sum technique to reduce unnecessary calculations, which helps in achieving better time complexity.

What is the time complexity of this problem?

The time complexity depends on the solution. A direct simulation is O(n * k), while prefix sum optimization can reduce it to O(n * k) with better performance.

How does prefix sum help in this problem?

Prefix sum allows for faster computation of each element's updated value by maintaining a cumulative sum, thus reducing the need for multiple iterations.

Is it possible to solve this problem without a prefix sum approach?

Yes, but it would be inefficient, especially for large values of n and k, and would lead to time limit exceeded errors.

#3179

Medium

auto_awesome数组·数学

LeetCode 题解工作台

K 秒后第 N 个元素的值

给你两个整数 n 和 k 。最初，你有一个长度为 n 的整数数组 a ，对所有 0 ，都有 a[i] = 1 。每过一秒，你会同时更新每个元素为其前面所有元素的和加上该元素本身。例如，一秒后， a[0] 保持不变， a[1] 变为 a[0] + a[1] ， a[2] 变为 a[0] + a[1]…

数组数学模拟组合数学前缀和

题目描述

给你两个整数 n 和 k。

最初，你有一个长度为 n 的整数数组 a，对所有 0 <= i <= n - 1，都有 a[i] = 1 。每过一秒，你会同时更新每个元素为其前面所有元素的和加上该元素本身。例如，一秒后，a[0] 保持不变，a[1] 变为 a[0] + a[1]，a[2] 变为 a[0] + a[1] + a[2]，以此类推。

返回 k 秒后 a[n - 1] 的值。

由于答案可能非常大，返回其对 10⁹ + 7 取余后的结果。

示例 1：

输入：n = 4, k = 5

输出：56

解释：

时间（秒）	数组状态
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]

示例 2：

输入：n = 5, k = 3

输出：35

解释：

时间（秒）	数组状态
0	[1,1,1,1,1]
1	[1,2,3,4,5]
2	[1,3,6,10,15]
3	[1,4,10,20,35]

提示：

1 <= n, k <= 1000

lightbulb

解题思路

方法一：模拟

我们注意到，整数 $n$ 的范围是 $1 \leq n \leq 1000$ ，因此我们可以直接模拟这个过程。

我们定义一个长度为 $n$ 的数组 $a$ ，并初始化所有元素为 $1$ 。然后我们模拟 $k$ 秒的过程，每一秒我们都更新数组 $a$ 的元素，直到 $k$ 秒结束。

最后，我们返回 $a[n - 1]$ 即可。

时间复杂度 $O(n \times k)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为数组 $a$ 的长度。

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

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Can the candidate optimize a brute-force simulation approach using mathematical tricks like prefix sums?
question_mark
Does the candidate understand the pattern of updating elements based on their predecessors?
question_mark
Is the candidate comfortable with trade-offs between time and space complexity?

warning

常见陷阱

外企场景

error
Overlooking the need for prefix sum optimization and resorting to a brute-force simulation, leading to inefficiency.
error
Not correctly handling the simulation for large values of k, resulting in time limit exceeded errors.
error
Misunderstanding the problem's structure and failing to implement the running sum technique effectively.

swap_horiz

进阶变体

外企场景

arrow_right_alt
What if the array starts with different initial values instead of 1?
arrow_right_alt
What if the problem only asks for the sum of the array instead of the last element after k seconds?
arrow_right_alt
Can this approach be adapted for multidimensional arrays or more complex data structures?

help

常见问题

外企场景

继续练习

#3250 单调数组对的数目 I #3251 单调数组对的数目 II #3312 查询排序后的最大公约数