Posts published in September 2018

花花酱 LeetCode 909. Snakes and Ladders

By zxi on September 25, 2018

Problem

On an N x N board, the numbers from 1 to N*N are written boustrophedonically starting from the bottom left of the board, and alternating direction each row. For example, for a 6 x 6 board, the numbers are written as follows:

You start on square 1 of the board (which is always in the last row and first column). Each move, starting from square x, consists of the following:

You choose a destination square S with number x+1, x+2, x+3, x+4, x+5, or x+6, provided this number is <= N*N.
- (This choice simulates the result of a standard 6-sided die roll: ie., there are always at most 6 destinations.)
If S has a snake or ladder, you move to the destination of that snake or ladder. Otherwise, you move to S.

A board square on row r and column c has a “snake or ladder” if board[r][c] != -1. The destination of that snake or ladder is board[r][c].

Note that you only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do not continue moving. (For example, if the board is [[4,-1],[-1,3]], and on the first move your destination square is 2, then you finish your first move at 3, because you do notcontinue moving to 4.)

Return the least number of moves required to reach square N*N. If it is not possible, return -1.

Example 1:

Input: [
[-1,-1,-1,-1,-1,-1],
[-1,-1,-1,-1,-1,-1],
[-1,-1,-1,-1,-1,-1],
[-1,35,-1,-1,13,-1],
[-1,-1,-1,-1,-1,-1],
[-1,15,-1,-1,-1,-1]]
Output: 4
Explanation: 
At the beginning, you start at square 1 [at row 5, column 0].
You decide to move to square 2, and must take the ladder to square 15.
You then decide to move to square 17 (row 3, column 5), and must take the snake to square 13.
You then decide to move to square 14, and must take the ladder to square 35.
You then decide to move to square 36, ending the game.
It can be shown that you need at least 4 moves to reach the N*N-th square, so the answer is 4.

Note:

2 <= board.length = board[0].length <= 20
board[i][j] is between 1 and N*N or is equal to -1.
The board square with number 1 has no snake or ladder.
The board square with number N*N has no snake or ladder.

Solution: BFS

Time complexity: O(n*n)

Space complexity: O(n*n)

C++

// Author: Huahua, 8 ms (beats 100%)

class Solution {

public:

int snakesAndLadders(vector<vector<int>>& board) {

const int N = board.size();

int steps = 0;

vector<int> seen(N * N + 1, 0);

queue<int> q;

q.push(1);

seen[1] = 1;

while (!q.empty()) {

int size = q.size();

++steps;

while (size--) {

int s = q.front(); q.pop();

for (int x = s + 1; x <= min(s + 6, N * N); ++x) {

int r, c;

getRC(x, N, &r, &c);

int nx = board[r][c] == -1 ? x : board[r][c];

if (seen[nx]) continue;

if (nx == N * N) return steps;

q.push(nx);

seen[nx] = 1;

}

return -1;

}

private:

void getRC(int s, int N, int* r, int* c) {

int y = (s - 1) / N;

int x = (s - 1) % N;

*r = N - 1 - y;

*c = (y % 2) ? N - 1 - x : x;

}

};

花花酱 LeetCode 911. Online Election

By zxi on September 25, 2018

Problem

n an election, the i-th vote was cast for persons[i] at time times[i].

Now, we would like to implement the following query function: TopVotedCandidate.q(int t) will return the number of the person that was leading the election at time t.

Votes cast at time t will count towards our query. In the case of a tie, the most recent vote (among tied candidates) wins.

Example 1:

Input: ["TopVotedCandidate","q","q","q","q","q","q"], [[[0,1,1,0,0,1,0],[0,5,10,15,20,25,30]],[3],[12],[25],[15],[24],[8]]
Output: [null,0,1,1,0,0,1]
Explanation: 
At time 3, the votes are [0], and 0 is leading.
At time 12, the votes are [0,1,1], and 1 is leading.
At time 25, the votes are [0,1,1,0,0,1], and 1 is leading (as ties go to the most recent vote.)
This continues for 3 more queries at time 15, 24, and 8.

Note:

1 <= persons.length = times.length <= 5000
0 <= persons[i] <= persons.length
times is a strictly increasing array with all elements in [0, 10^9].
TopVotedCandidate.q is called at most 10000 times per test case.
TopVotedCandidate.q(int t) is always called with t >= times[0].

Solution: HashTable + Binary Search

Compute the leads for each t in times using a hash table.

binary search the upper bound of t, and return the lead of previous entry.

Time complexity: Constructor O(n), Query: O(logn)

Space complexity: O(n)

C++

// Author: Huahua

class TopVotedCandidate {

public:

TopVotedCandidate(vector<int> persons, vector<int> times) {

vector<int> votes(persons.size() + 1);

int last_lead = persons.front();

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

if (++votes[persons[i]] >= votes[last_lead])

last_lead = persons[i];

leads_[times[i]] = last_lead;

}

int q(int t) {

return prev(leads_.upper_bound(t))->second;

}

private:

map<int, int> leads_; // time -> lead

};

花花酱 LeetCode 908. Smallest Range I

By zxi on September 23, 2018

Problem

Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add xto A[i].

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B.

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3
Output: 0
Explanation: B = [3,3,3] or B = [4,4,4]

Note:

1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000

Solution 0: Brute Force (TLE)

Try all pairs

Time complexity: O(n^2)

Space complexity: O(1)

Solution 1: Math

Time complexity: O(n)

Space complexity: O(1)

Find the min/max element of the array.

min + k v.s. max – k

ans = max(0, (max – min) – 2 * k))

C++

// Author: Huahua

class Solution {

public:

int smallestRangeI(vector<int>& A, int K) {

int a_min = *min_element(begin(A), end(A));

int a_max = *max_element(begin(A), end(A));

return max(0, (a_max - a_min) - 2 * K);

}

};

Python3

# Author: Huahua, 72 ms

class Solution:

def smallestRangeI(self, A, K):

return max(0, max(A) - min(A) - 2 * K);

花花酱 LeetCode 19. Remove Nth Node From End of List

By zxi on September 19, 2018

Problem

Given a linked list, remove the n-th node from the end of list and return its head.

