Description Link to heading

1218. Longest Arithmetic Subsequence of Given Difference (Medium)

  1. Longest Arithmetic Subsequence of Given Difference (Medium)

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

Solution Link to heading

We can use a hash table ump[nums[i]] to record the length of longest arithmetic subsequence in which nums[i] is the last element.

We have

if (ump.find(num - difference) != ump.end()) {
    ump[num] = ump[num - difference] + 1;
} else {
    ump[num] = 1;
}

Code Link to heading 

class Solution {
  public:
    int longestSubsequence(vector<int> &arr, int difference) {
        unordered_map<int, int> ump;
        for (auto &num : arr) {
            if (ump.find(num - difference) != ump.end()) {
                ump[num] = ump[num - difference] + 1;
            } else {
                ump[num] = 1;
            }
        }
        int res = 0;
        for (auto &num : ump) {
            res = res < num.second ? num.second : res;
        }
        return res;
    }
};