花花酱 LeetCode 1. Two Sum

By zxi on November 23, 2017

Problem:

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Idea:

Brute force / Hashtable

Solution1:

Brute force / C++

Time complexity: O(n^2)

Space complexity: O(1)

// Author: Huahua

// Runtime: 166 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

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

for (int j = i + 1;j < nums.size(); ++j) {

int sum = nums[i] + nums[j];

if (sum == target) {

return {i, j};

}

return {};

}

};

Solution 2:

Hashtable / C++

Time complexity: O(n)

Space complexity: O(n)

// Author: Huahua

// Runtime: 6 ms

class Solution {

public:

vector<int> twoSum(vector<int>& nums, int target) {

unordered_map<int, int> indies;

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

indies[nums[i]] = i;

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

int left = target - nums[i];

if (indies.count(left) && indies[left] != i) {

return {i, indies[left]};

}

return {};

}

};

Playlists by difficulties

By zxi on November 22, 2017

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花花酱 LeetCode 730. Count Different Palindromic Subsequences

By zxi on November 20, 2017

Problem:

Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7.

A subsequence of a string S is obtained by deleting 0 or more characters from S.

A sequence is palindromic if it is equal to the sequence reversed.

Two sequences A_1, A_2, ... and B_1, B_2, ... are different if there is some i for which A_i != B_i.

Example 1:

Input: 
S = 'bccb'
Output: 6
Explanation: 
The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'.
Note that 'bcb' is counted only once, even though it occurs twice.

Example 2:

Input: 
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
Output: 104860361
Explanation: 
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.

Note:

The length of S will be in the range [1, 1000].
Each character S[i] will be in the set {'a', 'b', 'c', 'd'}.

Idea:

Solution 1: Recursion with memoization

Time complexity: O(n^2)
Space complexity: O(n^2)

C++

// Author: Huahua

// Runtime: 72 ms

class Solution {

public:

int countPalindromicSubsequences(const string& S) {

const int n = S.length();

m_ = vector<vector<int>>(n + 1, vector<int>(n + 1, 0));

return count(S, 0, S.length() - 1);

}

private:

static constexpr long kMod = 1000000007;

long count(const string& S, int s, int e) {

if (s > e) return 0;

if (s == e) return 1;

if (m_[s][e] > 0) return m_[s][e];

long ans = 0;

if (S[s] == S[e]) {

int l = s + 1;

int r = e - 1;

while (l <= r && S[l] != S[s]) ++l;

while (l <= r && S[r] != S[s]) --r;

if (l > r)

ans = count(S, s + 1, e - 1) * 2 + 2;

else if (l == r)

ans = count(S, s + 1, e - 1) * 2 + 1;

else

ans = count(S, s + 1, e - 1) * 2

- count(S, l + 1, r - 1);

} else {

ans = count(S, s, e - 1)

+ count(S, s + 1, e)

- count(S, s + 1, e - 1);

}

return m_[s][e] = (ans + kMod) % kMod;

}

vector<vector<int>> m_;

};

Python

"""

Author: Huahua

Runtime: 3582 ms

"""

class Solution:

def countPalindromicSubsequences(self, S):

def count(S, i, j):

if i > j: return 0

if i == j: return 1

if self.m_[i][j]: return self.m_[i][j]

if S[i] == S[j]:

ans = count(S, i + 1, j - 1) * 2

l = i + 1

r = j - 1

while l <= r and S[l] != S[i]: l += 1

while l <= r and S[r] != S[i]: r -= 1

if l > r: ans += 2

elif l == r: ans += 1

else: ans -= count(S, l + 1, r - 1)

else:

ans = count(S, i + 1, j) + count(S, i, j - 1) - count(S, i + 1, j - 1)

self.m_[i][j] = ans % 1000000007

return self.m_[i][j]

n = len(S)

self.m_ = [[None for _ in range(n)] for _ in range(n)]

return count(S, 0, n - 1)

Java

// Author: Huahua

// Runtime: 107 ms

class Solution {

private int[][] m_;

private static final int kMod = 1000000007;

public int countPalindromicSubsequences(String S) {

int n = S.length();

m_ = new int[n][n];

return count(S.toCharArray(), 0, n - 1);

