July 15, 2023
Summary: Input size / sub-problem size / sub-problem complexity / time complexity / space complexity
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Input: O(n) Sub-problems: O(n) Each sub-problem depends on O(1) smaller problems Time complexity: O(n) Space complexity: O(n) -> O(1) dp[i] := solution of A[1->i] // prefix |
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dp = new int[n + 1] for i = 1 to n: dp[i] = f(A[i], dp[i-1], dp[i-2], ...) return dp[n] |
Summary and slides
Given a square array of integers A, we want the minimum sum of a falling path through A.
A falling path starts at any element in the first row, and chooses one element from each row. The next row’s choice must be in a column that is different from the previous row’s column by at most one.
Example 1:
Input: [[1,2,3],[4,5,6],[7,8,9]] Output: 12 Explanation: The possible falling paths are:
[1,4,7], [1,4,8], [1,5,7], [1,5,8], [1,5,9][2,4,7], [2,4,8], [2,5,7], [2,5,8], [2,5,9], [2,6,8], [2,6,9][3,5,7], [3,5,8], [3,5,9], [3,6,8], [3,6,9]The falling path with the smallest sum is [1,4,7], so the answer is 12.
Note:
1 <= A.length == A[0].length <= 100-100 <= A[i][j] <= 100Time complexity: O(mn)
Space complexity: O(mn)
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// Author: Huahua, 12 ms class Solution { public: int minFallingPathSum(vector<vector<int>>& A) { int n = A.size(); int m = A[0].size(); vector<vector<int>> dp(n + 2, vector<int>(m + 2)); for (int i = 1; i <= n; ++i) { dp[i][0] = dp[i][m + 1] = INT_MAX; for (int j = 1; j <= m; ++j) dp[i][j] = *min_element(dp[i - 1].begin() + j - 1, dp[i - 1].begin() + j + 2) + A[i - 1][j - 1]; } return *min_element(dp[n].begin() + 1, dp[n].end() - 1); } }; |
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// Author: Huahua, 12 ms class Solution { public: int minFallingPathSum(vector<vector<int>>& A) { int n = A.size(); int m = A[0].size(); for (int i = 1; i < n; ++i) for (int j = 0; j < m; ++j) { int sum = A[i - 1][j]; if (j > 0) sum = min(sum, A[i - 1][j - 1]); if (j < m - 1) sum = min(sum, A[i - 1][j + 1]); A[i][j] += sum; } return *min_element(begin(A.back()), end(A.back())); } }; |
A string of '0's and '1's is monotone increasing if it consists of some number of '0's (possibly 0), followed by some number of '1's (also possibly 0.)
We are given a string S of '0's and '1's, and we may flip any '0' to a '1' or a '1' to a '0'.
Return the minimum number of flips to make S monotone increasing.
Example 1:
Input: "00110" Output: 1 Explanation: We flip the last digit to get 00111.
Example 2:
Input: "010110" Output: 2 Explanation: We flip to get 011111, or alternatively 000111.
Example 3:
Input: "00011000" Output: 2 Explanation: We flip to get 00000000.
Note:
1 <= S.length <= 20000S only consists of '0' and '1' characters.
l[i] := number of flips to make S[0] ~ S[i] become all 0s.
r[i] := number of flips to make S[i] ~ S[n – 1] become all 1s.
Try all possible break point, S[0] ~ S[i – 1] are all 0s and S[i] ~ S[n-1] are all 1s.
ans = min{l[i – 1] + r[i]}
Time complexity: O(n)
Space complexity: O(n)
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// Author: Huahua class Solution { public: int minFlipsMonoIncr(string S) { const int n = S.size(); vector<int> l(n + 1); // 1 -> 0 vector<int> r(n + 1); // 0 -> 1 l[0] = S[0] - '0'; r[n - 1] = '1' - S[n - 1]; for (int i = 1; i < n; ++i) l[i] = l[i - 1] + S[i] - '0'; for (int i = n - 2; i >= 0; --i) r[i] = r[i + 1] + '1' - S[i]; int ans = r[0]; // all 1s. for (int i = 1; i <= n; ++i) ans = min(ans, l[i - 1] + r[i]); return ans; } }; |
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// Author: Huahua class Solution { public: int minFlipsMonoIncr(string S) { const int n = S.length(); vector<vector<int>> dp(n + 1, vector<int>(2)); for (int i = 1; i <= n; ++i) { if (S[i - 1] == '0') { dp[i][0] = dp[i - 1][0]; dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]) + 1; } else { dp[i][0] = dp[i - 1][0] + 1; dp[i][1] = min(dp[i - 1][0], dp[i - 1][1]); } } return min(dp[n][0], dp[n][1]); } }; |
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// Author: Huahua class Solution { public: int minFlipsMonoIncr(string S) { const int n = S.length(); int dp0 = 0; int dp1 = 0; for (int i = 1; i <= n; ++i) { int tmp0 = dp0 + (S[i - 1] == '1'); dp1 = min(dp0, dp1) + (S[i - 1] == '0'); dp0 = tmp0; } return min(dp0, dp1); } }; |
Your music player contains N different songs and she wants to listen to L (not necessarily different) songs during your trip. You create a playlist so that:
K other songs have been playedReturn the number of possible playlists. As the answer can be very large, return it modulo 10^9 + 7.
