Monday, November 7, 2016

LeetCode 446 - Arithmetic Slices II - Subsequence


LeetCode 413 - Arithmetic Slices
https://leetcode.com/problems/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.
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]
X. DP
https://discuss.leetcode.com/topic/67413/detailed-explanation-for-java-o-n-2-solution
At first glance of the problem description, I had a strong feeling that the solution to the original problem can be built through its subproblems, i.e., the total number of arithmetic subsequence slices of the whole input array can be constructed from those of the subarrays of the input array. While I was on the right track to the final solution, it's not so easy to figure out the relations between the original problem and its subproblems.
To begin with, let's be ambitious and reformulate our problem as follows: let T(i) denote the total number of arithmetic subsequence slices that can be formed within subarray A[0, i], where A is the input array and 0 <= i < n with n = A.length. Then our original problem will be T(n - 1), and the base case is T(0) = 0.
To make the above idea work, we need to relate T(i) to all T(j) with 0 <= j < i. Let's take some specific j as an example. If we want to incorporate element A[i] into the subarray A[0, j], what information do we need? As far as I can see, we need to know at least the total number of arithmetic subsequence slices ending at each index k with difference d where 0 <= k <= j and d = A[i] - A[k], (i.e., for each such slice, its last element is A[k] and the difference between every two consecutive elements is d), so that adding A[i] to the end of each such slice will make a new arithmetic subsequence slice.
However, our original formulation of T(i) says nothing about the the total number of arithmetic subsequence slices ending at some particular index and with some particular difference. This renders it impossible to relate T(i) to all T(j)As a rule of thumb, when there is difficulty relating original problem to its subproblems, it usually indicates something goes wrong with your formulation for the original problem.
From our analyses above, each intermediate solution should at least contain information about the total number of arithmetic subsequence slices ending at some particular index with some particular difference. So let's go along this line and reformulate our problem as T(i, d), which denotes the total number of arithmetic subsequence slices ending at index i with difference d. The base case and recurrence relation are as follows:
  1. Base case: T(0, d) = 0 (This is true for any d).
  2. Recurrence relation: T(i, d) = summation of (1 + T(j, d)) as long as 0 <= j < i && d == A[i] - A[j].
For the recurrence relation, it's straightforward to understand the T(j, d) part: for each slice ending at index j with difference d == A[i] - A[j], adding A[i] to the end of the slice will make a new arithmetic subsequence slice, therefore the total number of such new slices will be the same as T(j, d). What you are probably wondering is: where does the 1 come from?
The point here is that to make our recurrence relation work properly, the meaning of arithmetic subsequence slice has to be extended to include slices with only two elements (of course we will make sure these "phony" slices won't contribute to our final count). This is because for each slice, we are adding A[i] to its end to form a new one. If the original slice is of length two, after adding we will have a valid arithmetic subsequence slice with three elements. Our T(i, d) will include all these "generalized" slices. And for each pair of elements (A[j], A[i]), they will form one such "generalized" slice (with only two elements) and thus contribute to one count of T(i, d).
Before jumping to the solution below, I'd like to point out that there are actually overlapping among our subproblems (for example, both T(i, d) and T(i + 1, d) require knowledge of T(j, d) with 0 <= j < i). This necessitates memorization of the intermediate results. Each intermediate result is characterized by two integers: i and d. The former is bounded (i.e., 0 <= i < n) since they are the indices of the element in the input array while the latter is not as d is the difference of two elements in the input array and can be any value. For bounded integers, we can use them to index arrays (or lists) while for unbounded ones, use of HashMap would be more appropriate. So we end up with an array of the same length as the input and whose element type is HashMap.
Here is the Java program (with a quick explanation given at the end). Both time and space complexity are O(n^2). Some minor points for improving time and space performance are:
  1. Define the type of the difference as Integer type instead of Long. This is because there is no valid arithmetic subsequence slice that can have difference out of the Integer value range. But we do need a long integer to filter out those invalid cases.
  2. Preallocate the HashMap to avoid reallocation to deal with extreme cases.
  3. Refrain from using lambda expressions inside loops.
public int numberOfArithmeticSlices(int[] A) {
    int res = 0;
    Map<Integer, Integer>[] map = new Map[A.length];
  
    for (int i = 0; i < A.length; i++) {
        map[i] = new HashMap<>(i);
         
        for (int j = 0; j < i; j++) {
            long diff = (long)A[i] - A[j];
            if (diff <= Integer.MIN_VALUE || diff > Integer.MAX_VALUE) continue;
          
            int d = (int)diff;
            int c1 = map[i].getOrDefault(d, 0);
            int c2 = map[j].getOrDefault(d, 0);
            res += c2;
            map[i].put(d, c1 + c2 + 1);
        }
    }
  
