// Source : https://leetcode.com/problems/arithmetic-slices/
// Author : Hao Chen
// Date   : 2016-11-13

/*************************************************************************************** 
 *
 * A sequence of number 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, these are arithmetic sequence:
 * 1, 3, 5, 7, 9
 * 7, 7, 7, 7
 * 3, -1, -5, -9
 * 
 * The following sequence is not arithmetic. 1, 1, 2, 5, 7 
 * 
 * A zero-indexed array A consisting of N numbers is given. A slice of that array is 
 * any pair of integers (P, Q) such that 0 
 * 
 * A slice (P, Q) of array A is called arithmetic if the sequence:
 *     A[P], A[p + 1], ..., A[Q - 1], A[Q] is arithmetic. In particular, this means 
 * that P + 1 
 * 
 * The function should return the number of arithmetic slices in the array A. 
 * 
 * Example:
 * 
 * A = [1, 2, 3, 4]
 * 
 * return: 3, for 3 arithmetic slices in A: [1, 2, 3], [2, 3, 4] and [1, 2, 3, 4] 
 * itself.
 ***************************************************************************************/

class Solution {
public:
    //
    // It's easy to find out how many 3-length slices in an arithmetic array with N length.
    //
    //   len = 3, then  1 slices, sum(1)
    //   len = 4, then  3 slices, sum(1,2) - TWO 3-length slices + ONE 4-length slice
    //   len = 5, then  6 slices, sum(1,2,3) - THREE 3-length slices + TWO 4-length slices + ONE 5-length slice
    //   len = 6, then 10 slices, sum(1,2,3,4) - ...
    //   len = 7, then 15 slices, sum(1,2,3,4,5) - ..
    //
    // So, with N length arithmetic array, there are Sum[1, N-3+1] 3-length slices
    //
    // And, we know the formula sum from 1 to n is:
    //
    //                  n * ( n + 1 )
    //    sum [1, n] = ---------------
    //                       2
    // Then, we could have the solution - O(n) Time with O(1) Space
    //
    
    int sum1toN(int n) {
        return n * (n+1) / 2;
    }

    int numberOfArithmeticSlices(vector<int>& A) {
        int result = 0;
        int len = 0; // the current length of arithmetic
        for (int i=2; i<A.size(); i++) {
            if (A[i] - A[i-1] == A[i-1] - A[i-2]) {
                len++;
            }else{
                result += sum1toN(len);
                len=0;
            }
        }
        return len==0 ? result : result + sum1toN(len);
    }
};