446. Arithmetic Slices II - Subsequence

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, these are arithmetic sequences:

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 subsequence slice of that array is any sequence of integers (P0, P1, ..., Pk) such that 0 ≤ P0 < P1 < ... < Pk < N.

A subsequence slice (P0, P1, ..., Pk) of array A is called arithmetic if the sequence A[P0], A[P1], ..., A[Pk-1], A[Pk] is arithmetic. In particular, this means that k ≥ 2.

The function should return the number of arithmetic subsequence slices in the array A.

The input contains N integers. Every integer is in the range of -231 and 231-1 and 0 ≤ N ≤ 1000. The output is guaranteed to be less than 231-1.

 

Example:

Input: [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]

 

Approach #1: DP. [C++]

class Solution {
public:
    int numberOfArithmeticSlices(vector<int>& A) {
        if (A.size() == 0) return 0;
        vector<map<int, int>> dp(A.size()+1);
        
        int res = 0;
        for (int i = 0; i < A.size(); ++i) {
            for (int j = 0; j < i; ++j) {
                long dif = (long)A[i] - A[j];
                if (dif < INT_MIN || dif > INT_MAX) continue;
                int d = (int)dif;
                dp[i][d] += 1;
                if (dp[j].find(d) != dp[j].end()) {
                    dp[i][d] += dp[j][d];
                    res += dp[j][d];
                }
            }
        }
        
        return res;
    }
};

　　

Analysis:

1. res is the final count of all valid  arithmetic subsequence slices;

2. dp will store the intermediate results [i, [dif, count]], with i indexed into the array and dif as the key. count is the number of result with the intermediate results.

3. for each index i, we find the total number of "generalized" arithmetic subsequence slices ending at it with all possible differences. This is done by attaching A[i] to all slices of dp[j][d] with j less than i.

4. Within the inner loop, we first use a long variable diff to filter out invalid cases, then get the count of all valid slices (with element >= 3) as dp[j][d] add it to the final count. At last we update the count of all "generalized" slices for dp[i][d] by adding the two parts together: the orginal value of dp[i][d], the counts from dp[j][d].

 

Reference:

https://leetcode.com/problems/arithmetic-slices-ii-subsequence/