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花花酱 LeetCode 1042. Flower Planting With No Adjacent

By zxi on May 14, 2019

You have N gardens, labelled 1 to N. In each garden, you want to plant one of 4 types of flowers.

paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y.

Also, there is no garden that has more than 3 paths coming into or leaving it.

Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.

Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)-th garden. The flower types are denoted 1, 2, 3, or 4. It is guaranteed an answer exists.

Example 1:

Input: N = 3, paths = [[1,2],[2,3],[3,1]]
Output: [1,2,3]

Example 2:

Input: N = 4, paths = [[1,2],[3,4]]
Output: [1,2,1,2]

Example 3:

Input: N = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
Output: [1,2,3,4]

Note:

1 <= N <= 10000
0 <= paths.size <= 20000
No garden has 4 or more paths coming into or leaving it.
It is guaranteed an answer exists.

Solution: Graph coloring, choose any available color

Time complexity: O(|V|+|E|)
Space complexity: O(|V|+|E|)

C++

class Solution {

public:

vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {

vector<int> ans(N, 0);

vector<vector<int>> G(N);

for (const auto& p : paths) {

G[p[0] - 1].push_back(p[1] - 1);

G[p[1] - 1].push_back(p[0] - 1);

}

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

int mask = 0;

for (const auto& j : G[i])

mask |= (1 << ans[j]);

for (int c = 1; c <= 4 && ans[i] == 0; ++c)

if (!(mask & (1 << c))) ans[i] = c;

}

return ans;

}

};

花花酱 LeetCode 38. Count and Say

By zxi on April 17, 2019

Problem

https://leetcode.com/problems/count-and-say/

The count-and-say sequence is the sequence of integers with the first five terms as following:

1.     1
2.     11
3.     21
4.     1211
5.     111221

1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.

Given an integer n where 1 ≤ n ≤ 30, generate the n^th term of the count-and-say sequence.

Note: Each term of the sequence of integers will be represented as a string.

Example 1:

Input: 1
Output: "1"

Example 2:

Input: 4
Output: "1211"

Solution: Recursion + Simulation

C++

// Author: Huahua, 4 ms, 8.6 MB

class Solution {

public:

string countAndSay(int n) {

string ans = "1";

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

ans = say(ans);

return ans;

}

private:

string say(const string& n){

string ans;

int s = 0, l = n.length();

for (int e = 1; e <= l; ++e)

if (e == l || n[s] != n[e]) {

int count = e - s;

ans += '0' + count;

ans += n[s];

s = e;

}

return ans;

}

};

花花酱 LeetCode 67. Add Binary

By zxi on April 10, 2019

Given two binary strings, return their sum (also a binary string).

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

Solution: Big integer

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

C++

// Author: Huahua

class Solution {

public:

string addBinary(string a, string b) {

int carry = 0;

int i = a.length() - 1;

int j = b.length() - 1;

string ans;

while (i >= 0 || j >= 0) {

int s = (i >= 0 ? a[i--] - '0' : 0) +

(j >= 0 ? b[j--] - '0' : 0) +

carry;

carry = s >> 1;

ans += '0' + (s & 1);

}

if (carry) ans += '1';

return {rbegin(ans), rend(ans)};

}

};

花花酱 LeetCode 125. Valid Palindrome

By zxi on April 10, 2019

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

Solution: Two pointers

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

C++

class Solution {

public:

bool isPalindrome(string s) {

int i = 0;

int j = s.length() - 1;

while (i < j) {

while (i < j && !isalnum(s[i])) ++i;

while (i < j && !isalnum(s[j])) --j;

if (tolower(s[i++]) != tolower(s[j--]))

return false;

}

return true;

}

};

花花酱 LeetCode 66. Plus One

By zxi on April 10, 2019

Given a non-empty array of digits representing a non-negative integer, plus one to the integer.

The digits are stored such that the most significant digit is at the head of the list, and each element in the array contain a single digit.

You may assume the integer does not contain any leading zero, except the number 0 itself.

Example 1:

Input: [1,2,3]
Output: [1,2,4]
Explanation: The array represents the integer 123.

Example 2:

Input: [4,3,2,1]
Output: [4,3,2,2]
Explanation: The array represents the integer 4321.

Solution: Big Integer

Process from right to left (lest significant to most signification). Use carry to indicate whether need +1 for next digit or not.

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

c++

class Solution {

public:

vector<int> plusOne(vector<int>& digits) {

int carry = 1;

for (int i = digits.size() - 1; i >= 0; --i) {

digits[i] += carry;

carry = digits[i] / 10;

digits[i] %= 10;

}

if (carry) digits.insert(begin(digits), carry);

return digits;

}

};

Posts tagged as “easy”

花花酱 LeetCode 1042. Flower Planting With No Adjacent

Solution: Graph coloring, choose any available color

C++

花花酱 LeetCode 38. Count and Say

Problem

C++

花花酱 LeetCode 67. Add Binary

C++

花花酱 LeetCode 125. Valid Palindrome

C++

花花酱 LeetCode 66. Plus One

c++