Posts tagged as “hard”

花花酱 LeetCode 10. Regular Expression Matching

By zxi on September 17, 2018

Problem

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

s could be empty and contains only lowercase letters a-z.
p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

Solution 1: Recursion

Time complexity: O((|s| + |p|) * 2 ^ (|s| + |p|))

Space complexity: O(|s| + |p|)

C++

// Author: Huahua, 14 ms

class Solution {

public:

bool isMatch(string s, string p) {

return isMatch(s.c_str(), p.c_str());

}

private:

bool isMatch(const char* s, const char* p) {

if (*p == '\0') return *s == '\0';

// normal case, e.g. 'a.b','aaa', 'a'

if (p[1] != '*' || p[1] == '\0') {

// no char to match

if (*s == '\0') return false;

if (*s == *p || *p == '.')

return isMatch(s + 1, p + 1);

else

return false;

}

else {

int i = -1;

while (i == -1 || s[i] == p[0] || p[0] == '.') {

if (isMatch(s + i + 1, p + 2)) return true;

if (s[++i] == '\0') break;

}

return false;

}

return false;

}

};

花花酱 LeetCode 907. Sum of Subarray Minimums

By zxi on September 16, 2018

Problem

Given an array of integers A, find the sum of min(B), where B ranges over every (contiguous) subarray of A.

Since the answer may be large, return the answer modulo 10^9 + 7.

Example 1:

Input: [3,1,2,4]
Output: 17
Explanation: Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.  Sum is 17.

Note:

1 <= A.length <= 30000
1 <= A[i] <= 30000

Idea

order matters, unlike 花花酱 LeetCode 898. Bitwise ORs of Subarrays we can not sort the numbers in this problem.
1. e.g. sumSubarrayMins([3, 1, 2, 4]) !=sumSubarrayMins([1, 2, 3, 4]) since the first one will not have a subarray of [3,4].
For A[i], assuming there are L numbers that are greater than A[i] in range A[0] ~ A[i – 1], and there are R numbers that are greater or equal than A[i] in the range of A[i+1] ~ A[n – 1]. Thus A[i] will be the min of (L + 1) * (R + 1) subsequences.
1. e.g. A = [3,1,2,4], A[1] = 1, L = 1, R = 2, there are (1 + 1) * (2 + 1) = 6 subsequences are 1 is the min number. [3,1], [3,1,2], [3,1,2,4], [1], [1,2], [1,2,4]

Solution 1: Brute Force

Time complexity: O(n^2)

Space complexity: O(1)

C++

// Author: Huahua, 1504 ms

class Solution {

public:

int sumSubarrayMins(vector<int>& A) {

constexpr int kMod = 1e9 + 7;

int ans = 0;

for (int i = 0; i < A.size(); ++i) {

int left = 0;

for (int j = i - 1; j >= 0 && A[j] > A[i]; --j, ++left);

int right = 0;

for (int j = i + 1; j < A.size() && A[j] >= A[i]; ++j, ++right);

ans = (ans + static_cast<long>(A[i]) * (left + 1) * (right + 1)) % kMod;

}

return ans;

}

};

Java

// Author: Huahua, 511 ms

class Solution {

public int sumSubarrayMins(int[] A) {

final int kMod = 1000000007;

final int n = A.length;

int ans = 0;

for (int i = 0; i < n; ++i) {

int left = 0;

for (int j = i - 1; j >= 0 && A[j] > A[i]; --j, ++left);

int right = 0;

for (int j = i + 1; j < n && A[j] >= A[i]; ++j, ++right);

ans = (int)((ans + (long)A[i] * (left + 1) * (right + 1)) % kMod);

}

return ans;

}

Python3 (TLE)

# Author: Huahua, 96/100 passed

class Solution:

def sumSubarrayMins(self, A):

kMod = 10 ** 9 + 7

n = len(A)

ans = 0

for i in range(n):

left, right = 1, 1

while i - left >= 0 and A[i - left] > A[i]:

left += 1

while i + right < n and A[i + right] >= A[i]:

right += 1

ans += A[i] * left * right

return ans % kMod

Solution2 : Monotonic Stack

Time complexity: O(n)

Space complexity: O(n)

We can use a monotonic stack to compute left[i] and right[i] similar to 花花酱 LeetCode 901. Online Stock Span

