446. Arithmetic Slices II - Subsequence
- Hardness: \(\color{red}\textsf{Hard}\)
- Ralated Topics:
Array、Dynamic Programming
一、題目
Given an integer array nums, return the number of all the arithmetic subsequences of nums.
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,
[1, 3, 5, 7, 9],[7, 7, 7, 7], and[3, -1, -5, -9]are arithmetic sequences. - For example,
[1, 1, 2, 5, 7]is not an arithmetic sequence. A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array. - For example,
[2,5,10]is a subsequence of[1,2,1,2,4,1,5,10]. The test cases are generated so that the answer fits in 32-bit integer.
Example 1:
- 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]
Example 2:
- Input: [7,7,7,7,7]
- Output: 16
- Explanation: Any subsequence of this array is arithmetic.
Constraints:
- 1 <= nums.length <= 1000
- -2^31 <= nums[i] <= 2^31 - 1
二、分析
- 定義
dp[i][j]為索引為i，且間隔為j時的子序列個數。 - 由於間隔可能或大或小，故改用
Hash Map記錄，故我們使用vector<unordered_map<int,int>> dp(n) - 注意數值的範圍，故
target = nums[j] - diff若超出範圍，需要剔除。
三、解題
1. DP
- Time complexity: \(O(n^2)\)
- Space complexity: \(O(n)\)
int numberOfArithmeticSlices(vector<int>& nums) {
int n = nums.size();
int res = 0;
vector<unordered_map<int,long long>> dp(n); // dp[index][diff]
for (int i = 2; i < n; i++) {
unordered_map<int,int> map;
for (int j = 0; j < i; j++) {
long long diff = (long long)nums[i] - (long long)nums[j];
long long target = nums[j] - diff; // check target is in the range
if (target >= INT_MIN && target <= INT_MAX){
if (map.count(target)) dp[i][diff] += map[target]; // 三個一組的子序列
if (dp[j].count(diff)) dp[i][diff] += dp[j][diff]; // 三個以上的子序列
}
map[nums[j]]++;
}
}
for (int i = 0; i < n; i++) {
for (auto m : dp[i]) {
res += m.second;
}
}
return res;
}