LeetCode: Longest Arithmetic Subsequence of Given Difference

Posted on August 5, 2019July 26, 2020 by braindenny

Longest Arithmetic Subsequence of Given Difference

Similar Problems:

Given an integer array arr and an integer difference, return the length of the longest subsequence in arr which is an arithmetic sequence 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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution: dp with hashmap: O(N) space + no need for initialization

// https://code.dennyzhang.com/longest-arithmetic-subsequence-of-given-difference
// Basic Ideas: dynamic programming + hashmap
//
// dp(i)
//     j: arr[i]-difference, dp(j)+1
//     1
// Complexity: Time O(n), Space O(n)
func longestSubsequence(arr []int, difference int) int {
    // m[k]: the width when the subsequence ends with k
    m := map[int]int{}
    res := 0
    for _, v := range arr {
        // dp[prev] gets 0, if not found
        m[v] = m[v-difference]+1
        // evaluate candidate
        if m[v] > res {
            res = m[v]
        }
    }
    return res
}

Solution: dp with hashmap: O(2N) space

// https://code.dennyzhang.com/longest-arithmetic-subsequence-of-given-difference
// Basic Ideas: dynamic programming + hashmap
//
// dp(i)
//     j: arr[i]-difference, dp(j)+1
//     1
// Complexity: Time O(n), Space O(n)
func longestSubsequence(arr []int, difference int) int {
    m := map[int]int{}
    dp := make([]int, len(arr))
    dp[0] = 1
    m[arr[0]] = 0
    res := 1
    for i:=1; i<len(arr); i++ {
        prev := arr[i]-difference
        if j, ok := m[prev]; ok {
            dp[i] = dp[j]+1
        } else {
            dp[i] = 1
        }
        // update index
        m[arr[i]] = i
        // evaluate candidate
        if dp[i] > res {
            res = dp[i]
        }
    }
    return res
}

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