Given an array nums
of integers, return the length of the longest arithmetic subsequence in nums
.
Recall that a subsequence of an array nums
is a list nums[i1], nums[i2], ..., nums[ik]
with 0 <= i1 < i2 < ... < ik <= nums.length - 1
, and that a sequence seq
is arithmetic if seq[i+1] - seq[i]
are all the same value (for 0 <= i < seq.length - 1
).
Example 1:
Input: nums = [3,6,9,12] Output: 4 Explanation: The whole array is an arithmetic sequence with steps of length = 3.
Example 2:
Input: nums = [9,4,7,2,10] Output: 3 Explanation: The longest arithmetic subsequence is [4,7,10].
Example 3:
Input: nums = [20,1,15,3,10,5,8] Output: 4 Explanation: The longest arithmetic subsequence is [20,15,10,5].
Constraints:
2 <= nums.length <= 1000
0 <= nums[i] <= 500
class Solution:
def longestArithSeqLength(self, nums: List[int]) -> int:
n = len(nums)
dp = [[1] * 1001 for _ in range(n)]
ans = 0
for i in range(1, n):
for j in range(i):
d = nums[i] - nums[j] + 500
dp[i][d] = max(dp[i][d], dp[j][d] + 1)
ans = max(ans, dp[i][d])
return ans
class Solution {
public int longestArithSeqLength(int[] nums) {
int n = nums.length;
int ans = 0;
int[][] dp = new int[n][1001];
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
int d = nums[i] - nums[j] + 500;
dp[i][d] = Math.max(dp[i][d], dp[j][d] + 1);
ans = Math.max(ans, dp[i][d]);
}
}
return ans + 1;
}
}
class Solution {
public:
int longestArithSeqLength(vector<int>& nums) {
int n = nums.size();
int ans = 0;
vector<vector<int>> dp(n, vector<int>(1001, 1));
for (int i = 1; i < n; ++i) {
for (int j = 0; j < i; ++j) {
int d = nums[i] - nums[j] + 500;
dp[i][d] = max(dp[i][d], dp[j][d] + 1);
ans = max(ans, dp[i][d]);
}
}
return ans;
}
};
func longestArithSeqLength(nums []int) int {
n := len(nums)
dp := make([][]int, n)
for i := range dp {
dp[i] = make([]int, 1001)
}
ans := 0
for i := 1; i < n; i++ {
for j := 0; j < i; j++ {
d := nums[i] - nums[j] + 500
dp[i][d] = max(dp[i][d], dp[j][d]+1)
ans = max(ans, dp[i][d])
}
}
return ans + 1
}
func max(a, b int) int {
if a > b {
return a
}
return b
}