Posts published in “Simulation”

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 8. String to Integer (atoi)

By zxi on September 12, 2018

Problem

Implement atoi which converts a string to an integer.

The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.

The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.

If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.

If no valid conversion could be performed, a zero value is returned.

Note:

Only the space character ' ' is considered as whitespace character.
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹− 1]. If the numerical value is out of the range of representable values, INT_MAX (2³¹− 1) or INT_MIN (−2³¹) is returned.

Example 1:

Input: "42"
Output: 42

Example 2:

Input: "   -42"
Output: -42
Explanation: The first non-whitespace character is '-', which is the minus sign.
             Then take as many numerical digits as possible, which gets 42.

Example 3:

Input: "4193 with words"
Output: 4193
Explanation: Conversion stops at digit '3' as the next character is not a numerical digit.

Example 4:

Input: "words and 987"
Output: 0
Explanation: The first non-whitespace character is 'w', which is not a numerical 
             digit or a +/- sign. Therefore no valid conversion could be performed.

Example 5:

Input: "-91283472332"
Output: -2147483648
Explanation: The number "-91283472332" is out of the range of a 32-bit signed integer.
             Thefore INT_MIN (−2³¹) is returned.

Solution: Simulation

You need to handle all corner cases in order to pass…

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

int myAtoi(const string& s) {

long long ans = 0;

int neg = 0;

int pos = 0;

bool p = false; // has '.'

bool a = false; // has letters

bool n = false; // has number

for (int i = 0; i < s.length(); ++i) {

if (s[i] == ' ') { if (n || neg || pos) break; }

else if (s[i] == '-') { if (n) break; else neg++; }

else if (s[i] == '+') { if (n) break; else pos++; }

else if (s[i] == '.') p = true;

else if (s[i] >= '0' && s[i] <= '9' && !p && !a) {

ans = ans * 10 + (s[i] - '0');

n = true;

} else {

a = true;

}

if (ans / 10 > INT_MAX) break; // ans can be > 64 bit

}

if (neg) ans = -ans;

if (ans > INT_MAX) ans = INT_MAX;

if (ans < INT_MIN) ans = INT_MIN;

if (pos + neg > 1) ans = 0;

return ans;

}

};

花花酱 LeetCode 7. Reverse Integer

By zxi on September 12, 2018

Problem

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321

Example 2:

Input: -123
Output: -321

Example 3:

Input: 120
Output: 21

Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2³¹, 2³¹− 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Solution: Simulation

Reverse digit by digit. Be careful about the overflow and negative numbers (especially in Python)

Time complexity: O(log(x)) ~ O(1)

Space complexity: O(log(x)) ~ O(1)

C++

// Author: Huahua

class Solution {

public:

int reverse(int x) {

int ans = 0;

while (x != 0) {

int r = x % 10;

if (ans > INT_MAX / 10 || ans < INT_MIN / 10) return 0;

ans = ans * 10 + r;

x /= 10;

}

return ans;

}

};

Java

class Solution {

public int reverse(int x) {

int ans = 0;

while (x != 0) {

int r = x % 10;

if (ans > Integer.MAX_VALUE / 10 || ans < Integer.MIN_VALUE / 10) return 0;

ans = ans * 10 + r;

x /= 10;

}

return ans;

}

Python3

class Solution:

def reverse(self, x):

min_val = -(2**31);

max_val = 2**31 - 1;

sign = -1 if x < 0 else 1

x = abs(x)

ans = 0

while x != 0:

ans = ans * 10 + (x % 10)

x //= 10

ans *= sign

if ans > max_val or ans < min_val: return 0

return ans

花花酱 LeetCode 6. ZigZag Conversion

By zxi on September 12, 2018

Problem

The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)

P   A   H   N
A P L S I I G
Y   I   R

And then read line by line: "PAHNAPLSIIGYIR"

Write the code that will take a string and make this conversion given a number of rows:

string convert(string s, int numRows);

Example 1:
<preclass=”crayon:false”>Input: s = “PAYPALISHIRING”, numRows = 3 Output: “PAHNAPLSIIGYIR”

Example 2:

Input: s = "PAYPALISHIRING", numRows = 4
Output: "PINALSIGYAHRPI"
Explanation:

P     I    N
A   L S  I G
Y A   H R
P     I

Solution: Simulation

Store the zigzag results for each row and then output row by row.

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

string convert(string s, int nRows) {

if (nRows == 1) return s;

vector<string> ss(nRows);

int l = s.length();

int x = 0, y = 0, i = 0;

bool down = true;

while (i < l) {

ss[y] += s[i];

if (down) {

++y;

if (y == nRows) {

down = false;

y -=2;

}

} else {

--y;

if (y < 0) {

down = true;

y = 1;

}

i++;

}

string ans;

for(const string& r : ss)

ans += r;

return ans;

}

};