    return res;
}
Quick explanation:
  1. res is the final count of all valid arithmetic subsequence slices; map will store the intermediate results T(i, d), with iindexed into the array and d as the key of the corresponding HashMap.
  2. 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 T(j, d) with j less than i.
  3. Within the inner loop, we first use a long variable diff to filter out invalid cases, then get the counts of all valid slices (with element >= 3) as c2 and add it to the final count. At last we update the count of all "generalized" slices for T(i, d) by adding the three parts together: the original value of T(i, d), which is c1 here, the counts from T(j, d), which is c2 and lastly the 1 count of the "two-element" slice (A[j], A[i]).
http://www.cnblogs.com/grandyang/p/6057934.html
这道题是之前那道Arithmetic Slices的延伸,但是比较简单是因为要求等差数列是连续的,而这道题让我们求是等差数列的子序列,可以跳过某些数字,不一定非得连续,那么难度就加大了,但还是需要用DP来做。我们建立一个一维数组dp,数组里的元素不是数字,而是放一个哈希表,建立等差数列的差值和其长度之间的映射。我们遍历数组中的所有数字,对于当前遍历到的数字,又从开头遍历到当前数字,计算两个数字之差diff,如果越界了不做任何处理,如果没越界,我们看dp[i]中diff的差值映射自增1,然后我们看dp[j]中是否有diff的映射,如果有的话,说明此时已经能构成等差数列了,将dp[j][d]加入结果res中,然后再更新dp[i][d],这样等遍历完数组,res即为所求,我们用题目中给的例子数字[2,4,6,8,10]来看:
2    4    6    8   10    
    2-1  4-1  6-1  8-1
         2-2  4-1  6-1 
              2-3  4-2

最终累计出来的结果是上面红色的数字1+2+1+3=7,分别对应着如下的等差数列:
1:[2,4,6]
2:[4,6,8] [2,4,6,8]
1:[2,6,10]
3:[6,8,10] [4,6,8,10] [2,4,6,8,10]
    int numberOfArithmeticSlices(vector<int>& A) {
        int res = 0, n = A.size();
        vector<unordered_map<int, int>> dp(n);
        for (int i = 0; i < n; ++i) {
            for (int j = 0; j < i; ++j) {
                long long diff = (long long)A[i] - (long long)A[j];
                if (diff > INT_MAX || diff < INT_MIN) continue;
                int d = (int)diff;
                if (!dp[i].count(d)) dp[i][d] = 0;
                ++dp[i][d];
                if (dp[j].count(d)) {
                    dp[i][d] += dp[j][d];
                    res += dp[j][d];
                }
            }
        }
        return res;
    }

http://blog.csdn.net/gqk289/article/details/62218365
  1.     public int numberOfArithmeticSlices(int[] A) {  
  2.         int res = 0;  
  3.         HashMap<Integer, Integer>[] map = new HashMap[A.length];  
  4.         for (int i = 0; i < A.length; i++) {  
  5.             map[i] = new HashMap();  
  6.             for (int j = 0; j < i; j++) {  
  7.                 long diff = (long)A[i] - A[j];  
  8.                 if (diff > Integer.MAX_VALUE || diff < Integer.MIN_VALUE) {  
  9.                     continue;  
  10.                 }  
  11.                 int d = (int)diff;  
  12.                 int c1 = map[i].getOrDefault(d, 0);  
  13.                 int c2 = map[j].getOrDefault(d, 0);  
  14.                 // c2保存以A[j]为尾,gap为d的数组个数,如果把A[i]加到A[j]后面,  
  15.                 // 那么以A[j]为尾,gap为d的数组个数也是c2  
  16.                 res += c2;  
  17.                 map[i].put(d, c1 + c2 + 1);  
  18.             }  
  19.         }  
  20.         return res;  
  21.     }  
http://bookshadow.com/weblog/2016/11/06/leetcode-arithmetic-slices-ii-subsequence/
如果一组长度至少为3的数,两两连续元素之差都相等,则称其为等差数列。
给定包含N个元素,起始下标为0的数组A。子序列切片是指任意整数序列(P0, P1, ..., Pk)满足0 ≤ P0 < P1 < ... < Pk < N。
如果A[P0], A[P1], ..., A[Pk-1], A[Pk]是等差序列,则数组A的子序列切片 (P0, P1, ..., Pk) 为等差子序列切片。特别的,k ≥ 2。
函数应当返回数组A的等差子序列切片个数。
输入包含N个整数。每一个整数范围为[ -2^31, 2^31 - 1 ],并且 0 ≤ N ≤ 1000。输出确保小于 2^31-1。

解题思路:

动态规划(Dynamic Programming)
状态转移方程:dp[x][delta] += dp[y][delta] + 1(y∈[0, x - 1])
dp[x][delta]表示以第x个元素结尾,且公差为delta的等差子序列切片个数。
def numberOfArithmeticSlices(self, A): """ :type A: List[int] :rtype: int """ size = len(A) ans = 0 dp = [collections.defaultdict(int) for x in range(size)] for x in range(size): for y in range(x): delta = A[x] - A[y] dp[x][delta] += 1 if delta in dp[y]: dp[x][delta] += dp[y][delta] ans += dp[y][delta] return ans
https://discuss.leetcode.com/topic/67012/java-15-lines-solution
I think the idea is based on each element to cal the number of all possible differences cases. ==>
[2,4,6,8,10]
Like based on 6, to calculate the difference value of 4,2,0,-2,-4, the map is
4:1
2:2
so when we move to element 8, the map will be
6:1
4:1
2:3
    public int numberOfArithmeticSlices(int[] A) {
        int re = 0;
        HashMap<Integer, Integer>[] maps = new HashMap[A.length];
        for(int i=0; i<A.length; i++) {
            maps[i] = new HashMap<>();
            int num = A[i];
            for(int j=0; j<i; j++) {
                if((long)num-A[j]>Integer.MAX_VALUE) continue;
                if((long)num-A[j]<Integer.MIN_VALUE) continue;
                int diff = num - A[j];//\\
                int count = maps[j].getOrDefault(diff, 0);
                maps[i].put(diff, maps[i].getOrDefault(diff,0)+count+1);
                re += count;
            }
        }
        return re;
    }

http://www.itdadao.com/articles/c15a715289p0.html

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