C++

// Author: Huahua, 60 ms

class Solution {

public:

int sumSubarrayMins(vector<int>& A) {

constexpr int kMod = 1e9 + 7;

const int n = A.size();

vector<int> left(n);

vector<int> right(n);

stack<pair<int, int>> s;

int ans = 0;

for (int i = 0; i < n; ++i) {

int len = 1;

while (!s.empty() && s.top().first > A[i]) {

len += s.top().second; s.pop();

}

s.emplace(A[i], len);

left[i] = len;

}

while (!s.empty()) s.pop();

for (int i = n - 1; i >= 0; --i) {

int len = 1;

while (!s.empty() && s.top().first >= A[i]) {

len += s.top().second; s.pop();

}

s.emplace(A[i], len);

right[i] = len;

}

for (int i = 0; i < n; ++i)

ans = (ans + static_cast<long>(left[i]) * A[i] * right[i]) % kMod;

return ans;

}

};

Java

// Author: Huahua, 154 ms

class Solution {

public int sumSubarrayMins(int[] A) {

final int kMod = 1000000007;

final int n = A.length;

Stack<Integer> nums = new Stack<>();

Stack<Integer> lens = new Stack<>();

int[] left = new int[n];

int[] right = new int[n];

int ans = 0;

for (int i = 0; i < n; ++i) {

int len = 1;

while (!nums.empty() && nums.peek() > A[i]) {

len += lens.pop(); nums.pop();

}

nums.push(A[i]);

lens.push(len);

left[i] = len;

}

nums.clear();

lens.clear();

for (int i = n - 1; i >= 0; --i) {

int len = 1;

while (!nums.empty() && nums.peek() >= A[i]) {

len += lens.pop(); nums.pop();

}

nums.push(A[i]);

lens.push(len);

right[i] = len;

}

for (int i = 0; i < n; ++i)

ans = (int)(ans + (long)A[i] * left[i] * right[i]) % kMod;

return ans;

}

Python3

# Author: Huahua, 376 ms

class Solution:

def sumSubarrayMins(self, A):

kMod = 10 ** 9 + 7

n = len(A)

s, left, right = [], [1] * n, [1] * n

for i in range(n):

while s and s[-1][0] > A[i]:

left[i] += s.pop()[1]

s.append((A[i], left[i]))

s = []

for i in range(n - 1, -1, -1):

while s and s[-1][0] >= A[i]:

right[i] += s.pop()[1]

s.append((A[i], right[i]))

ans = sum(a * l * r for a, l, r in zip(A, left, right)) % kMod

return ans

Python3 V2

# Author: Huahua, 366 ms

class Solution:

def sumSubarrayMins(self, A):

kMod = 10 ** 9 + 7

n = len(A)

left, right = [1] * n, [1] * n

def fillLength(counts, start, end, step):

s = []

for i in range(start, end, step):

num = A[i] if step > 0 else A[i] - 1

while s and s[-1][0] > num:

counts[i] += s.pop()[1]

s.append((A[i], counts[i]))

fillLength(left, 0, n, 1)

fillLength(right, n - 1, -1, -1)

ans = sum(a * l * r for a, l, r in zip(A, left, right)) % kMod

return ans

花花酱 LeetCode 901. Online Stock Span

By zxi on September 14, 2018

Problem

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:

Calls to StockSpanner.next(int price) will have 1 <= price <= 10^5.
There will be at most 10000 calls to StockSpanner.next per test case.
There will be at most 150000 calls to StockSpanner.next across all test cases.
The total time limit for this problem has been reduced by 75% for C++, and 50% for all other languages.

Solution 1: Brute Force (TLE)

Time complexity: O(n) per next call

Space complexity: O(n)

Solution 2: DP

dp[i] := span of prices[i]

j = i – 1
while j >= 0 and prices[i] >= prices[j]: j -= dp[j]
dp[i] = i – j

C++

// 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_;

};

Solution 3: Monotonic Stack

Maintain a monotonic stack whose element are pairs of <price, span>, sorted by price from high to low.

When a new price comes in

If it’s less than top price, add a new pair (price, 1) to the stack, return 1
If it’s greater than top element, collapse the stack and accumulate the span until the top price is higher than the new price. return the total span

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.

C++

// 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.

};

Java

// 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;

}

Python3

# 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

Problem

We have a sorted set of digits D, a non-empty subset of {'1','2','3','4','5','6','7','8','9'}. (Note that '0' is not included.)

Now, we write numbers using these digits, using each digit as many times as we want. For example, if D = {'1','3','5'}, we may write numbers such as '13', '551', '1351315'.

