README_EN.md

comments	difficulty	edit_url	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/0400-0499/0413.Arithmetic%20Slices/README_EN.md	Array Dynamic Programming

413. Arithmetic Slices

中文文档

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

Python3

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        ans, cnt = 0, 2
        d = 3000
        for a, b in pairwise(nums):
            if b - a == d:
                cnt += 1
            else:
                d = b - a
                cnt = 2
            ans += max(0, cnt - 2)
        return ans

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Solution 2

Python3

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