Posts tagged as “simulation”

花花酱 LeetCode 1560. Most Visited Sector in a Circular Track

By zxi on August 23, 2020

Given an integer n and an integer array rounds. We have a circular track which consists of n sectors labeled from 1 to n. A marathon will be held on this track, the marathon consists of m rounds. The i^th round starts at sector rounds[i - 1] and ends at sector rounds[i]. For example, round 1 starts at sector rounds[0] and ends at sector rounds[1]

Return an array of the most visited sectors sorted in ascending order.

Notice that you circulate the track in ascending order of sector numbers in the counter-clockwise direction (See the first example).

Example 1:

Input: n = 4, rounds = [1,3,1,2]
Output: [1,2]
Explanation: The marathon starts at sector 1. The order of the visited sectors is as follows:
1 --> 2 --> 3 (end of round 1) --> 4 --> 1 (end of round 2) --> 2 (end of round 3 and the marathon)
We can see that both sectors 1 and 2 are visited twice and they are the most visited sectors. Sectors 3 and 4 are visited only once.

Example 2:

Input: n = 2, rounds = [2,1,2,1,2,1,2,1,2]
Output: [2]

Example 3:

Input: n = 7, rounds = [1,3,5,7]
Output: [1,2,3,4,5,6,7]

Constraints:

2 <= n <= 100
1 <= m <= 100
rounds.length == m + 1
1 <= rounds[i] <= n
rounds[i] != rounds[i + 1] for 0 <= i < m

Solution: Simulation

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

C++

class Solution {

public:

vector<int> mostVisited(int n, vector<int>& rounds) {

vector<int> counts(n);

counts[rounds[0] - 1] = 1;

for (int i = 1; i < rounds.size(); ++i)

for (int s = rounds[i - 1]; ; ++s) {

++counts[s %= n];

if (s == rounds[i] - 1) break;

}

const int max_count = *max_element(begin(counts), end(counts));

vector<int> ans;

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

if (counts[i] == max_count) ans.push_back(i + 1);

return ans;

}

};

花花酱 LeetCode 1535. Find the Winner of an Array Game

By zxi on August 1, 2020

Given an integer array arr of distinct integers and an integer k.

A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0 and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds.

Return the integer which will win the game.

It is guaranteed that there will be a winner of the game.

Example 1:

Input: arr = [2,1,3,5,4,6,7], k = 2
Output: 5
Explanation: Let's see the rounds of the game:
Round |       arr       | winner | win_count
  1   | [2,1,3,5,4,6,7] | 2      | 1
  2   | [2,3,5,4,6,7,1] | 3      | 1
  3   | [3,5,4,6,7,1,2] | 5      | 1
  4   | [5,4,6,7,1,2,3] | 5      | 2
So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games.

Example 2:

Input: arr = [3,2,1], k = 10
Output: 3
Explanation: 3 will win the first 10 rounds consecutively.

Example 3:

Input: arr = [1,9,8,2,3,7,6,4,5], k = 7
Output: 9

Example 4:

Input: arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
Output: 99

Constraints:

2 <= arr.length <= 10^5
1 <= arr[i] <= 10^6
arr contains distinct integers.
1 <= k <= 10^9

Solution 1: Simulation with a List

Observation : if k >= n – 1, ans = max(arr)

Time complexity: O(n+k)
Space complexity: O(n)

C++

class Solution {

public:

int getWinner(vector<int>& arr, int k) {

if (k >= arr.size()) return *max_element(begin(arr), end(arr));

int win = 0;

list<int> a(begin(arr), end(arr));

while (true) {

auto it0 = begin(a);

auto it1 = next(it0);

if (*it0 > *it1) {

a.splice(a.end(), a, it1);

++win;

} else {

swap(it0, it1);

a.splice(a.end(), a, it1);

win = 1;

}

if (win == k) return *it0;

}

return -1;

}

};

Solution 2: One pass

Since smaller numbers will be put to the end of the array, we must compare the rest of array before meeting those used numbers again. And the winner is monotonically increasing. So we can do it in one pass, just keep the largest number seen so far. If we reach to the end of the array, arr[0] will be max(arr) and it will always win no matter what k is.

