Posts tagged as “easy”

花花酱 LeetCode 2667. Create Hello World Function

By zxi on May 7, 2023

Write a function createHelloWorld. It should return a new function that always returns "Hello World".

Example 1:

Input: args = []
Output: "Hello World"
Explanation:
const f = createHelloWorld();
f(); // "Hello World"
The function returned by createHelloWorld should always return "Hello World".

Example 2:

Input: args = [{},null,42]
Output: "Hello World"
Explanation:
const f = createHelloWorld();
f({}, null, 42); // "Hello World"
Any arguments could be passed to the function but it should still always return "Hello World".

Constraints:

0 <= args.length <= 10

Solution:

JavaScript

// Author: Huahua

/**

* @return {Function}

var createHelloWorld = function() {

return function(...args) {

return "Hello World";

}

};

花花酱 LeetCode 2660. Determine the Winner of a Bowling Game

By zxi on April 30, 2023

You are given two 0-indexed integer arrays player1 and player2, that represent the number of pins that player 1 and player 2 hit in a bowling game, respectively.

The bowling game consists of n turns, and the number of pins in each turn is exactly 10.

Assume a player hit x_i pins in the i^th turn. The value of the i^th turn for the player is:

2x_i if the player hit 10 pins in any of the previous two turns.
Otherwise, It is x_i.

The score of the player is the sum of the values of their n turns.

Return

1 if the score of player 1 is more than the score of player 2,
2 if the score of player 2 is more than the score of player 1, and
0 in case of a draw.

Example 1:

Input: player1 = [4,10,7,9], player2 = [6,5,2,3]
Output: 1
Explanation: The score of player1 is 4 + 10 + 2*7 + 2*9 = 46.
The score of player2 is 6 + 5 + 2 + 3 = 16.
Score of player1 is more than the score of player2, so, player1 is the winner, and the answer is 1.

Example 2:

Input: player1 = [3,5,7,6], player2 = [8,10,10,2]
Output: 2
Explanation: The score of player1 is 3 + 5 + 7 + 6 = 21.
The score of player2 is 8 + 10 + 2*10 + 2*2 = 42.
Score of player2 is more than the score of player1, so, player2 is the winner, and the answer is 2.

Example 3:

Input: player1 = [2,3], player2 = [4,1]
Output: 0
Explanation: The score of player1 is 2 + 3 = 5
The score of player2 is 4 + 1 = 5
The score of player1 equals to the score of player2, so, there is a draw, and the answer is 0.

Constraints:

n == player1.length == player2.length
1 <= n <= 1000
0 <= player1[i], player2[i] <= 10

Solution: Simulation

We can write a function to compute the score of a player.

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

C++

// Author: Huahua

class Solution {

public:

int isWinner(vector<int>& player1, vector<int>& player2) {

auto score = [] (const vector<int>& player) {

int sum = 0;

int twice = 0;

for (int x : player) {

sum += twice-- > 0 ? x * 2 : x;

if (x == 10) twice = 2;

}

return sum;

};

int s1 = score(player1);

int s2 = score(player2);

if (s1 == s2) return 0;

return s1 > s2 ? 1 : 2;

}

};

花花酱 LeetCode 2656. Maximum Sum With Exactly K Elements

By zxi on April 30, 2023

You are given a 0-indexed integer array nums and an integer k. Your task is to perform the following operation exactly k times in order to maximize your score:

Select an element m from nums.
Remove the selected element m from the array.
Add a new element with a value of m + 1 to the array.
Increase your score by m.

Return the maximum score you can achieve after performing the operation exactly k times.

Example 1:

Input: nums = [1,2,3,4,5], k = 3
Output: 18
Explanation: We need to choose exactly 3 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [1,2,3,4,6]
For the second iteration, we choose 6. Then sum is 5 + 6 and nums = [1,2,3,4,7]
For the third iteration, we choose 7. Then sum is 5 + 6 + 7 = 18 and nums = [1,2,3,4,8]
So, we will return 18.
It can be proven, that 18 is the maximum answer that we can achieve.

