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 :
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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