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Formatted question description: https://leetcode.ca/all/1027.html

# 1027. Longest Arithmetic Sequence (Medium)

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:

1. 2 <= A.length <= 2000
2. 0 <= A[i] <= 10000

Related Topics:
Dynamic Programming

## Solution 1. DP

// OJ: https://leetcode.com/problems/longest-arithmetic-sequence/
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
int longestArithSeqLength(vector<int>& A) {
unordered_map<int, unordered_map<int, int>> m;
int ans = 0;
for (int n : A) {
m[n] = {};
for (auto &p : m) {
int d = n - p.first;
ans = max(ans, m[n][d] = p.second.count(d) ? p.second[d] + 1 : 2);
}
}
return ans;
}
};

• class Solution {
public int longestArithSeqLength(int[] A) {
if (A == null)
return 0;
int length = A.length;
if (length <= 2)
return length;
int maxLength = 2;
Map<Integer, TreeSet<Integer>> numIndicesMap = new HashMap<Integer, TreeSet<Integer>>();
for (int i = 0; i < length; i++) {
int num = A[i];
TreeSet<Integer> indices = numIndicesMap.getOrDefault(num, new TreeSet<Integer>());
indices.add(i);
numIndicesMap.put(num, indices);
if (indices.size() > maxLength)
maxLength = indices.size();
}
for (int i = 0; i < length; i++) {
int num1 = A[i];
for (int j = i + 1; j < length; j++) {
int num2 = A[j];
int difference = num2 - num1;
if (difference == 0)
continue;
int curLength = 2;
int prevIndex = j;
int prevNum = num2;
while (true) {
int curNum = prevNum + difference;
TreeSet<Integer> indices = numIndicesMap.getOrDefault(curNum, new TreeSet<Integer>());
Integer curIndex = indices.ceiling(prevIndex);
if (curIndex == null)
break;
curLength++;
prevIndex = curIndex;
prevNum = curNum;
}
maxLength = Math.max(maxLength, curLength);
}
}
return maxLength;
}
}

############

class Solution {
public int longestArithSeqLength(int[] nums) {
int n = nums.length;
int ans = 0;
int[][] f = new int[n][1001];
for (int i = 1; i < n; ++i) {
for (int k = 0; k < i; ++k) {
int j = nums[i] - nums[k] + 500;
f[i][j] = Math.max(f[i][j], f[k][j] + 1);
ans = Math.max(ans, f[i][j]);
}
}
return ans + 1;
}
}

• // OJ: https://leetcode.com/problems/longest-arithmetic-sequence/
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
int longestArithSeqLength(vector<int>& A) {
unordered_map<int, unordered_map<int, int>> m;
int ans = 0;
for (int n : A) {
m[n] = {};
for (auto &p : m) {
int d = n - p.first;
ans = max(ans, m[n][d] = p.second.count(d) ? p.second[d] + 1 : 2);
}
}
return ans;
}
};

• 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

############

# 1027. Longest Arithmetic Subsequence
# https://leetcode.com/problems/longest-arithmetic-subsequence/

class Solution:
def longestArithSeqLength(self, nums: List[int]) -> int:
dp = {}
n = len(nums)

for i in range(n):
for j in range(i + 1, n):
dp[(j, nums[j] - nums[i])] = dp.get((i, nums[j] - nums[i]), 1) + 1

return max(dp.values())


• func longestArithSeqLength(nums []int) int {
n := len(nums)
f := make([][]int, n)
for i := range f {
f[i] = make([]int, 1001)
}
ans := 0
for i := 1; i < n; i++ {
for k := 0; k < i; k++ {
j := nums[i] - nums[k] + 500
f[i][j] = max(f[i][j], f[k][j]+1)
ans = max(ans, f[i][j])
}
}
return ans + 1
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

• function longestArithSeqLength(nums: number[]): number {
const n = nums.length;
let ans = 0;
const f: number[][] = Array.from({ length: n }, () =>
new Array(1001).fill(0),
);
for (let i = 1; i < n; ++i) {
for (let k = 0; k < i; ++k) {
const j = nums[i] - nums[k] + 500;
f[i][j] = Math.max(f[i][j], f[k][j] + 1);
ans = Math.max(ans, f[i][j]);
}
}
return ans + 1;
}