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1218. Longest Arithmetic Subsequence of Given Difference

SourceWeekly Contest 157 Q2DifficultyMediumRating1597

Description

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.

A subsequence is a sequence that can be derived from arr by deleting some or no elements without changing the order of the remaining elements.

 

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 <= 105
  • -104 <= arr[i], difference <= 104

Solutions

Solution 1: Dynamic Programming

Thinking

The subsequence must keep order and use a fixed \(difference\). \(n \le 10^5\) rules out subset enumeration. The longest chain ending at \(x\) depends only on the chain ending at \(x-difference\), which, if it exists, has already been seen.

Scanning left to right, set \(f[x]=f[x-difference]+1\). A hash map indexes the predecessor by value; one pass yields every ending length, and we take the maximum.

We can use a hash table \(f\) to store the length of the longest arithmetic subsequence ending with \(x\).

Traverse the array \(\textit{arr}\), and for each element \(x\), update \(f[x]\) to be \(f[x - \textit{difference}] + 1\).

After the traversal, return the maximum value in \(f\) as the answer.

The time complexity is \(O(n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the array \(\textit{arr}\).

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class Solution:
    def longestSubsequence(self, arr: List[int], difference: int) -> int:
        f = defaultdict(int)
        for x in arr:
            f[x] = f[x - difference] + 1
        return max(f.values())

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class Solution {
    public int longestSubsequence(int[] arr, int difference) {
        Map<Integer, Integer> f = new HashMap<>();
        int ans = 0;
        for (int x : arr) {
            f.put(x, f.getOrDefault(x - difference, 0) + 1);
            ans = Math.max(ans, f.get(x));
        }
        return ans;
    }
}

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class Solution {
public:
    int longestSubsequence(vector<int>& arr, int difference) {
        unordered_map<int, int> f;
        int ans = 0;
        for (int x : arr) {
            f[x] = f[x - difference] + 1;
            ans = max(ans, f[x]);
        }
        return ans;
    }
};

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func longestSubsequence(arr []int, difference int) (ans int) {
    f := map[int]int{}
    for _, x := range arr {
        f[x] = f[x-difference] + 1
        ans = max(ans, f[x])
    }
    return
}

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function longestSubsequence(arr: number[], difference: number): number {
    const f: Map<number, number> = new Map();
    for (const x of arr) {
        f.set(x, (f.get(x - difference) ?? 0) + 1);
    }
    return Math.max(...f.values());
}

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use std::collections::HashMap;

impl Solution {
    pub fn longest_subsequence(arr: Vec<i32>, difference: i32) -> i32 {
        let mut f = HashMap::new();
        let mut ans = 0;
        for &x in &arr {
            let count = f.get(&(x - difference)).unwrap_or(&0) + 1;
            f.insert(x, count);
            ans = ans.max(count);
        }
        ans
    }
}

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/**
 * @param {number[]} arr
 * @param {number} difference
 * @return {number}
 */
var longestSubsequence = function (arr, difference) {
    const f = new Map();
    for (const x of arr) {
        f.set(x, (f.get(x - difference) || 0) + 1);
    }
    return Math.max(...f.values());
};

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