Longest Arithmetic Subsequence with Difference Constraint - Google Top Interview Questions


Problem Statement :


Given a list of integers nums and an integer diff, return the length of the longest arithmetic subsequence where the difference between each consecutive numbers in the subsequence is diff.

Constraints

0 ≤ n ≤ 100,000 where n is the length of nums

Example 1

Input

nums = [-2, 0, 3, 6, 1, 9]

diff = 3

Output
4

Explanation

We can pick the subsequence [0, 3, 6, 9].



Example 2

Input

nums = [9, 8, 7, 5, 3]

diff = -2

Output

4

Explanation

We can pick the subsequence [9, 7, 5, 3].



Solution :



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                        Solution in C++ :

int solve(vector<int>& nums, int diff) {
    int res = 0;
    unordered_map<int, int> hash;

    for (const auto& num : nums) res = max(res, hash[num] = hash[num - diff] + 1);
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] nums, int diff) {
        int n = nums.length;
        if (n == 0)
            return 0;

        // Maximum length is atleast 1
        int maxLen = 1;
        // Init map of processed values
        Map<Integer, Integer> numsMap = new HashMap<>();
        for (int i = 0; i < n; i++) {
            int comp = nums[i] - diff;
            if (numsMap.containsKey(comp))
                numsMap.put(nums[i], numsMap.get(comp) + 1);
            else
                numsMap.put(nums[i], 1);
            maxLen = Math.max(numsMap.get(nums[i]), maxLen);
        }
        return maxLen;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, nums, diff):
        seen = defaultdict(int)
        mx = 0
        for x in nums:
            if x - diff in seen:
                seen[x] = seen[x - diff] + 1
            else:
                seen[x] = 1
            mx = max(mx, seen[x])
        return mx
                    


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