Problem

Given an integer array nums, return the number of all the arithmetic subsequences of nums.

A sequence of numbers is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same.

  • For example, [1, 3, 5, 7, 9], [7, 7, 7, 7], and [3, -1, -5, -9] are arithmetic sequences.
  • For example, [1, 1, 2, 5, 7] is not an arithmetic sequence.

A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array.

  • For example, [2,5,10] is a subsequence of [1,2,1,**2**,4,1,**5**,**10**].

The test cases are generated so that the answer fits in 32-bit integer.

Examples

Example 1:

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Input:
nums = [2,4,6,8,10]
Output:
 7
Explanation: All arithmetic subsequence slices are:
[2,4,6]
[4,6,8]
[6,8,10]
[2,4,6,8]
[4,6,8,10]
[2,4,6,8,10]
[2,6,10]

Example 2:

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Input:
nums = [7,7,7,7,7]
Output:
 16
Explanation: Any subsequence of this array is arithmetic.

Solution

Method 1 – DP with Hash Maps (Count by Difference)

Intuition

To count all arithmetic subsequences (length ≥ 3), we can use dynamic programming. For each pair of indices (j, i), if nums[i] - nums[j] = d, then any arithmetic subsequence ending at j with difference d can be extended by nums[i].

We use an array of hash maps, where f[i][d] is the number of arithmetic subsequences ending at index i with difference d (including pairs, i.e., length 2). We only count those of length ≥ 3 in the answer.

Approach

  1. For each index i, loop over all previous indices j < i.
  2. Compute the difference d = nums[i] - nums[j].
  3. The number of subsequences ending at j with difference d is f[j][d].
  4. All those can be extended by nums[i], so add f[j][d] to the answer.
  5. Also, every pair (nums[j], nums[i]) forms a new sequence of length 2, so increment f[i][d] by f[j][d] + 1.
  6. At the end, the answer is the total number of arithmetic subsequences of length ≥ 3.

Code

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##### Python
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from collections import defaultdict
from typing import List

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        f = [defaultdict(int) for _ in nums]
        ans = 0
        for i, x in enumerate(nums):
            for j, y in enumerate(nums[:i]):
                d = x - y
                ans += f[j][d]
                f[i][d] += f[j][d] + 1
        return ans
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import java.util.*;

class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int n = nums.length;
        Map<Long, Integer>[] f = new Map[n];
        Arrays.setAll(f, k -> new HashMap<>());
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                long d = 1L * nums[i] - nums[j];
                int cnt = f[j].getOrDefault(d, 0);
                ans += cnt;
                f[i].put(d, f[i].getOrDefault(d, 0) + cnt + 1);
            }
        }
        return ans;
    }
}
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#include <vector>
#include <unordered_map>
using namespace std;

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& nums) {
        int n = nums.size();
        vector<unordered_map<long long, int>> f(n);
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                long long d = 1LL * nums[i] - nums[j];
                int cnt = f[j][d];
                ans += cnt;
                f[i][d] += cnt + 1;
            }
        }
        return ans;
    }
};
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func numberOfArithmeticSlices(nums []int) int {
    n := len(nums)
    f := make([]map[int]int, n)
    for i := range f {
        f[i] = map[int]int{}
    }
    ans := 0
    for i, x := range nums {
        for j, y := range nums[:i] {
            d := x - y
            cnt := f[j][d]
            ans += cnt
            f[i][d] += cnt + 1
        }
    }
    return ans
}
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function numberOfArithmeticSlices(nums: number[]): number {
    const n = nums.length;
    const f: Map<number, number>[] = Array.from({ length: n }, () => new Map());
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        for (let j = 0; j < i; ++j) {
            const d = nums[i] - nums[j];
            const cnt = f[j].get(d) || 0;
            ans += cnt;
            f[i].set(d, (f[i].get(d) || 0) + cnt + 1);
        }
    }
    return ans;
}

Complexity

  • ⏰ Time complexity: O(n²). For each index i, we loop over all previous indices j < i.
  • 🧺 Space complexity: O(n²). We store a hash map for each index, and in the worst case, each map can have up to O(n) entries.