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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;
    }
    
    

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