Posts tagged as “easy”

花花酱 LeetCode 2389. Longest Subsequence With Limited Sum

By zxi on September 1, 2022

You are given an integer array nums of length n, and an integer array queries of length m.

Return an array answer of length m where answer[i] is the maximum size of a subsequence that you can take from nums such that the sum of its elements is less than or equal to queries[i].

A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [4,5,2,1], queries = [3,10,21]
Output: [2,3,4]
Explanation: We answer the queries as follows:
- The subsequence [2,1] has a sum less than or equal to 3. It can be proven that 2 is the maximum size of such a subsequence, so answer[0] = 2.
- The subsequence [4,5,1] has a sum less than or equal to 10. It can be proven that 3 is the maximum size of such a subsequence, so answer[1] = 3.
- The subsequence [4,5,2,1] has a sum less than or equal to 21. It can be proven that 4 is the maximum size of such a subsequence, so answer[2] = 4.

Example 2:

Input: nums = [2,3,4,5], queries = [1]
Output: [0]
Explanation: The empty subsequence is the only subsequence that has a sum less than or equal to 1, so answer[0] = 0.

Constraints:

n == nums.length
m == queries.length
1 <= n, m <= 1000
1 <= nums[i], queries[i] <= 10⁶

Solution: Sort + PrefixSum + Binary Search

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

C++

// Author: Huahua

class Solution {

public:

vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {

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

partial_sum(begin(nums), end(nums), begin(nums));

vector<int> ans;

for (int q : queries)

ans.push_back(upper_bound(begin(nums), end(nums), q) - begin(nums));

return ans;

}

};

花花酱 LeetCode 2373. Largest Local Values in a Matrix

By zxi on August 14, 2022

You are given an n x n integer matrix grid.

Generate an integer matrix maxLocal of size (n - 2) x (n - 2) such that:

maxLocal[i][j] is equal to the largest value of the 3 x 3 matrix in grid centered around row i + 1 and column j + 1.

In other words, we want to find the largest value in every contiguous 3 x 3 matrix in grid.

Return the generated matrix.

Example 1:

Input: grid = [[9,9,8,1],[5,6,2,6],[8,2,6,4],[6,2,2,2]]
Output: [[9,9],[8,6]]
Explanation: The diagram above shows the original matrix and the generated matrix.
Notice that each value in the generated matrix corresponds to the largest value of a contiguous 3 x 3 matrix in grid.

Example 2:

Input: grid = [[1,1,1,1,1],[1,1,1,1,1],[1,1,2,1,1],[1,1,1,1,1],[1,1,1,1,1]]
Output: [[2,2,2],[2,2,2],[2,2,2]]
Explanation: Notice that the 2 is contained within every contiguous 3 x 3 matrix in grid.

Constraints:

n == grid.length == grid[i].length
3 <= n <= 100
1 <= grid[i][j] <= 100

Solution: Brute Force

Time complexity: O(n*n*9)
Space complexity: O(n*n)

C++

// Author: Huahua

class Solution {

public:

vector<vector<int>> largestLocal(vector<vector<int>>& grid) {

const int m = grid.size() - 2;

vector<vector<int>> ans(m, vector<int>(m));

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

for (int j = 0; j < m; ++j)

for (int dy = 0; dy <= 2; ++dy)

for (int dx = 0; dx <= 2; ++dx)

ans[i][j] = max(ans[i][j], grid[i + dy][j + dx]);

return ans;

}

};

花花酱 LeetCode 2315. Count Asterisks

By zxi on June 25, 2022

You are given a string s, where every two consecutive vertical bars '|' are grouped into a pair. In other words, the 1^st and 2^nd '|' make a pair, the 3^rd and 4^th '|' make a pair, and so forth.

Return the number of '*' in s, excluding the '*' between each pair of '|'.

Note that each '|' will belong to exactly one pair.

Example 1:

Input: s = "l|*e*et|c**o|*de|"
Output: 2
Explanation: The considered characters are underlined: "l|*e*et|c**o|*de|".
The characters between the first and second '|' are excluded from the answer.
Also, the characters between the third and fourth '|' are excluded from the answer.
There are 2 asterisks considered. Therefore, we return 2.

Example 2:

Input: s = "iamprogrammer"
Output: 0
Explanation: In this example, there are no asterisks in s. Therefore, we return 0.

Example 3:

Input: s = "yo|uar|e**|b|e***au|tifu|l"
Output: 5
Explanation: The considered characters are underlined: "yo|uar|e**|b|e***au|tifu|l". There are 5 asterisks considered. Therefore, we return 5.

