Skip to content

413. Arithmetic Slices

Description

An integer array 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.

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

A subarray is a contiguous subsequence of the array.

 

Example 1:

Input: nums = [1,2,3,4]
Output: 3
Explanation: We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] itself.

Example 2:

Input: nums = [1]
Output: 0

 

Constraints:

  • 1 <= nums.length <= 5000
  • -1000 <= nums[i] <= 1000

Solutions

Solution 1: Iteration and Counting

We use \(d\) to represent the current difference between two adjacent elements, and \(cnt\) to represent the length of the current arithmetic sequence. Initially, \(d = 3000\), \(cnt = 2\).

We iterate through the array nums. For two adjacent elements \(a\) and \(b\), if \(b - a = d\), it means that the current element \(b\) also belongs to the current arithmetic sequence, and we increment \(cnt\) by 1. Otherwise, it means that the current element \(b\) does not belong to the current arithmetic sequence, and we update \(d = b - a\), and \(cnt = 2\). If \(cnt \ge 3\), it means that the length of the current arithmetic sequence is at least 3, and the number of arithmetic sequences is \(cnt - 2\), which we add to the answer.

After the iteration, we can get the answer.

In the code implementation, we can also initialize \(cnt\) to \(0\), and when resetting \(cnt\), we directly set \(cnt\) to \(0\). When adding to the answer, we directly add \(cnt\).

The time complexity is \(O(n)\), and the space complexity is \(O(1)\). Where \(n\) is the length of the array nums.

Similar problems:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        ans = cnt = 0
        d = 3000
        for a, b in pairwise(nums):
            if b - a == d:
                cnt += 1
            else:
                d = b - a
                cnt = 0
            ans += cnt
        return ans

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int ans = 0, cnt = 0;
        int d = 3000;
        for (int i = 0; i < nums.length - 1; ++i) {
            if (nums[i + 1] - nums[i] == d) {
                ++cnt;
            } else {
                d = nums[i + 1] - nums[i];
                cnt = 0;
            }
            ans += cnt;
        }
        return ans;
    }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& nums) {
        int ans = 0, cnt = 0;
        int d = 3000;
        for (int i = 0; i < nums.size() - 1; ++i) {
            if (nums[i + 1] - nums[i] == d) {
                ++cnt;
            } else {
                d = nums[i + 1] - nums[i];
                cnt = 0;
            }
            ans += cnt;
        }
        return ans;
    }
};

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
func numberOfArithmeticSlices(nums []int) (ans int) {
    cnt, d := 0, 3000
    for i, b := range nums[1:] {
        a := nums[i]
        if b-a == d {
            cnt++
        } else {
            d = b - a
            cnt = 0
        }
        ans += cnt
    }
    return
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
function numberOfArithmeticSlices(nums: number[]): number {
    let ans = 0;
    let cnt = 0;
    let d = 3000;
    for (let i = 0; i < nums.length - 1; ++i) {
        const a = nums[i];
        const b = nums[i + 1];
        if (b - a == d) {
            ++cnt;
        } else {
            d = b - a;
            cnt = 0;
        }
        ans += cnt;
    }
    return ans;
}

Comments