}

private int count(char[] s, int i, int j) {

if (i > j) return 0;

if (i == j) return 1;

if (m_[i][j] > 0) return m_[i][j];

long ans = 0;

if (s[i] == s[j]) {

ans += count(s, i + 1, j - 1) * 2;

int l = i + 1;

int r = j - 1;

while (l <= r && s[l] != s[i]) ++l;

while (l <= r && s[r] != s[i]) --r;

if (l > r) ans += 2;

else if (l == r) ans += 1;

else ans -= count(s, l + 1, r - 1);

} else {

ans = count(s, i, j - 1)

+ count(s, i + 1, j)

- count(s, i + 1, j - 1);

}

m_[i][j] = (int)((ans + kMod) % kMod);

return m_[i][j];

}

Solution 2: DP

C++

// Author: Huahua

// Runtime: 79 ms

class Solution {

public:

int countPalindromicSubsequences(const string& S) {

int n = S.length();

vector<vector<int>> dp(n, vector<int>(n, 0));

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

dp[i][i] = 1;

for (int len = 1; len <= n; ++len) {

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

const int j = i + len;

if (S[i] == S[j]) {

dp[i][j] = dp[i + 1][j - 1] * 2;

int l = i + 1;

int r = j - 1;

while (l <= r && S[l] != S[i]) ++l;

while (l <= r && S[r] != S[i]) --r;

if (l == r) dp[i][j] += 1;

else if (l > r) dp[i][j] += 2;

else dp[i][j] -= dp[l + 1][r - 1];

} else {

dp[i][j] = dp[i][j - 1] + dp[i + 1][j] - dp[i + 1][j - 1];

}

dp[i][j] = (dp[i][j] + kMod) % kMod;

}

return dp[0][n - 1];

}

private:

static constexpr long kMod = 1000000007;

};

Pyhton

"""

Author: Huahua

Runtime: 3639 ms

"""

class Solution:

def countPalindromicSubsequences(self, S):

n = len(S)

if n == 0: return 0

dp = [[0 for _ in range(n)] for _ in range(n)]

for i in range(n): dp[i][i] = 1

for size in range(2, n + 1):

for i in range(n - size + 1):

j = i + size - 1

if S[i] == S[j]:

dp[i][j] = dp[i + 1][j - 1] * 2

l = i + 1

r = j - 1

while l <= r and S[l] != S[i]: l += 1

while l <= r and S[r] != S[i]: r -= 1

if l > r: dp[i][j] += 2

elif l == r: dp[i][j] += 1

else: dp[i][j] -= dp[l + 1][r - 1]

else:

dp[i][j] = dp[i + 1][j] + dp[i][j - 1] - dp[i + 1][j - 1]

dp[i][j] %= 1000000007

return dp[0][n - 1]

Related Problems:

[解题报告] LeetCode 664. Strange Printer

花花酱 LeetCode 731. My Calendar II – 花花酱

By zxi on November 19, 2017

Problem:

Implement a MyCalendarTwo class to store your events. A new event can be added if adding the event will not cause a triple booking.

Your class will have one method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

A triple booking happens when three events have some non-empty intersection (ie., there is some time that is common to all 3 events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a triple booking. Otherwise, return false and do not add the event to the calendar.

Your class will be called like this: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)

Example 1:

MyCalendar();

MyCalendar.book(10, 20); // returns true

MyCalendar.book(50, 60); // returns true

MyCalendar.book(10, 40); // returns true

MyCalendar.book(5, 15); // returns false

MyCalendar.book(5, 10); // returns true

MyCalendar.book(25, 55); // returns true

Explanation<b>:</b>

The first two events can be booked. The third event can be double booked.

The fourth event (5, 15) can't be booked, because it would result in a triple booking.

The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked.

The sixth event (25, 55) can be booked, as the time in [25, 40) will be double booked with the third event;

the time [40, 50) will be single booked, and the time [50, 55) will be double booked with the second event.

Note:

The number of calls to MyCalendar.book per test case will be at most 1000.
In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

Idea:

Brute Force

Solution1:

Brute Force

Time Complexity: O(n^2)

Space Complexity: O(n)

// Author: Huahua

// Runtime: 82 ms

class MyCalendarTwo {

public:

MyCalendarTwo() {}

bool book(int start, int end) {

for (const auto& kv : overlaps_)

if (max(start, kv.first) < min(end, kv.second)) return false;

for (const auto& kv : booked_) {

const int ss = max(start, kv.first);

const int ee = min(end, kv.second);

if (ss < ee) overlaps_.emplace_back(ss, ee);

}

booked_.emplace_back(start, end);

return true;

}

private:

vector<pair<int, int>> booked_;

vector<pair<int, int>> overlaps_;

};

