{ "data": { "question": { "questionId": "3962", "questionFrontendId": "3699", "categoryTitle": "Algorithms", "boundTopicId": 3793313, "title": "Number of ZigZag Arrays I", "titleSlug": "number-of-zigzag-arrays-i", "content": "
You are given three integers n
, l
, and r
.
A ZigZag array of length n
is defined as follows:
[l, r]
.Return the total number of valid ZigZag arrays.
\n\nSince the answer may be large, return it modulo 109 + 7
.
A sequence is said to be strictly increasing if each element is strictly greater than its previous one (if exists).
\n\nA sequence is said to be strictly decreasing if each element is strictly smaller than its previous one (if exists).
\n\n\n
Example 1:
\n\nInput: n = 3, l = 4, r = 5
\n\nOutput: 2
\n\nExplanation:
\n\nThere are only 2 valid ZigZag arrays of length n = 3
using values in the range [4, 5]
:
[4, 5, 4]
[5, 4, 5]
Example 2:
\n\nInput: n = 3, l = 1, r = 3
\n\nOutput: 10
\n\nExplanation:
\n\nThere are 10 valid ZigZag arrays of length n = 3
using values in the range [1, 3]
:
[1, 2, 1]
, [1, 3, 1]
, [1, 3, 2]
[2, 1, 2]
, [2, 1, 3]
, [2, 3, 1]
, [2, 3, 2]
[3, 1, 2]
, [3, 1, 3]
, [3, 2, 3]
All arrays meet the ZigZag conditions.
\n\n
Constraints:
\n\n3 <= n <= 2000
1 <= l < r <= 2000
给你 三个整数 n
、l
和 r
。
长度为 n
的 ZigZag 数组定义如下:
[l, r]
。返回满足条件的 ZigZag 数组的总数。
\n\n由于答案可能很大,请将结果对 109 + 7
取余数。
序列 被称为 严格递增 需要满足:当且仅当每个元素都严格大于它的前一个元素(如果存在)。
\n\n序列 被称为 严格递减 需要满足,当且仅当每个元素都严格小于它的前一个元素(如果存在)。
\n\n\n\n
示例 1:
\n\n输入:n = 3, l = 4, r = 5
\n\n输出:2
\n\n解释:
\n\n在取值范围 [4, 5]
内,长度为 n = 3
的 ZigZag 数组只有 2 种:
[4, 5, 4]
[5, 4, 5]
示例 2:
\n\n输入:n = 3, l = 1, r = 3
\n\n输出:10
\n\n解释:
\n\n在取值范围 [1, 3]
内,长度为 n = 3
的 ZigZag 数组共有 10 种:
[1, 2, 1]
, [1, 3, 1]
, [1, 3, 2]
[2, 1, 2]
, [2, 1, 3]
, [2, 3, 1]
, [2, 3, 2]
[3, 1, 2]
, [3, 1, 3]
, [3, 2, 3]
所有数组均符合 ZigZag 条件。
\n\n\n
提示:
\n\n3 <= n <= 2000
1 <= l < r <= 2000
dp[i][dir][x]
be the count of length-i
sequences ending at value x
where dir
is the required next comparison (0 = down, 1 = up).",
"If the required move is up
(dir=1) do dp[i+1][0][y] += sum(dp[i][1][x]) for x < y
; if the required move is down
(dir=0) do dp[i+1][1][y] += sum(dp[i][0][x]) for x > y
.",
"Speed up with prefix/suffix sums so each layer updates in O(m
) instead of O(m
2); take values mod 109+7
."
],
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