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

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

All of the nodes’ values will be unique.
p and q are different and both values will exist in the binary tree.

Solution 1: Recursion

Time complexity: O(n)

Space complexity: O(h)

For a given root, recursively call LCA(root.left, p, q) and LCA(root.right, p, q)

if both returns a valid node which means p, q are in different subtrees, then root will be their LCA.

if only one valid node returns, which means p, q are in the same subtree, return that valid node as their LCA.

C++

// Author: Huahua

class Solution {

public:

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

if (!root || root == p || root == q) return root;

TreeNode* l = lowestCommonAncestor(root->left, p, q);

TreeNode* r = lowestCommonAncestor(root->right, p, q);

if (l && r) return root;

return l ? l : r;

}

};

Java

// Author: Huahua

class Solution {

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

if (root == null || root == p || root == q) return root;

TreeNode l = lowestCommonAncestor(root.left, p, q);

TreeNode r = lowestCommonAncestor(root.right, p, q);

if (l == null || r == null) return l == null ? r : l;

return root;

}

Python3

# Author: Huahua

class Solution:

def lowestCommonAncestor(self, root, p, q):

if any((not root, root == p, root == q)): return root

l = self.lowestCommonAncestor(root.left, p, q)

r = self.lowestCommonAncestor(root.right, p, q)

if not l or not r: return l if l else r

return root

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 900. RLE Iterator

By zxi on September 13, 2018

Problem

Write an iterator that iterates through a run-length encoded sequence.

The iterator is initialized by RLEIterator(int[] A), where A is a run-length encoding of some sequence. More specifically, for all even i, A[i] tells us the number of times that the non-negative integer value A[i+1] is repeated in the sequence.

The iterator supports one function: next(int n), which exhausts the next n elements (n >= 1) and returns the last element exhausted in this way. If there is no element left to exhaust, next returns -1 instead.

For example, we start with A = [3,8,0,9,2,5], which is a run-length encoding of the sequence [8,8,8,5,5]. This is because the sequence can be read as “three eights, zero nines, two fives”.

Example 1:

Input: ["RLEIterator","next","next","next","next"], [[[3,8,0,9,2,5]],[2],[1],[1],[2]]
Output: [null,8,8,5,-1]
Explanation: 
RLEIterator is initialized with RLEIterator([3,8,0,9,2,5]).
This maps to the sequence [8,8,8,5,5].
RLEIterator.next is then called 4 times:

.next(2) exhausts 2 terms of the sequence, returning 8.  The remaining sequence is now [8, 5, 5].

.next(1) exhausts 1 term of the sequence, returning 8.  The remaining sequence is now [5, 5].

.next(1) exhausts 1 term of the sequence, returning 5.  The remaining sequence is now [5].

.next(2) exhausts 2 terms, returning -1.  This is because the first term exhausted was 5,
but the second term did not exist.  Since the last term exhausted does not exist, we return -1.

Note:

0 <= A.length <= 1000
A.length is an even integer.
0 <= A[i] <= 10^9
There are at most 1000 calls to RLEIterator.next(int n) per test case.
Each call to RLEIterator.next(int n) will have 1 <= n <= 10^9.

Solution: Simulation

Time complexity: O(|A|)

Space complexity: O(|A|)

C++

// Author: Huahua, 4 ms

class RLEIterator {

public:

RLEIterator(vector<int> A): A_(std::move(A)), i_(0) {}

int next(int n) {

while (n && i_ < A_.size()) {

if (n >= A_[i_]) {

n -= A_[i_];

i_ += 2;

if (n == 0) return A_[i_ - 1];

} else {

A_[i_] -= n;

return A_[i_ + 1];

}

return -1;

}

private:

int i_;

vector<int> A_;

};

Java

// Author: Huahua, 73 ms

class RLEIterator {

private int[] A;

private int i;

public RLEIterator(int[] A) {

this.A = A;

this.i = 0;

}

public int next(int n) {

while (n >= 0 && i < A.length) {

if (n >= A[i]) {

n -= A[i];

i += 2;

if (n == 0) return A[i - 1];

} else {

A[i] -= n;

return A[i + 1];

}

return -1;

}

Python3

# Author: Huahua, 44 ms

class RLEIterator:

def __init__(self, A):

self.A = A

self.i = 0

def next(self, n):

while n != 0 and self.i < len(self.A):

if n >= self.A[self.i]:

n -= self.A[self.i]

self.i += 2

if n == 0: return self.A[self.i - 1]

else:

self.A[self.i] -= n

return self.A[self.i + 1]

return -1

花花酱 LeetCode 12. Integer to Roman

By zxi on September 12, 2018

Problem

Roman numerals are represented by seven different symbols: I, V, X, L, C, D and M.

Symbol       Value
I             1
V             5
X             10
L             50
C             100
D             500
M             1000

For example, two is written as II in Roman numeral, just two one’s added together. Twelve is written as, XII, which is simply X + II. The number twenty seven is written as XXVII, which is XX + V + II.

Roman numerals are usually written largest to smallest from left to right. However, the numeral for four is not IIII. Instead, the number four is written as IV. Because the one is before the five we subtract it making four. The same principle applies to the number nine, which is written as IX. There are six instances where subtraction is used:

I can be placed before V (5) and X (10) to make 4 and 9.
X can be placed before L (50) and C (100) to make 40 and 90.
C can be placed before D (500) and M (1000) to make 400 and 900.

Given an integer, convert it to a roman numeral. Input is guaranteed to be within the range from 1 to 3999.

Example 1:

Input: 3
Output: "III"

Example 2:

Input: 4
Output: "IV"

Example 3:

Input: 9
Output: "IX"

Example 4:

Input: 58
Output: "LVIII"
Explanation: C = 100, L = 50, XXX = 30 and III = 3.

Example 5:

Input: 1994
Output: "MCMXCIV"
Explanation: M = 1000, CM = 900, XC = 90 and IV = 4.

Solution: HashTable + Simulation

Map integer 1,4,5,9,10,40,50,90, …, 1000 to Romain

Start from the largest number y,

if x >= y:
ans += Roman[y]
x -= y

Time complexity: O(x)

Space complexity: O(x)

C++

// Author: Huahua

class Solution {

public:

string intToRoman(int num) {

vector<pair<int,string>> v{{1000, "M"}, {900, "CM"}, {500, "D"}, {400, "CD"},

{ 100, "C"}, { 90, "XC"}, { 50, "L"}, { 40, "XL"},

{ 10, "X"}, { 9, "IX"}, { 5, "V"}, { 4, "IV"},

{ 1, "I"}};

string ans;

auto it = cbegin(v);

while (num) {

if (num >= it->first) {

num -= it->first;

ans += it->second;

} else {

++it;

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def intToRoman(self, num):

m = [[1000, "M"], [900, "CM"], [500, "D"], [400, "CD"],

[ 100, "C"], [ 90, "XC"], [ 50, "L"], [ 40, "XL"],

[ 10, "X"], [ 9, "IX"], [ 5, "V"], [ 4, "IV"],

[ 1, "I"]]

index = 0

ans = ''

while num > 0:

if num >= m[index][0]:

ans += m[index][1]

num -= m[index][0]

else:

index += 1

return ans

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

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