花花酱 LeetCode 1840. Maximum Building Height

By zxi on April 25, 2021

You want to build n new buildings in a city. The new buildings will be built in a line and are labeled from 1 to n.

However, there are city restrictions on the heights of the new buildings:

The height of each building must be a non-negative integer.
The height of the first building must be 0.
The height difference between any two adjacent buildings cannot exceed 1.

Additionally, there are city restrictions on the maximum height of specific buildings. These restrictions are given as a 2D integer array restrictions where restrictions[i] = [id_i, maxHeight_i] indicates that building id_i must have a height less than or equal to maxHeight_i.

It is guaranteed that each building will appear at most once in restrictions, and building 1 will not be in restrictions.

Return the maximum possible height of the tallest building.

Example 1:

Input: n = 5, restrictions = [[2,1],[4,1]]
Output: 2
Explanation: The green area in the image indicates the maximum allowed height for each building.
We can build the buildings with heights [0,1,2,1,2], and the tallest building has a height of 2.

Example 2:

Input: n = 6, restrictions = []
Output: 5
Explanation: The green area in the image indicates the maximum allowed height for each building.
We can build the buildings with heights [0,1,2,3,4,5], and the tallest building has a height of 5.

Example 3:

Input: n = 10, restrictions = [[5,3],[2,5],[7,4],[10,3]]

Output: 5

Explanation: The green area in the image indicates the maximum allowed height for each building.

We can build the buildings with heights [0,1,2,3,3,4,4,5,4,3], and the tallest building has a height of 5.

Constraints:

2 <= n <= 10⁹
0 <= restrictions.length <= min(n - 1, 10⁵)
2 <= id_i <= n
id_i is unique.
0 <= maxHeight_i <= 10⁹

Solution: Two Passes

Trim the max heights based on neighboring max heights.
Two passes: left to right, right to left.

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

C++

// Author: Huahua

class Solution {

public:

int maxBuilding(int n, vector<vector<int>>& rs) {

rs.push_back({1, 0});

sort(begin(rs), end(rs));

if (rs.back()[0] != n)

rs.push_back({n, n - 1});

const int m = rs.size();

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

rs[i][1] = min(rs[i][1], rs[i - 1][1] + rs[i][0] - rs[i - 1][0]);

for (int i = m - 2; i >= 0; --i)

rs[i][1] = min(rs[i][1], rs[i + 1][1] - rs[i][0] + rs[i + 1][0]);

int ans = 0;

for (int i = 1; i < m; ++i) {

const int l = rs[i - 1][1];

const int r = rs[i][1];

ans = max(ans, max(l, r) + (rs[i][0] - rs[i - 1][0] - abs(l - r)) / 2);

}

return ans;

}

};

花花酱 LeetCode 1839. Longest Substring Of All Vowels in Order

By zxi on April 25, 2021

A string is considered beautiful if it satisfies the following conditions:

Each of the 5 English vowels ('a', 'e', 'i', 'o', 'u') must appear at least once in it.
The letters must be sorted in alphabetical order (i.e. all 'a's before 'e's, all 'e's before 'i's, etc.).

For example, strings "aeiou" and "aaaaaaeiiiioou" are considered beautiful, but "uaeio", "aeoiu", and "aaaeeeooo" are not beautiful.

Given a string word consisting of English vowels, return the length of the longest beautiful substring of word. If no such substring exists, return 0.

A substring is a contiguous sequence of characters in a string.

Example 1:

Input: word = "aeiaaioaaaaeiiiiouuuooaauuaeiu"
Output: 13
Explanation: The longest beautiful substring in word is "aaaaeiiiiouuu" of length 13.

Example 2:

Input: word = "aeeeiiiioooauuuaeiou"
Output: 5
Explanation: The longest beautiful substring in word is "aeiou" of length 5.

Example 3:

Input: word = "a"
Output: 0
Explanation: There is no beautiful substring, so return 0.

Constraints:

1 <= word.length <= 5 * 10⁵
word consists of characters 'a', 'e', 'i', 'o', and 'u'.

Solution: Counter

Use a counter to track how many unique vowels we saw so far. Reset the counter whenever the s[i] < s[i-1]. Incase the counter if s[i] > s[i – 1]. When counter becomes 5, we know we found a valid substring.

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

C++

// Author: Huahua

class Solution {

public:

int longestBeautifulSubstring(string word) {

const int n = word.length();

int ans = 0;

char p = 'a' - 1;

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

if (word[i] < p) {

vowels = (word[i] == 'a');

l = i;

} else if (word[i] > p) {

++vowels;

}

if (vowels == 5)

ans = max(ans, i - l + 1);

p = word[i];

}

return ans;

}

};

花花酱 LeetCode 1838. Frequency of the Most Frequent Element

By zxi on April 25, 2021

The frequency of an element is the number of times it occurs in an array.

