Posts tagged as “array”

花花酱 LeetCode 2079. Watering Plants

By zxi on November 20, 2021

You want to water n plants in your garden with a watering can. The plants are arranged in a row and are labeled from 0 to n - 1 from left to right where the i^th plant is located at x = i. There is a river at x = -1 that you can refill your watering can at.

Each plant needs a specific amount of water. You will water the plants in the following way:

Water the plants in order from left to right.
After watering the current plant, if you do not have enough water to completely water the next plant, return to the river to fully refill the watering can.
You cannot refill the watering can early.

You are initially at the river (i.e., x = -1). It takes one step to move one unit on the x-axis.

Given a 0-indexed integer array plants of n integers, where plants[i] is the amount of water the i^th plant needs, and an integer capacity representing the watering can capacity, return the number of steps needed to water all the plants.

Example 1:

Input: plants = [2,2,3,3], capacity = 5
Output: 14
Explanation: Start at the river with a full watering can:
- Walk to plant 0 (1 step) and water it. Watering can has 3 units of water.
- Walk to plant 1 (1 step) and water it. Watering can has 1 unit of water.
- Since you cannot completely water plant 2, walk back to the river to refill (2 steps).
- Walk to plant 2 (3 steps) and water it. Watering can has 2 units of water.
- Since you cannot completely water plant 3, walk back to the river to refill (3 steps).
- Walk to plant 3 (4 steps) and water it.
Steps needed = 1 + 1 + 2 + 3 + 3 + 4 = 14.

Example 2:

Input: plants = [1,1,1,4,2,3], capacity = 4
Output: 30
Explanation: Start at the river with a full watering can:
- Water plants 0, 1, and 2 (3 steps). Return to river (3 steps).
- Water plant 3 (4 steps). Return to river (4 steps).
- Water plant 4 (5 steps). Return to river (5 steps).
- Water plant 5 (6 steps).
Steps needed = 3 + 3 + 4 + 4 + 5 + 5 + 6 = 30.

Example 3:

Input: plants = [7,7,7,7,7,7,7], capacity = 8
Output: 49
Explanation: You have to refill before watering each plant.
Steps needed = 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 + 7 = 49.

Constraints:

n == plants.length
1 <= n <= 1000
1 <= plants[i] <= 10⁶
max(plants[i]) <= capacity <= 10⁹

Solution: Simulation

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

C++

// Author: Huahua

class Solution {

public:

int wateringPlants(vector<int>& plants, int capacity) {

const int n = plants.size();

int ans = 0;

for (int i = 0, c = capacity; i < n; c -= plants[i++], ++ans) {

if (c >= plants[i]) continue;

ans += i * 2;

c = capacity;

}

return ans;

}

};

花花酱 LeetCode 2078. Two Furthest Houses With Different Colors

By zxi on November 20, 2021

There are n houses evenly lined up on the street, and each house is beautifully painted. You are given a 0-indexed integer array colors of length n, where colors[i] represents the color of the i^th house.

Return the maximum distance between two houses with different colors.

The distance between the i^th and j^th houses is abs(i - j), where abs(x) is the absolute value of x.

Example 1:

Input: colors = [1,1,1,6,1,1,1]
Output: 3
Explanation: In the above image, color 1 is blue, and color 6 is red.
The furthest two houses with different colors are house 0 and house 3.
House 0 has color 1, and house 3 has color 6. The distance between them is abs(0 - 3) = 3.
Note that houses 3 and 6 can also produce the optimal answer.

Example 2:

Input: colors = [1,8,3,8,3]
Output: 4
Explanation: In the above image, color 1 is blue, color 8 is yellow, and color 3 is green.
The furthest two houses with different colors are house 0 and house 4.
House 0 has color 1, and house 4 has color 3. The distance between them is abs(0 - 4) = 4.

Example 3:

Input: colors = [0,1]
Output: 1
Explanation: The furthest two houses with different colors are house 0 and house 1.
House 0 has color 0, and house 1 has color 1. The distance between them is abs(0 - 1) = 1.

Constraints:

n == colors.length
2 <= n <= 100
0 <= colors[i] <= 100
Test data are generated such that at least two houses have different colors.

Solution 1: Brute Force

Try all pairs.
Time complexity: O(n²)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int maxDistance(vector<int>& colors) {

const int n = colors.size();

for (int d = n - 1; d > 0; --d)

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

if (colors[i] != colors[i + d])

return d;

return 0;

}

};

Solution 2: Greedy / One pass

First house or last house must be involved in the ans.

Scan the house and check with first and last house.

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

C++

// Author: Huahua

class Solution {

public:

int maxDistance(vector<int>& colors) {

int ans = 0;

const int n = colors.size();

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

ans = max({ans,

i * (colors[i] != colors[0]),

(n - i - 1) * (colors[i] != colors[n - 1])});

return ans;

}

};

花花酱 LeetCode 2057. Smallest Index With Equal Value

By zxi on November 3, 2021

Given a 0-indexed integer array nums, return the smallest index i of nums such that i mod 10 == nums[i], or -1 if such index does not exist.

x mod y denotes the remainder when x is divided by y.

