Longest Arithmetic Subsequence of Given Difference

class Solution {
public:
    int longestSubsequence(vector<int>& sequence, int diff) {
        unordered_map<int, int> sequenceDP;
        int maxLen = 1;
        
        for (int value : sequence) {
            int priorValue = sequenceDP.count(value - diff) ? sequenceDP[value - diff] : 0;
            sequenceDP[value] = priorValue + 1;
            maxLen = max(maxLen, sequenceDP[value]);
        }
        
        return maxLen;
    }
};

This C++ solution addresses the problem of finding the longest arithmetic subsequence in an array where each pair of consecutive elements has a given difference. The method employed utilizes dynamic programming (DP) to efficiently compute the solution. The use of an unordered map sequenceDP tracks the length of the arithmetic subsequences each element can form.

To solve the problem:

• Initialize maxLen to 1, which is the smallest possible length of any subsequence, and sequenceDP to store the lengths of the subsequences.
• Iterate through each element value in the array sequence.
• For each value, compute priorValue, which is the subsequence length from the previous element (i.e., value - diff) if it exists in sequenceDP.
• Update sequenceDP[value] by adding 1 to priorValue. This step essentially extends the subsequence ending with the previous element to include the current element.
• Update maxLen to keep track of the maximum subsequence length encountered so far.

The final value of maxLen represents the length of the longest arithmetic subsequence with the given difference. By leveraging dynamic programming with a hash map, this approach ensures an efficient solution that avoids the quadratic time complexity of naive methods.

class Solution {
    public int maxSubsequenceLength(int[] sequence, int diff) {
        Map<Integer, Integer> lengthMap = new HashMap<>();
        int maxLength = 1;

        for (int num : sequence) {
            int prevLength = lengthMap.getOrDefault(num - diff, 0);
            lengthMap.put(num, prevLength + 1);
            maxLength = Math.max(maxLength, lengthMap.get(num));
        }

        return maxLength;
    }
}

This guide covers how to find the longest arithmetic subsequence of a given difference in an array of integers using Java. The provided method, maxSubsequenceLength, achieves this by employing a HashMap to track the lengths of subsequences ending at each element in the sequence.

• Initialize a HashMap named lengthMap where keys represent numbers from the array and values represent the maximum subsequence length ending at that number.
• Set maxLength to 1, which accounts for the minimum possible subsequence length.
• Loop through each integer num in the sequence array:
  • Compute prevLength using the HashMap for the number num minus the given diff. If not found in the map, default to 0.
  • Update the lengthMap for the number num to prevLength + 1.
  • Update maxLength whenever a longer subsequence is found by comparing the current value with the value stored in lengthMap for num.
• Return maxLength, which now contains the length of the longest arithmetic subsequence with the given difference.

This solution adeptly leverages the HashMap data structure to store interim results, effectively reducing the computational complexity in comparison to the naive approach, making it efficient for larger datasets.

class Solution:
    def longestSubseq(self, sequence: List[int], diff: int) -> int:
        memo = {}
        max_length = 1
        for number in sequence:
            prev_val = memo.get(number - diff, 0)
            memo[number] = prev_val + 1
            max_length = max(max_length, memo[number])
        
        return max_length

This Python solution tackles the problem of finding the longest arithmetic subsequence in an array with a given difference. Utilize a dictionary memo to store sequences with the number as the key and the sequence length as the value. As you iterate through each number in sequence, compute the previous number number - diff and fetch its value from memo with a default of 0. Store or update the current number's sequence length in memo by incrementing the fetched value. Simultaneously, update max_length to ensure it holds the value of the longest found subsequence at any point. Return max_length as the desired result, reflecting the length of the longest arithmetic subsequence for the provided difference. This approach ensures efficiency by avoiding redundant recalculations.