Return the number of positive integers that can be written (using the digits of D) that are less than or equal to N.

Example 1:

Input: D = ["1","3","5","7"], N = 100
Output: 20
Explanation: 
The 20 numbers that can be written are:
1, 3, 5, 7, 11, 13, 15, 17, 31, 33, 35, 37, 51, 53, 55, 57, 71, 73, 75, 77.

Example 2:

Input: D = ["1","4","9"], N = 1000000000
Output: 29523
Explanation: 
We can write 3 one digit numbers, 9 two digit numbers, 27 three digit numbers,
81 four digit numbers, 243 five digit numbers, 729 six digit numbers,
2187 seven digit numbers, 6561 eight digit numbers, and 19683 nine digit numbers.
In total, this is 29523 integers that can be written using the digits of D.

Note:

D is a subset of digits '1'-'9' in sorted order.
1 <= N <= 10^9

Solution -1: DFS (TLE)

Time complexity: O(|D|^log10(N))

Space complexity: O(n)

// Author: Huahua

class Solution {

public:

int atMostNGivenDigitSet(vector<string>& D, int N) {

int ans = 0;

dfs(D, 0, N, ans);

return ans;

}

private:

void dfs(const vector<string>& D, int cur, int N, int& ans) {

if (cur > N) return;

if (cur > 0 && cur <= N) ++ans;

for (const string& d : D)

dfs(D, cur * 10 + d[0] - '0', N, ans);

}

};

Solution 1: Math

Time complexity: O(log10(N))

Space complexity: O(1)

Suppose N has n digits.

less than n digits

We can use all the numbers from D to construct numbers of with length 1,2,…,n-1 which are guaranteed to be less than N.

e.g. n = 52125, D = [1, 2, 5]

format X: e.g. 1, 2, 5 counts = |D| ^ 1

format XX: e.g. 11,12,15,21,22,25,51,52,55, counts = |D|^2

format XXX: counts = |D|^3

format XXXX: counts = |D|^4

exact n digits

if all numbers in D != N[0], counts = |d < N[0] | d in D| * |D|^(n-1), and we are done.

e.g. N = 34567, D = [1,2,8]

we can make:

X |3|^1
XX |3| ^ 2
XXX |3| ^ 3
XXXX |3| ^ 3
1XXXX, |3|^4
2XXXX, |3|^4
we can’t do 8XXXX

Total = (3^1 + 3^2 + 3^3 + 3^4) + 2 * |3|^ 4 = 120 + 162 = 282

N = 52525, D = [1,2,5]

However, if d = N[i], we need to check the next digit…

X |3|^1
XX |3| ^ 2
XXX |3| ^ 3
XXXX |3| ^ 3
1XXXX, |3|^4
2XXXX, |3|^4
5????
- 51XXX |3|^3
- 52???
  - 521XX |3|^2
  - 522XX |3|^2
  - 525??
    - 5251X |3|^1
    - 5252?
      - 52521 |3|^0
      - 52522 |3|^0
      - 52525 +1

total = (120) + 2 * |3|^4 + |3|^3 + 2*|3|^2 + |3|^1 + 2 * |3|^0 + 1 = 120 + 213 = 333

if every digit of N is from D, then we also have a valid solution, thus need to + 1.

C++

class Solution {

public:

int atMostNGivenDigitSet(vector<string>& D, int N) {

const string s = to_string(N);

const int n = s.length();

int ans = 0;

for (int i = 1; i < n; ++i)

ans += pow(D.size(), i);

for (int i = 0; i < n; ++i) {

bool prefix = false;

for (const string& d : D) {

if (d[0] < s[i]) {

ans += pow(D.size(), n - i - 1);

} else if (d[0] == s[i]) {

prefix = true;

break;

}

if (!prefix) return ans;

}

// plus one for solution N itself, all the digits are from D.

return ans + 1;

}

};

Java

// Author: Huahua, 4 ms

class Solution {

public int atMostNGivenDigitSet(String[] D, int N) {

char[] s = String.valueOf(N).toCharArray();

int n = s.length;

int ans = 0;

for (int i = 1; i < n; ++i)

ans += (int)Math.pow(D.length, i);

for (int i = 0; i < n; ++i) {

boolean prefix = false;

for (String d : D) {

if (d.charAt(0) < s[i]) {

ans += Math.pow(D.length, n - i - 1);

