Given an array A of integers, return the length of the longest arithmetic subsequence in A.
Recall that a subsequence of A is a list A[i_1], A[i_2], ..., A[i_k] with 0 <= i_1 < i_2 < ... < i_k <= A.length - 1, and that a sequence B is arithmetic if B[i+1] - B[i] are all the same value (for 0 <= i < B.length - 1).
Example 1:
Input: [3,6,9,12]
Output: 4
Explanation:
The whole array is an arithmetic sequence with steps of length = 3.
Example 2:
Input: [9,4,7,2,10]
Output: 3
Explanation:
The longest arithmetic subsequence is [4,7,10].
Example 3:
Input: [20,1,15,3,10,5,8]
Output: 4
Explanation:
The longest arithmetic subsequence is [20,15,10,5].
Note:
2 <= A.length <= 2000
0 <= A[i] <= 10000
class Solution {
// The main idea is to maintain a map of differences seen at each index.
public int longestArithSeqLength(int[] A) {
if (A.length <= 1)
return A.length;
int longest = 0;
HashMap<Integer, Integer>[] dp = new HashMap[A.length];
for (int i = 0; i < A.length; ++i) {
// Initialize the map.
dp[i] = new HashMap<Integer, Integer>();
}
for (int i = 1; i < A.length; ++i) {
int x = A[i];
for (int j = 0; j < i; ++j) {
int y = A[j];
int d = x - y;
// We at least have a minimum chain length of 2 now,
// given that (A[j], A[i]) with the difference d,
// by default forms a chain of length 2.
int len = 2;
if (dp[j].containsKey(d)) {
// At index j, if we had already seen a difference d,
// then potentially, we can add A[i] to the same chain
// and extend it by length 1.
len = dp[j].get(d) + 1;
}
// Update the max chain length for difference d at index i.
dp[i].put(d, len);
// Update the global max.
longest = Math.max(longest, dp[i].get(d));
}
}
return longest;
}
}