Constraints:

1 <= s.length <= 1000
s consists of lowercase English letters, vertical bars '|', and asterisks '*'.
s contains an even number of vertical bars '|'.

Solution: Counting

Count the number of bars so far, and only count ‘*’ when there are even number of bars on the left.

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

C++

// Author: Huahua

class Solution {

public:

int countAsterisks(string s) {

int bars = 0;

int ans = 0;

for (char c : s) {

if (c == '*' && bars % 2 == 0)

++ans;

if (c == '|') ++bars;

}

return ans;

}

};

花花酱 LeetCode 2303. Calculate Amount Paid in Taxes

By zxi on June 14, 2022

You are given a 0-indexed 2D integer array brackets where brackets[i] = [upper_i, percent_i] means that the i^th tax bracket has an upper bound of upper_i and is taxed at a rate of percent_i. The brackets are sorted by upper bound (i.e. upper_i-1 < upper_i for 0 < i < brackets.length).

Tax is calculated as follows:

The first upper₀ dollars earned are taxed at a rate of percent₀.
The next upper₁ - upper₀ dollars earned are taxed at a rate of percent₁.
The next upper₂ - upper₁ dollars earned are taxed at a rate of percent₂.
And so on.

You are given an integer income representing the amount of money you earned. Return the amount of money that you have to pay in taxes. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

Input: brackets = [[3,50],[7,10],[12,25]], income = 10
Output: 2.65000
Explanation:
The first 3 dollars you earn are taxed at 50%. You have to pay $3 * 50% = $1.50 dollars in taxes.
The next 7 - 3 = 4 dollars you earn are taxed at 10%. You have to pay $4 * 10% = $0.40 dollars in taxes.
The final 10 - 7 = 3 dollars you earn are taxed at 25%. You have to pay $3 * 25% = $0.75 dollars in taxes.
You have to pay a total of $1.50 + $0.40 + $0.75 = $2.65 dollars in taxes.

Example 2:

Input: brackets = [[1,0],[4,25],[5,50]], income = 2
Output: 0.25000
Explanation:
The first dollar you earn is taxed at 0%. You have to pay $1 * 0% = $0 dollars in taxes.
The second dollar you earn is taxed at 25%. You have to pay $1 * 25% = $0.25 dollars in taxes.
You have to pay a total of $0 + $0.25 = $0.25 dollars in taxes.

Example 3:

Input: brackets = [[2,50]], income = 0
Output: 0.00000
Explanation:
You have no income to tax, so you have to pay a total of $0 dollars in taxes.

Constraints:

1 <= brackets.length <= 100
1 <= upper_i <= 1000
0 <= percent_i <= 100
0 <= income <= 1000
upper_i is sorted in ascending order.
All the values of upper_i are unique.
The upper bound of the last tax bracket is greater than or equal to income.

Solution: Follow the rules

“Nothing is certain except death and taxes” – Benjamin Franklin

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

C++

// Author: Huahua

class Solution {

public:

double calculateTax(vector<vector<int>>& A, int income) {

double ans = 0;

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

ans += (min(income, A[i][0]) - (i ? A[i-1][0] : 0)) * A[i][1] / 100.0;

if (A[i][0] >= income) break;

}

return ans;

}

};

花花酱 LeetCode 2269. Find the K-Beauty of a Number

By zxi on May 15, 2022

The k-beauty of an integer num is defined as the number of substrings of num when it is read as a string that meet the following conditions:

It has a length of k.
It is a divisor of num.

Given integers num and k, return the k-beauty of num.

Note:

Leading zeros are allowed.
0 is not a divisor of any value.

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

Example 1:

Input: num = 240, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
- "24" from "240": 24 is a divisor of 240.
- "40" from "240": 40 is a divisor of 240.
Therefore, the k-beauty is 2.

Example 2:

Input: num = 430043, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
- "43" from "430043": 43 is a divisor of 430043.
- "30" from "430043": 30 is not a divisor of 430043.
- "00" from "430043": 0 is not a divisor of 430043.
- "04" from "430043": 4 is not a divisor of 430043.
- "43" from "430043": 43 is a divisor of 430043.
Therefore, the k-beauty is 2.

Constraints:

1 <= num <= 10⁹
1 <= k <= num.length (taking num as a string)

Solution: Substring

Note: the substring can be 0, e.g. “00”

Time complexity: O((l-k)*k)
Space complexity: O(l + k) -> O(1)

C++

// Author: Huahua

class Solution {

public:

int divisorSubstrings(int num, int k) {

string s = to_string(num);

int ans = 0;

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

const int t = stoi(s.substr(i, k));

ans += t && (num % t == 0);

}

return ans;

}

};