Approach: DP on ending index

def numberOfArithmeticSlices(nums):
    total = cur = 0
    for i in range(2, len(nums)):
        if nums[i]-nums[i-1] == nums[i-1]-nums[i-2]:
            cur += 1; total += cur
        else:
            cur = 0
    return total

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def numberOfArithmeticSlices(self, nums):
        total = cur = 0
        for i in range(2, len(nums)):
            if nums[i] - nums[i-1] == nums[i-1] - nums[i-2]:
                cur += 1
                total += cur
            else:
                cur = 0
        return total

import java.util.*;
class Solution {
    public int numberOfArithmeticSlices(int[] nums) {
        int total = 0, cur = 0;
        for (int i = 2; i < nums.length; i++) {
            if (nums[i] - nums[i-1] == nums[i-1] - nums[i-2]) {
                cur++;
                total += cur;
            } else cur = 0;
        }
        return total;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int numberOfArithmeticSlices(vector<int>& nums) {
        int total = 0, cur = 0;
        for (int i = 2; i < (int)nums.size(); i++) {
            if (nums[i] - nums[i-1] == nums[i-1] - nums[i-2]) {
                cur++;
                total += cur;
            } else cur = 0;
        }
        return total;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var numberOfArithmeticSlices = function(nums) {
    let total = 0, cur = 0;
    for (let i = 2; i < nums.length; i++) {
        if (nums[i] - nums[i-1] === nums[i-1] - nums[i-2]) {
            cur++;
            total += cur;
        } else cur = 0;
    }
    return total;
};

Arithmetic Slices

Example 1

Example 2

Example 3

Constraints