Example 2:

Input: nums = [5,5,5], k = 2
Output: 11
Explanation: We need to choose exactly 2 elements from nums to maximize the sum.
For the first iteration, we choose 5. Then sum is 5 and nums = [5,5,6]
For the second iteration, we choose 6. Then sum is 5 + 6 = 11 and nums = [5,5,7]
So, we will return 11.
It can be proven, that 11 is the maximum answer that we can achieve.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100
1 <= k <= 100

Solution: Greedy

Always to chose the largest element from the array.

We can find the largest element of the array m, then the total score will be
m + (m + 1) + (m + 2) + … + (m + k – 1),
We can use summation formula of arithmetic sequence to compute that in O(1)
ans = (m + (m + k – 1)) * k / 2

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

C++

// Author: Huahua

class Solution {

public:

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

int m = *max_element(begin(nums), end(nums));

return (m + m + k - 1) * k / 2;

}

};

花花酱 LeetCode 2639. Find the Width of Columns of a Grid

By zxi on April 29, 2023

You are given a 0-indexed m x n integer matrix grid. The width of a column is the maximum length of its integers.

For example, if grid = [[-10], [3], [12]], the width of the only column is 3 since -10 is of length 3.

Return an integer array ans of size n where ans[i] is the width of the i^th column.

The length of an integer x with len digits is equal to len if x is non-negative, and len + 1 otherwise.

Example 1:

Input: grid = [[1],[22],[333]]
Output: [3]
Explanation: In the 0^th column, 333 is of length 3.

Example 2:

Input: grid = [[-15,1,3],[15,7,12],[5,6,-2]]
Output: [3,1,2]
Explanation: 
In the 0^th column, only -15 is of length 3.
In the 1^st column, all integers are of length 1. 
In the 2^nd column, both 12 and -2 are of length 2.

Constraints:

m == grid.length
n == grid[i].length
1 <= m, n <= 100
-10⁹ <= grid[r][c] <= 10⁹

Solution: Simulation

Note: width of ‘0’ is 1.

Time complexity: O(m*n*log(x))
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

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

vector<int> ans;

for (int j = 0; j < grid[0].size(); ++j) {

int maxWidth = 1;

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

int x = grid[i][j];

int width = 0;

if (x < 0) {

x = -x;

++width;

}

while (x) {

x /= 10;

++width;

}

maxWidth = max(maxWidth, width);

}

ans.push_back(maxWidth);

}

return ans;

}

};

花花酱 LeetCode 2643. Row With Maximum Ones

By zxi on April 28, 2023

Given a m x n binary matrix mat, find the 0-indexed position of the row that contains the maximum count of ones, and the number of ones in that row.

In case there are multiple rows that have the maximum count of ones, the row with the smallest row number should be selected.

Return an array containing the index of the row, and the number of ones in it.

Example 1:

Input: mat = [[0,1],[1,0]]
Output: [0,1]
Explanation: Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1].

Example 2:

Input: mat = [[0,0,0],[0,1,1]]
Output: [1,2]
Explanation: The row indexed 1 has the maximum count of ones (2). So we return its index, 1, and the count. So, the answer is [1,2].

Example 3:

Input: mat = [[0,0],[1,1],[0,0]]
Output: [1,2]
Explanation: The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].

Constraints:

m == mat.length
n == mat[i].length
1 <= m, n <= 100
mat[i][j] is either 0 or 1.

Solution: Counting

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

C++

// Author: Huahua

class Solution {

public:

vector<int> rowAndMaximumOnes(vector<vector<int>>& mat) {

vector<int> ans(2);

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

int ones = accumulate(begin(mat[i]), end(mat[i]), 0);

if (ones > ans[1]) {

ans[0] = i;

ans[1] = ones;

}

return ans;

}

};