Algorithms Archives - Page 27 of 43

花花酱 LeetCode 1287. Element Appearing More Than 25% In Sorted Array

By zxi on December 14, 2019

Given an integer array sorted in non-decreasing order, there is exactly one integer in the array that occurs more than 25% of the time.

Return that integer.

Example 1:

1 2	<strong>Input:</strong> arr = [1,2,2,6,6,6,6,7,10] <strong>Output:</strong> 6

Constraints:

1 <= arr.length <= 10^4
0 <= arr[i] <= 10^5

Solution 1: Linear Scan

if arr[i] == arr[i + len/4] => arr[i] is the special integer.

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

C++

// Author: Huahua

class Solution {

public:

int findSpecialInteger(vector<int>& arr) {

int s = arr.size() / 4;

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

if (arr[i] == arr[i + s]) return arr[i];

return -1;

}

};

Solution 2: Binary Search

The answer must be one of (s[0], s[l/4], s[l/2], s[l*3/4])
Using binary search to find the range of each number, the one has more than 1/4 of total elements is the answer.

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

C++

// Author: Huahua

class Solution {

public:

int findSpecialInteger(vector<int>& arr) {

const int s = arr.size() / 4;

for (int i = 0; i < 4; ++i) {

const int x = arr[i * s];

const auto [it1, it2] = equal_range(begin(arr), end(arr), x);

if (it2 - it1 > s) return x;

}

return -1;

}

};

花花酱 LeetCode 1283. Find the Smallest Divisor Given a Threshold

By zxi on December 8, 2019

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor and divide all the array by it and sum the result of the division. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

It is guaranteed that there will be an answer.

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1. 
If the divisor is 4 we can get a sum to 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

Input: nums = [2,3,5,7,11], threshold = 11
Output: 3

Example 3:

Input: nums = [19], threshold = 5
Output: 4

Solution: Binary Search

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

C++

// Author: Huahua

class Solution {

public:

int smallestDivisor(vector<int>& nums, int threshold) {

auto sums = [&](int d) {

int s = 0;

for (int n : nums)

s += (n + (d - 1)) / d;

return s;

};

int l = 1;

int r = 1e6;

while (l < r) {

int m = l + (r - l) / 2;

if (sums(m) <= threshold)

r = m;

else

l = m + 1;

}

return l;

}

};

花花酱 LeetCode 1268. Search Suggestions System

By zxi on November 24, 2019

Given an array of strings products and a string searchWord. We want to design a system that suggests at most three product names from products after each character of searchWord is typed. Suggested products should have common prefix with the searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products.

Return list of lists of the suggested products after each character of searchWord is typed.

Example 1:

Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse"
Output: [
["mobile","moneypot","monitor"],
["mobile","moneypot","monitor"],
["mouse","mousepad"],
["mouse","mousepad"],
["mouse","mousepad"]
]
Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"]
After typing m and mo all products match and we show user ["mobile","moneypot","monitor"]
After typing mou, mous and mouse the system suggests ["mouse","mousepad"]

Example 2:

Input: products = ["havana"], searchWord = "havana"
Output: [["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]

Example 3:

Input: products = ["bags","baggage","banner","box","cloths"], searchWord = "bags"
Output: [["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]

Example 4:

Input: products = ["havana"], searchWord = "tatiana"
Output: [[],[],[],[],[],[],[]]

Constraints:

1 <= products.length <= 1000
1 <= Σ products[i].length <= 2 * 10^4
All characters of products[i] are lower-case English letters.
1 <= searchWord.length <= 1000
All characters of searchWord are lower-case English letters.