Example 1:
Input: N = 3, L = 3, K = 1 Output: 6 Explanation: There are 6 possible playlists. [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].
Example 2:
Input: N = 2, L = 3, K = 0 Output: 6 Explanation: There are 6 possible playlists. [1, 1, 2], [1, 2, 1], [2, 1, 1], [2, 2, 1], [2, 1, 2], [1, 2, 2]
Example 3:
Input: N = 2, L = 3, K = 1 Output: 2 Explanation: There are 2 possible playlists. [1, 2, 1], [2, 1, 2]
Note:
0 <= K < N <= L <= 100dp[i][j] := # of playlists of length i using j different songs.
dp[i][j] = dp[i – 1][j – 1] * (N – (j – 1)) + // Adding a new song. j – 1 used, choose any one from (N – (j – 1)) unused.
dp[i -1][j] * max(j – K, 0) // Reuse an existing song.
Time complexity: O(LN)
Space complexity: O(LN) -> O(N)
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// Author: Huahua class Solution { public: int numMusicPlaylists(int N, int L, int K) { constexpr long kMod = 1e9 + 7; vector<vector<long>> dp(L + 1, vector<long>(N + 1, 0)); dp[0][0] = 1; for (int i = 1; i <= L; ++i) for (int j = 1; j <= min(i, N); ++j) dp[i][j] = (dp[i - 1][j - 1] * (N - (j - 1)) + dp[i - 1][j] * max(j - K, 0)) % kMod; return dp[L][N]; } }; |
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// Author: Huahua, 0 ms class Solution { public: int numMusicPlaylists(int N, int L, int K) { constexpr long kMod = 1e9 + 7; vector<long> dp(N + 1, 0); for (int i = 1; i <= L; ++i) { dp[0] = i == 1; for (int j = min(i, N); j >= 1; --j) dp[j] = (dp[j - 1] * (N - (j - 1)) + dp[j] * max(j - K, 0)) % kMod; } return dp[N]; } }; |
Write a class StockSpanner which collects daily price quotes for some stock, and returns the span of that stock’s price for the current day.
The span of the stock’s price today is defined as the maximum number of consecutive days (starting from today and going backwards) for which the price of the stock was less than or equal to today’s price.
For example, if the price of a stock over the next 7 days were [100, 80, 60, 70, 60, 75, 85], then the stock spans would be [1, 1, 1, 2, 1, 4, 6].
Example 1:
Input: ["StockSpanner","next","next","next","next","next","next","next"], [[],[100],[80],[60],[70],[60],[75],[85]] Output: [null,1,1,1,2,1,4,6] Explanation: First, S = StockSpanner() is initialized. Then: S.next(100) is called and returns 1, S.next(80) is called and returns 1, S.next(60) is called and returns 1, S.next(70) is called and returns 2, S.next(60) is called and returns 1, S.next(75) is called and returns 4, S.next(85) is called and returns 6. Note that (for example) S.next(75) returned 4, because the last 4 prices (including today's price of 75) were less than or equal to today's price.
Note:
StockSpanner.next(int price) will have 1 <= price <= 10^5.10000 calls to StockSpanner.next per test case.150000 calls to StockSpanner.next across all test cases.
Time complexity: O(n) per next call
Space complexity: O(n)
dp[i] := span of prices[i]
j = i – 1
while j >= 0 and prices[i] >= prices[j]: j -= dp[j]
dp[i] = i – j
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// Author: Huahua, 144 ms class StockSpanner { public: StockSpanner() {} int next(int price) { if (prices_.empty() || price < prices_.back()) { dp_.push_back(1); } else { int j = prices_.size() - 1; while (j >= 0 && price >= prices_[j]) { j -= dp_[j]; } dp_.push_back(prices_.size() - j); } prices_.push_back(price); return dp_.back(); } private: vector<int> dp_; vector<int> prices_; }; |
Maintain a monotonic stack whose element are pairs of <price, span>, sorted by price from high to low.
When a new price comes in
e.g. prices: 10, 6, 5, 4, 3, 7
after 3, the stack looks [(10,1), (6,1), (5,1), (4,1), (3, 1)],
when 7 arrives, [(10,1), (6,1), (5,1), (4,1), (3, 1), (7, 4 + 1)] = [(10, 1), (7, 5)]
Time complexity: O(1) amortized, each element will be pushed on to stack once, and pop at most once.
Space complexity: O(n), in the worst case, the prices is in descending order.
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// Author: Huahua class StockSpanner { public: StockSpanner() {} int next(int price) { int span = 1; while (!s_.empty() && price >= s_.top().first) { span += s_.top().second; s_.pop(); } s_.emplace(price, span); return span; } private: stack<pair<int, int>> s_; // {price, span}, ordered by price DESC. }; |
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// Author: Huahua class StockSpanner { private Stack<Integer> prices; private Stack<Integer> spans; public StockSpanner() { prices = new Stack<>(); spans = new Stack<>(); } public int next(int price) { int span = 1; while (!prices.empty() && price >= prices.peek()) { span += spans.pop(); prices.pop(); } prices.push(price); spans.push(span); return span; } } |
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# Author: Huahua class StockSpanner: def __init__(self): self.s = list() def next(self, price): span = 1 while self.s and price >= self.s[-1][0]: span += self.s[-1][1] self.s.pop() self.s.append((price, span)) return span |