Example:

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

Solution 0: Cheating! store the nodes in an array

C++

// Author: Huahua

class Solution {

public:

ListNode *removeNthFromEnd(ListNode *head, int n) {

if (!head) return nullptr;

vector<ListNode *> nodes;

ListNode *cur = head;

while (cur) {

nodes.push_back(cur);

cur = cur->next;

}

if (n == nodes.size()) return head->next;

ListNode* nodes_to_remove = nodes[nodes.size()-n];

ListNode* parent = nodes[nodes.size() - n - 1];

parent->next = nodes_to_remove->next;

delete nodes_to_remove;

return head;

}

};

Solution 1: Two passes

Time complexity: O(L)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode* removeNthFromEnd(ListNode* head, int n) {

int l = 0;

ListNode* cur = head;

while (cur) {

++l;

cur = cur->next;

}

if (n == l) {

ListNode* ans = head->next;

delete head;

return ans;

}

l -= n;

cur = head;

while (--l) cur = cur->next;

ListNode* node = cur->next;

cur->next = node->next;

delete node;

return head;

}

};

Solution 2: Fast/Slow Pointers + Dummy Head / Prev

Fast pointer moves n steps first, and then slow pointer starts moving.

When fast pointer reaches tail, slow pointer is n-th node from the end.

Time complexity: O(L)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

ListNode* removeNthFromEnd(ListNode* head, int n) {

ListNode* fast = head;

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

fast = fast->next;

ListNode dummy(0);

dummy.next = head;

ListNode* prev = &dummy;

while (fast) {

fast = fast->next;

prev = prev->next;

}

ListNode* node = prev->next;

prev->next = node->next;

delete node;

return dummy.next;

}

};

Java

// Author: Huahua

class Solution {

public ListNode removeNthFromEnd(ListNode head, int n) {

ListNode dummy = new ListNode(0);

dummy.next = head;

ListNode fast = head;

while (n-- > 0) fast = fast.next;

ListNode prev = dummy;

while (fast != null) {

fast = fast.next;

prev = prev.next;

}

prev.next = prev.next.next;

return dummy.next;

}

Python3

# Author: Huahua

class Solution:

def removeNthFromEnd(self, head, n):

dummy = ListNode(0)

dummy.next = head

fast, prev = head, dummy

for _ in range(n):

fast = fast.next

while fast:

fast, prev = fast.next, prev.next

prev.next = prev.next.next

return dummy.next

花花酱 LeetCode 18. 4Sum

By zxi on September 19, 2018

Problem

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

Solution 1: Sorting + Binary Search

Time complexity: O(n^3 log n + klogk)

Space complexity: O(k)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> fourSum(vector<int> &num, int target) {

set<vector<int>> h;

sort(num.begin(), num.end());

int n = num.size();

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

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

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

int t = target - num[i] - num[j] - num[k];

if (t < num[k]) break;

if (!std::binary_search(num.begin() + k + 1, num.end(), t)) continue;

h.insert({num[i], num[j], num[k], t});

}

return vector<vector<int>>(begin(h), end(h));

}

};

C++ opt

// Author: Huahua

class Solution {

public:

vector<vector<int>> fourSum(vector<int> &num, int target) {

sort(num.begin(), num.end());

if (target > 0 && target > 4 * num.back()) return {};

if (target < 0 && target < 4 * num.front()) return {};

set<vector<int>> h;

int n = num.size();

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

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

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

int t = target - num[i] - num[j] - num[k];

if (t < num[k]) break;

if (!std::binary_search(num.begin() + k + 1, num.end(), t)) continue;

h.insert({num[i], num[j], num[k], t});

}

return vector<vector<int>>(begin(h), end(h));

}

};

Solution 2: Sorting + HashTable

Time complexity: O(n^3 + klogk)

Space complexity: O(n + k)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> fourSum(vector<int> &num, int target) {

sort(num.begin(), num.end());

if (target > 0 && target > 4 * num.back()) return {};

if (target < 0 && target < 4 * num.front()) return {};

unordered_map<int, int> index;

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

index[num[i]] = i;

set<vector<int>> h;

int n = num.size();

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

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

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

int t = target - num[i] - num[j] - num[k];

if (t < num[k]) break;

auto it = index.find(t);

if (it == index.end() || it->second <= k) continue;

h.insert({num[i], num[j], num[k], t});

}

return vector<vector<int>>(begin(h), end(h));

}

};

Posts published in September 2018

花花酱 LeetCode 909. Snakes and Ladders

Problem

Solution: BFS

C++

花花酱 LeetCode 911. Online Election

Problem

Solution: HashTable + Binary Search

C++

花花酱 LeetCode 908. Smallest Range I

Problem

Solution 0: Brute Force (TLE)

Solution 1: Math

C++

Python3

花花酱 LeetCode 19. Remove Nth Node From End of List

Problem

Solution 0: Cheating! store the nodes in an array

C++

Solution 1: Two passes

C++

Solution 2: Fast/Slow Pointers + Dummy Head / Prev

C++

Java

Python3

花花酱 LeetCode 18. 4Sum

Problem

Solution 1: Sorting + Binary Search

C++

C++ opt

Solution 2: Sorting + HashTable

C++

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