You are given an integer array nums and an integer k. In one operation, you can choose an index of nums and increment the element at that index by 1.

Return the maximum possible frequency of an element after performing at most k operations.

Example 1:

Input: nums = [1,2,4], k = 5
Output: 3
Explanation: Increment the first element three times and the second element two times to make nums = [4,4,4].
4 has a frequency of 3.

Example 2:

Input: nums = [1,4,8,13], k = 5
Output: 2
Explanation: There are multiple optimal solutions:
- Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
- Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.
- Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.

Example 3:

Input: nums = [3,9,6], k = 2
Output: 1

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵
1 <= k <= 10⁵

Solution: Sliding Window

Sort the elements, maintain a window such that it takes at most k ops to make the all the elements equal to nums[i].

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

C++

class Solution {

public:

int maxFrequency(vector<int>& nums, int k) {

sort(begin(nums), end(nums));

int l = 0;

long sum = 0;

int ans = 0;

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

sum += nums[r];

while (l < r &&

sum + k < static_cast<long>(nums[r]) * (r - l + 1))

sum -= nums[l++];

ans = max(ans, r - l + 1);

}

return ans;

}

};

花花酱 LeetCode 1837. Sum of Digits in Base K

By zxi on April 25, 2021

Given an integer n (in base 10) and a base k, return the sum of the digits of n after converting n from base 10 to base k.

After converting, each digit should be interpreted as a base 10 number, and the sum should be returned in base 10.

Example 1:

Input: n = 34, k = 6
Output: 9
Explanation: 34 (base 10) expressed in base 6 is 54. 5 + 4 = 9.

Example 2:

Input: n = 10, k = 10
Output: 1
Explanation: n is already in base 10. 1 + 0 = 1.

Constraints:

1 <= n <= 100
2 <= k <= 10

Solution: Base Conversion

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

C++

// Author: Huahua

class Solution {

public:

int sumBase(int n, int k) {

int ans = 0;

while (n) {

ans += n % k;

n /= k;

}

return ans;

}

};

花花酱 LeetCode 1835. Find XOR Sum of All Pairs Bitwise AND

By zxi on April 18, 2021

The XOR sum of a list is the bitwise XOR of all its elements. If the list only contains one element, then its XOR sum will be equal to this element.

For example, the XOR sum of [1,2,3,4] is equal to 1 XOR 2 XOR 3 XOR 4 = 4, and the XOR sum of [3] is equal to 3.

You are given two 0-indexed arrays arr1 and arr2 that consist only of non-negative integers.

Consider the list containing the result of arr1[i] AND arr2[j] (bitwise AND) for every (i, j) pair where 0 <= i < arr1.length and 0 <= j < arr2.length.

Return the XOR sum of the aforementioned list.

Example 1:

Input: arr1 = [1,2,3], arr2 = [6,5]
Output: 0
Explanation: The list = [1 AND 6, 1 AND 5, 2 AND 6, 2 AND 5, 3 AND 6, 3 AND 5] = [0,1,2,0,2,1].
The XOR sum = 0 XOR 1 XOR 2 XOR 0 XOR 2 XOR 1 = 0.

Example 2:

Input: arr1 = [12], arr2 = [4]
Output: 4
Explanation: The list = [12 AND 4] = [4]. The XOR sum = 4.

Constraints:

1 <= arr1.length, arr2.length <= 10⁵
0 <= arr1[i], arr2[j] <= 10⁹

Solution: Bit

(a[0] & b[i]) ^ (a[1] & b[i]) ^ … ^ (a[n-1] & b[i]) = (a[0] ^ a[1] ^ … ^ a[n-1]) & b[i]

We can pre compute that xor sum of array A.

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

C++

class Solution {

public:

int getXORSum(vector<int>& A, vector<int>& B) {

int ans = 0;

int xora = 0;

for (int a : A) xora ^= a;

for (int b : B) ans ^= (xora & b);

return ans;

}

};

Huahua's Tech Road

花花酱 LeetCode 1840. Maximum Building Height

Solution: Two Passes

C++

花花酱 LeetCode 1839. Longest Substring Of All Vowels in Order

Solution: Counter

C++

花花酱 LeetCode 1838. Frequency of the Most Frequent Element

Solution: Sliding Window

C++

花花酱 LeetCode 1837. Sum of Digits in Base K

Solution: Base Conversion

C++

花花酱 LeetCode 1835. Find XOR Sum of All Pairs Bitwise AND

Solution: Bit

C++