} else if (d.charAt(0) == s[i]) {

prefix = true;

break;

}

if (!prefix) return ans;

}

return ans + 1;

}

Python3

# Author: Huahua, 36 ms

class Solution:

def atMostNGivenDigitSet(self, D, N):

s = str(N)

n = len(s)

ans = sum(len(D) ** i for i in range(1, n))

for i, c in enumerate(s):

ans += len(D) ** (n - i - 1) * sum(d < c for d in D)

if c not in D: return ans

return ans + 1

花花酱 LeetCode 668. Kth Smallest Number in Multiplication Table

By zxi on September 10, 2018

Problem

Nearly every one have used the Multiplication Table. But could you find out the k-th smallest number quickly from the multiplication table?

Given the height m and the length n of a m * n Multiplication Table, and a positive integer k, you need to return the k-th smallest number in this table.

Example 1:

Input: m = 3, n = 3, k = 5
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6
3	6	9

The 5-th smallest number is 3 (1, 2, 2, 3, 3).

Example 2:

Input: m = 2, n = 3, k = 6
Output: 
Explanation: 
The Multiplication Table:
1	2	3
2	4	6

The 6-th smallest number is 6 (1, 2, 2, 3, 4, 6).

Note:

The m and n will be in the range [1, 30000].
The k will be in the range [1, m * n]

Solution 1: Nth element (MLE)

Time complexity: O(mn)

Space complexity: O(mn)

C++

// 59 / 69 test cases passed.

class Solution {

public:

int findKthNumber(int m, int n, int k) {

vector<int> s(m * n);

auto it = begin(s);

for (int i = 1; i <= m; ++i)

for (int j = 1; j <= n; ++j)

*it++ = i * j;

nth_element(begin(s), begin(s) + k - 1, end(s));

return s[k - 1];

}

};

Solution2 : Binary Search

Find first x such that there are k elements less or equal to x in the table.

Time complexity: O(m*log(m*n))

Space complexity: O(1)

C++

// Author: Huahua, 12 ms

class Solution {

public:

int findKthNumber(int m, int n, int k) {

int l = 1;

int r = m * n + 1;

while (l < r) {

int x = l + (r - l) / 2;

if (LEX(m, n, x) >= k)

r = x;

else

l = x + 1;

}

return l;

}

private:

// Returns # of elements in the m*n table that are <= x

int LEX(int m, int n, int x) {

int count = 0;

for (int i = 1; i <= m; ++i)

count += min(n, x / i);

return count;

}

};

Java

// Author: Huahua, 16 ms

class Solution {

public int findKthNumber(int m, int n, int k) {

int l = 1;

int r = m * n + 1;

while (l < r) {

int x = l + (r - l) / 2;

if (LEX(m, n, x) >= k)

r = x;

else

l = x + 1;

}

return l;

}

// Returns # of elements in the m*n table that are <= x

private int LEX(int m, int n, int x) {

int count = 0;

for (int i = 1; i <= m; ++i)

count += Math.min(n, x / i);

return count;

}

Python3

# Author: Huahua, 976 ms

class Solution:

def findKthNumber(self, m, n, k):

def LEX(m, n, x, k):

count = 0

for i in range(1, min(m + 1, x + 1)):

count += min(n, x // i)

if count >= k: return count

return count

l = 1

r = m * n + 1

while l < r:

x = l + (r - l) // 2

if LEX(m, n, x, k) >= k:

r = x

else:

l = x + 1

return l

Posts tagged as “hard”

花花酱 LeetCode 10. Regular Expression Matching

Problem

Solution 1: Recursion

C++

花花酱 LeetCode 907. Sum of Subarray Minimums

Problem

Idea

Solution 1: Brute Force

C++

Java

Python3 (TLE)

Solution2 : Monotonic Stack

C++

Java

Python3

Python3 V2

花花酱 LeetCode 901. Online Stock Span

Problem

Solution 1: Brute Force (TLE)

Solution 2: DP

C++

Solution 3: Monotonic Stack

C++

Java

Python3

Related Problems

花花酱 LeetCode 902. Numbers At Most N Given Digit Set

Problem

Solution -1: DFS (TLE)

Solution 1: Math

less than n digits

exact n digits

C++

Java

Python3

花花酱 LeetCode 668. Kth Smallest Number in Multiplication Table

Problem

Solution 1: Nth element (MLE)

C++

Solution2 : Binary Search

C++

Java

Python3

Related Problems