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Given an integer array arr
and an integer difference
, return the length of the longest arithmetic subsequence in arr
such that the difference between adjacent elements in the subsequence equals difference
.
Example 1:
Input: arr = [1,2,3,4], difference = 1 Output: 4 Explanation: The longest arithmetic subsequence is [1,2,3,4].
Example 2:
Input: arr = [1,3,5,7], difference = 1 Output: 1 Explanation: The longest arithmetic subsequence is any single element.
Example 3:
Input: arr = [1,5,7,8,5,3,4,2,1], difference = -2 Output: 4 Explanation: The longest arithmetic subsequence is [7,5,3,1].
Constraints:
1 <= arr.length <= 10^5
-10^4 <= arr[i], difference <= 10^4
给你一个整数数组 arr
和一个整数 difference
,请你找出 arr
中所有相邻元素之间的差等于给定 difference
的等差子序列,并返回其中最长的等差子序列的长度。
示例 1:
输入:arr = [1,2,3,4], difference = 1 输出:4 解释:最长的等差子序列是 [1,2,3,4]。
示例 2:
输入:arr = [1,3,5,7], difference = 1 输出:1 解释:最长的等差子序列是任意单个元素。
示例 3:
输入:arr = [1,5,7,8,5,3,4,2,1], difference = -2 输出:4 解释:最长的等差子序列是 [7,5,3,1]。
提示:
1 <= arr.length <= 10^5
-10^4 <= arr[i], difference <= 10^4
Runtime: 748 ms
Memory Usage: 23.9 MB
1 class Solution { 2 func longestSubsequence(_ arr: [Int], _ difference: Int) -> Int { 3 var dp:[Int:Int] = [Int:Int]() 4 var longest:Int = 0 5 for i in 0..<arr.count 6 { 7 dp[arr[i]] = dp[arr[i]-difference,default:0] + 1 8 longest = max(longest, dp[arr[i],default:0]) 9 } 10 return longest 11 } 12 }