Example 1:

Input: nums = [0,1,2]
Output: 0
Explanation: 
i=0: 0 mod 10 = 0 == nums[0].
i=1: 1 mod 10 = 1 == nums[1].
i=2: 2 mod 10 = 2 == nums[2].
All indices have i mod 10 == nums[i], so we return the smallest index 0.

Example 2:

Input: nums = [4,3,2,1]
Output: 2
Explanation: 
i=0: 0 mod 10 = 0 != nums[0].
i=1: 1 mod 10 = 1 != nums[1].
i=2: 2 mod 10 = 2 == nums[2].
i=3: 3 mod 10 = 3 != nums[3].
2 is the only index which has i mod 10 == nums[i].

Example 3:

Input: nums = [1,2,3,4,5,6,7,8,9,0]
Output: -1
Explanation: No index satisfies i mod 10 == nums[i].

Example 4:

Input: nums = [2,1,3,5,2]
Output: 1
Explanation: 1 is the only index with i mod 10 == nums[i].

Constraints:

1 <= nums.length <= 100
0 <= nums[i] <= 9

Solution: Brute Force

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

C++

// Author: Huahua

class Solution {

public:

int smallestEqual(vector<int>& nums) {

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

if (i % 10 == nums[i]) return i;

return -1;

}

};

花花酱 LeetCode 2053. Kth Distinct String in an Array

By zxi on November 1, 2021

A distinct string is a string that is present only once in an array.

Given an array of strings arr, and an integer k, return the k^th distinct string present in arr. If there are fewer than k distinct strings, return an empty string "".

Note that the strings are considered in the order in which they appear in the array.

Example 1:

Input: arr = ["d","b","c","b","c","a"], k = 2
Output: "a"
Explanation:
The only distinct strings in arr are "d" and "a".
"d" appears 1^st, so it is the 1^st distinct string.
"a" appears 2^nd, so it is the 2^nd distinct string.
Since k == 2, "a" is returned.

Example 2:

Input: arr = ["aaa","aa","a"], k = 1
Output: "aaa"
Explanation:
All strings in arr are distinct, so the 1^st string "aaa" is returned.

Example 3:

Input: arr = ["a","b","a"], k = 3
Output: ""
Explanation:
The only distinct string is "b". Since there are fewer than 3 distinct strings, we return an empty string "".

Constraints:

1 <= k <= arr.length <= 1000
1 <= arr[i].length <= 5
arr[i] consists of lowercase English letters.

Solution: Hashtable

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

C++

// Author: Huahua

class Solution {

public:

string kthDistinct(vector<string>& arr, int k) {

unordered_map<string, int> m;

for (const string& s : arr)

++m[s];

for (const string& s : arr)

if (m[s] == 1 && --k == 0) return s;

return "";

}

};

花花酱 LeetCode 2022. Convert 1D Array Into 2D Array

By zxi on October 23, 2021

You are given a 0-indexed 1-dimensional (1D) integer array original, and two integers, m and n. You are tasked with creating a 2-dimensional (2D) array with m rows and n columns using all the elements from original.

The elements from indices 0 to n - 1 (inclusive) of original should form the first row of the constructed 2D array, the elements from indices n to 2 * n - 1 (inclusive) should form the second row of the constructed 2D array, and so on.

Return an m x n 2D array constructed according to the above procedure, or an empty 2D array if it is impossible.

Example 1:

Input: original = [1,2,3,4], m = 2, n = 2
Output: [[1,2],[3,4]]
Explanation:
The constructed 2D array should contain 2 rows and 2 columns.
The first group of n=2 elements in original, [1,2], becomes the first row in the constructed 2D array.
The second group of n=2 elements in original, [3,4], becomes the second row in the constructed 2D array.

Example 2:

Input: original = [1,2,3], m = 1, n = 3
Output: [[1,2,3]]
Explanation:
The constructed 2D array should contain 1 row and 3 columns.
Put all three elements in original into the first row of the constructed 2D array.

Example 3:

Input: original = [1,2], m = 1, n = 1
Output: []
Explanation:
There are 2 elements in original.
It is impossible to fit 2 elements in a 1x1 2D array, so return an empty 2D array.

Example 4:

Input: original = [3], m = 1, n = 2
Output: []
Explanation:
There is 1 element in original.
It is impossible to make 1 element fill all the spots in a 1x2 2D array, so return an empty 2D array.

Constraints:

1 <= original.length <= 5 * 10⁴
1 <= original[i] <= 10⁵
1 <= m, n <= 4 * 10⁴

Solution: Brute Force

the i-th element in original array will have index (i//n, i % n) in the 2D array.

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

C++

class Solution {

public:

vector<vector<int>> construct2DArray(vector<int>& original, int m, int n) {

if (original.size() != m * n) return {};

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

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

ans[i / n][i % n] = original[i];

return ans;

}

};