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446. Arithmetic Slices II - Subsequence

Description

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.

 

Example 1:

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:

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

 

Constraints:

  • 1  <= nums.length <= 1000
  • -231 <= nums[i] <= 231 - 1

Solutions

Solution 1

Thinking

Subsequences need not be contiguous and the difference can be huge, so listing every length-\(\ge 3\) subsequence is impossible. \(n\le 1000\) allows \(O(n^2)\).

Let \(f[i][d]\) be the number of weak arithmetic subsequences (at least two terms) ending at \(i\) with difference \(d\). For \(j<i\) and \(d=nums[i]-nums[j]\), the \(f[j][d]\) sequences become real slices once \(nums[i]\) is appended, and \(f[i][d]\) grows by \(f[j][d]+1\) (including the new pair).

Differences live in hash maps. Add \(f[j][d]\) to the answer before updating \(f[i][d]\), so weak sequences and true slices are handled in one double loop.

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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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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].merge(d, cnt + 1, Integer::sum);
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& nums) {
        int n = nums.size();
        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) (ans int) {
    f := make([]map[int]int, len(nums))
    for i := range f {
        f[i] = map[int]int{}
    }
    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
}

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function numberOfArithmeticSlices(nums: number[]): number {
    const n = nums.length;
    const f: Map<number, number>[] = new Array(n).fill(0).map(() => 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;
}

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