Solution 1: Binary Search

Sort the input array and do two binary searches.
One for prefix of the search word as lower bound, another for prefix + ‘~’ as upper bound.
‘~’ > ‘z’

Time complexity: O(nlogn + l * logn)
Space complexity: O(1)

C++

// Author: Huahua, 36ms, 35MB

class Solution {

public:

vector<vector<string>> suggestedProducts(vector<string>& products, string searchWord) {

vector<vector<string>> ans;

std::sort(begin(products), end(products));

string key;

for (char c : searchWord) {

key += c;

auto l = lower_bound(begin(products), end(products), key);

auto r = upper_bound(begin(products), end(products), key += '~');

if (l == r) break; // early return

key.pop_back();

ans.emplace_back(l, min(l + 3, r));

}

while (ans.size() != searchWord.length()) ans.push_back({});

return ans;

}

};

Solution 2: Trie

Initialization: Sum(len(products[i]))
Query: O(len(searchWord))

C++

// Author: Huahua, 148 ms, 98.8 MB

struct Trie {

Trie(): nodes(26) {}

~Trie() {

for (auto* node : nodes)

delete node;

}

vector<Trie*> nodes;

vector<const string*> words;

static void addWord(Trie* root, const string& word) {

for (char c : word) {

if (root->nodes[c - 'a'] == nullptr) root->nodes[c - 'a'] = new Trie();

root = root->nodes[c - 'a'];

if (root->words.size() < 3)

root->words.push_back(&word);

}

static vector<vector<string>> getWords(Trie* root, const string& prefix) {

vector<vector<string>> ans(prefix.size());

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

char c = prefix[i];

root = root->nodes[c - 'a'];

if (root == nullptr) break;

for (auto* word : root->words)

ans[i].push_back(*word);

}

return ans;

}

};

class Solution {

public:

vector<vector<string>> suggestedProducts(vector<string>& products, string searchWord) {

std::sort(begin(products), end(products));

Trie root;

for (const auto& product : products)

Trie::addWord(&root, product);

return Trie::getWords(&root, searchWord);

}

};

花花酱 LeetCode 1267. Count Servers that Communicate

By zxi on November 23, 2019

You are given a map of a server center, represented as a m * n integer matrix grid, where 1 means that on that cell there is a server and 0 means that it is no server. Two servers are said to communicate if they are on the same row or on the same column.

Return the number of servers that communicate with any other server.

Example 1:

Input: grid = [[1,0],[0,1]]
Output: 0
Explanation: No servers can communicate with others.

Example 2:

Input: grid = [[1,0],[1,1]]
Output: 3
Explanation: All three servers can communicate with at least one other server.

Example 3:

Input: grid = [[1,1,0,0],[0,0,1,0],[0,0,1,0],[0,0,0,1]]
Output: 4
Explanation: The two servers in the first row can communicate with each other. The two servers in the third column can communicate with each other. The server at right bottom corner can't communicate with any other server.

Constraints:

m == grid.length
n == grid[i].length
1 <= m <= 250
1 <= n <= 250
grid[i][j] == 0 or 1

Solution: Counting

Two passes:
First pass, count number of computers for each row and each column.
Second pass, count grid[i][j] if rows[i] or cols[j] has more than 1 computer.

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

C++

// Author: Huahua

class Solution {

public:

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

const size_t m = grid.size();

const size_t n = grid[0].size();

vector<int> rows(m);

vector<int> cols(n);

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

for (size_t j = 0; j < n; ++j) {

rows[i] += grid[i][j];

cols[j] += grid[i][j];

}

int ans = 0;

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

for (size_t j = 0; j < n; ++j)

ans += (rows[i] > 1 || cols[j] > 1) ? grid[i][j] : 0;

return ans;

}

};

花花酱 LeetCode 48. Rotate Image

By zxi on October 2, 2019

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Note:

You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

Solution: 2 Passes

First pass: mirror around diagonal
Second pass: mirror around y axis

Time complexity: O(n^2)
Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

void rotate(vector<vector<int>>& matrix) {

const int n = matrix.size();

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

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

swap(matrix[i][j], matrix[j][i]);

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

reverse(begin(matrix[i]), end(matrix[i]));

}

};

Posts published in “Algorithms”

花花酱 LeetCode 1287. Element Appearing More Than 25% In Sorted Array

Solution 1: Linear Scan

C++

Solution 2: Binary Search

C++

花花酱 LeetCode 1283. Find the Smallest Divisor Given a Threshold

Solution: Binary Search

C++

花花酱 LeetCode 1268. Search Suggestions System

Solution 1: Binary Search

C++

Solution 2: Trie

C++

花花酱 LeetCode 1267. Count Servers that Communicate

Solution: Counting

C++

花花酱 LeetCode 48. Rotate Image

Solution: 2 Passes

C++