Java

// Author: Huahua

// Runtime: 156 ms

class MyCalendarTwo {

private List<int[]> booked_;

private List<int[]> overlaps_;

public MyCalendarTwo() {

booked_ = new ArrayList<>();

overlaps_ = new ArrayList<>();

}

public boolean book(int start, int end) {

for (int[] range : overlaps_)

if (Math.max(range[0], start) < Math.min(range[1], end))

return false;

for (int[] range : booked_) {

int ss = Math.max(range[0], start);

int ee = Math.min(range[1], end);

if (ss < ee) overlaps_.add(new int[]{ss, ee});

}

booked_.add(new int[]{start, end});

return true;

}

Python

"""

Author: Huahua

Runtime: 638 ms

"""

class MyCalendarTwo:

def __init__(self):

self.booked_ = []

self.overlaps_ = []

def book(self, start, end):

for s, e in self.overlaps_:

if start < e and end > s: return False

for s, e in self.booked_:

if start < e and end > s:

self.overlaps_.append([max(start, s), min(end, e)])

self.booked_.append([start, end])

return True

Solution 2:

Counting

Time Complexity: O(n^2logn)

Space Complexity: O(n)

// Author: Huahua

// Runtime: 212 ms

class MyCalendarTwo {

public:

MyCalendarTwo() {}

bool book(int start, int end) {

++delta_[start];

--delta_[end];

int count = 0;

for (const auto& kv : delta_) {

count += kv.second;

if (count == 3) {

--delta_[start];

++delta_[end];

return false;

}

if (kv.first > end) break;

}

return true;

}

private:

map<int, int> delta_;

};

Related Problems:

花花酱 LeetCode 729. My Calendar I

By zxi on November 19, 2017

Problem:

Implement a MyCalendar class to store your events. A new event can be added if adding the event will not cause a double booking.

Your class will have the method, book(int start, int end). Formally, this represents a booking on the half open interval [start, end), the range of real numbers x such that start <= x < end.

A double booking happens when two events have some non-empty intersection (ie., there is some time that is common to both events.)

For each call to the method MyCalendar.book, return true if the event can be added to the calendar successfully without causing a double booking. Otherwise, return false and do not add the event to the calendar.

Your class will be called like this: MyCalendar cal = new MyCalendar(); MyCalendar.book(start, end)

Example 1:

MyCalendar();

MyCalendar.book(10, 20); // returns true

MyCalendar.book(15, 25); // returns false

MyCalendar.book(20, 30); // returns true

Explanation:

The first event can be booked. The second can't because time 15 is already booked by another event.

The third event can be booked, as the first event takes every time less than 20, but not including 20.

Note:

The number of calls to MyCalendar.book per test case will be at most 1000.
In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9].

Idea:

Binary Search

Solution1:

Brute Force: O(n^2)

C++

// Author: Huahua

// Runtime: 99 ms

class MyCalendar {

public:

MyCalendar() {}

bool book(int start, int end) {

for (const auto& event : booked_) {

int s = event.first;

int e = event.second;

if (max(s, start) < min(e, end)) return false;

}

booked_.emplace_back(start, end);

return true;

}

private:

vector<pair<int, int>> booked_;

};

Solution 2:

Binary Search O(nlogn)

C++

// Author: Huahua

// Runtime: 82 ms

class MyCalendar {

public:

MyCalendar() {}

bool book(int start, int end) {

auto it = booked_.lower_bound(start);

if (it != booked_.cend() && it->first < end)

return false;

if (it != booked_.cbegin() && (--it)->second > start)

return false;

booked_[start] = end;

return true;

}

private:

map<int, int> booked_;

};

Java

// Author: Huahua

// Runtime: 170 ms

class MyCalendar {

TreeMap<Integer, Integer> booked_;

public MyCalendar() {

booked_ = new TreeMap<>();

}

public boolean book(int start, int end) {

Integer lb = booked_.floorKey(start);

if (lb != null && booked_.get(lb) > start) return false;

Integer ub = booked_.ceilingKey(start);

if (ub != null && ub < end) return false;

booked_.put(start, end);

return true;

}

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Huahua's Tech Road

花花酱 LeetCode 1. Two Sum

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花花酱 LeetCode 730. Count Different Palindromic Subsequences

Solution 1: Recursion with memoization

C++

Python

Java

Solution 2: DP

C++

Pyhton

花花酱 LeetCode 731. My Calendar II – 花花酱

花花酱 LeetCode 729. My Calendar I