Time complexity: O(n)
Space complexity: O(1)

C++

class Solution {

public:

int getWinner(vector<int>& arr, int k) {

int winner = arr[0];

int win = 0;

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

if (arr[i] > winner) {

winner = arr[i];

win = 0;

}

if (++win == k) break;

}

return winner;

}

};

Java

class Solution {

public int getWinner(int[] arr, int k) {

int winner = arr[0];

int win = 0;

for (int i = 1; i < arr.length && win < k; ++i, ++win)

if (arr[i] > winner) {

winner = arr[i];

win = 0;

}

return winner;

}

Python3

class Solution:

def getWinner(self, arr: List[int], k: int) -> int:

winner = arr[0]

win = 0

for x in arr[1:]:

if x > winner:

winner, win = x, 0

win += 1

if win == k: break

return winner

花花酱 1528. Shuffle String

By zxi on July 25, 2020

Given a string s and an integer array indices of the same length.

The string s will be shuffled such that the character at the i^th position moves to indices[i] in the shuffled string.

Return the shuffled string.

Example 1:

Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
Output: "leetcode"
Explanation: As shown, "codeleet" becomes "leetcode" after shuffling.

Example 2:

Input: s = "abc", indices = [0,1,2]
Output: "abc"
Explanation: After shuffling, each character remains in its position.

Example 3:

Input: s = "aiohn", indices = [3,1,4,2,0]
Output: "nihao"

Example 4:

Input: s = "aaiougrt", indices = [4,0,2,6,7,3,1,5]
Output: "arigatou"

Example 5:

Input: s = "art", indices = [1,0,2]
Output: "rat"

Constraints:

s.length == indices.length == n
1 <= n <= 100
s contains only lower-case English letters.
0 <= indices[i] < n
All values of indices are unique (i.e. indices is a permutation of the integers from 0 to n - 1).

Solution: Simulation

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

C++

// Author: Huahua

class Solution {

public:

string restoreString(string s, vector<int>& indices) {

string ans(s);

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

ans[indices[i]] = s[i];

return ans;

}

};

Java

// Author: Huahua

class Solution {

public String restoreString(String s, int[] indices) {

char[] ans = new char[s.length()];

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

ans[indices[i]] = s.charAt(i);

return new String(ans);

}

Pyhton3

class Solution:

def restoreString(self, s: str, indices: List[int]) -> str:

ans = [None] * len(s)

for i, idx in enumerate(indices):

ans[idx] = s[i]

return ''.join(ans)

花花酱 LeetCode 1505. Minimum Possible Integer After at Most K Adjacent Swaps On Digits

By zxi on July 5, 2020

Given a string num representing the digits of a very large integer and an integer k.

You are allowed to swap any two adjacent digits of the integer at most k times.

Return the minimum integer you can obtain also as a string.

Example 1:

Input: num = "4321", k = 4
Output: "1342"
Explanation: The steps to obtain the minimum integer from 4321 with 4 adjacent swaps are shown.

Example 2:

Input: num = "100", k = 1
Output: "010"
Explanation: It's ok for the output to have leading zeros, but the input is guaranteed not to have any leading zeros.

Example 3:

Input: num = "36789", k = 1000
Output: "36789"
Explanation: We can keep the number without any swaps.

Example 4:

Input: num = "22", k = 22
Output: "22"

Example 5:

Input: num = "9438957234785635408", k = 23
Output: "0345989723478563548"

Constraints:

1 <= num.length <= 30000
num contains digits only and doesn’t have leading zeros.
1 <= k <= 10^9

Solution: Greedy + Recursion (Update: TLE after 7/6/2020)

Move the smallest number to the left and recursion on the right substring with length equals to n -= 1.

4321 k = 4 => 1 + solve(432, 4-3) = 1 + solve(432, 1) = 1 + 3 + solve(42, 0) = 1 + 3 + 42 = 1342.

Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

string minInteger(string num, int k) {

const int n = num.size();

if (k >= n * (n - 1) / 2) {

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

return num;

}

int s = 0;

while (k > 0 && s < n) {

auto bit = begin(num);

auto it = min_element(bit + s, bit + min(s + k + 1, n));

k -= distance(bit + s, it);

rotate(bit + (s++), it, next(it));

}

return num;

}

};

Solution 2: Binary Indexed Tree / Fenwick Tree

Moving elements in a string is a very expensive operation, basically O(n) per op. Actually, we don’t need to move the elements physically, instead we track how many elements before i has been moved to the “front”. Thus we know the cost to move the i-th element to the “front”, which is i – elements_moved_before_i or prefix_sum(0~i-1) if we mark moved element as 1.

We know BIT / Fenwick Tree is good for dynamic prefix sum computation which helps to reduce the time complexity to O(nlogn).

Time complexity: O(nlogn)
Space complexity: O(n)

C++

// Author: Huahua

class Fenwick {

public:

explicit Fenwick(int n) : sums_(n + 1) {}

void update(int i, int delta) {

++i;

while (i < sums_.size()) {

sums_[i] += delta;

i += i & -i;

}

int query(int i) {

++i;

int ans = 0;

while (i > 0) {

ans += sums_[i];

i -= i & -i;

}

return ans;

}

private:

vector<int> sums_;

};

class Solution {

public:

string minInteger(string num, int k) {

const int n = num.length();

vector<queue<int>> pos(10);

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

pos[num[i] - '0'].push(i);

Fenwick tree(n);

vector<int> used(n);

string ans;

while (k > 0 & ans.length() < n) {

for (int d = 0; d < 10; ++d) {

if (pos[d].empty()) continue;

const int i = pos[d].front();

const int cost = i - tree.query(i - 1);

if (cost > k) continue;

k -= cost;

ans += ('0' + d);

tree.update(i, 1);

used[i] = true;

pos[d].pop();

break;

}

};

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

if (!used[i]) ans += num[i];

return ans;

}

};

Java

// Author: Huahua

class Solution {

class Fenwick {

private int[] sums;

public Fenwick(int n) {

this.sums = new int[n + 1];

}

public void update(int i, int delta) {

++i;

while (i < sums.length) {

this.sums[i] += delta;

i += i & -i;

}

public int query(int i) {

int ans = 0;

++i;

while (i > 0) {

ans += this.sums[i];

i -= i & -i;

}

return ans;

}

public String minInteger(String num, int k) {

int n = num.length();

var used = new boolean[n];

var pos = new ArrayList<ArrayDeque<Integer>>();

for (int i = 0; i <= 9; ++i)

pos.add(new ArrayDeque<Integer>());

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

pos.get(num.charAt(i) - '0').offer(i);

var tree = new Fenwick(n);

var sb = new StringBuilder();

while (k > 0 && sb.length() < n) {

for (int d = 0; d <= 9; ++d) {

Integer i = pos.get(d).peek();

if (i == null) continue;

int cost = i - tree.query(i - 1);

if (cost > k) continue;

sb.append((char)(d + '0'));

k -= cost;

pos.get(d).removeFirst();

tree.update(i, 1);

used[i] = true;

break;

}

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

if (!used[i]) sb.append(num.charAt(i));

return sb.toString();

}

Python3

# Author: Huahua

class Fenwick:

def __init__(self, n):

self.sums = [0] * (n + 1)

def query(self, i):

ans = 0

i += 1

while i > 0:

ans += self.sums[i];

i -= i & -i

return ans

def update(self, i, delta):

i += 1

while i < len(self.sums):

self.sums[i] += delta

i += i & -i

class Solution:

def minInteger(self, num: str, k: int) -> str:

n = len(num)

used = [False] * n

pos = [deque() for _ in range(10)]

for i, c in enumerate(num):

pos[ord(c) - ord("0")].append(i)

tree = Fenwick(n)

ans = []

while k > 0 and len(ans) < n:

for d in range(10):

if not pos[d]: continue

i = pos[d][0]

cost = i - tree.query(i - 1)

if cost > k: continue

k -= cost

ans.append(chr(d + ord("0")))

tree.update(i, 1)

used[i] = True

pos[d].popleft()

break

for i in range(n):

if not used[i]: ans.append(num[i])

return "".join(ans)

花花酱 LeetCode 1496. Path Crossing

By zxi on June 28, 2020

Given a string path, where path[i] = 'N', 'S', 'E' or 'W', each representing moving one unit north, south, east, or west, respectively. You start at the origin (0, 0) on a 2D plane and walk on the path specified by path.

Return True if the path crosses itself at any point, that is, if at any time you are on a location you’ve previously visited. Return False otherwise.

Example 1:

Input: path = "NES"
Output: false 
Explanation: Notice that the path doesn't cross any point more than once.

Example 2:

Input: path = "NESWW"
Output: true
Explanation: Notice that the path visits the origin twice.

Constraints:

1 <= path.length <= 10^4
path will only consist of characters in {'N', 'S', 'E', 'W}

Solution: Simulation + Hashtable

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

C++

// Author: Huahua

class Solution {

public:

bool isPathCrossing(string path) {

unordered_set<int> s;

int x = 0;

int y = 0;

s.insert(x * 10000 + y);

for (char d : path) {

if (d == 'N') ++y;

else if (d == 'S') --y;

else if (d == 'E') ++x;

else if (d == 'W') --x;

if (!s.insert(x * 10000 + y).second)

return true;

}

return false;

}

};