Posts published in “Algorithms”

花花酱 LeetCode 496. Next Greater Element I

By zxi on March 5, 2018

题目大意：给你一个组数A里面每个元素都不相同。再给你一个数组B，元素是A的子集，问对于B中的每个元素，在A数组中相同元素之后第一个比它的元素是多少。

Problem:

You are given two arrays (without duplicates) nums1 and nums2 where nums1’s elements are subset of nums2. Find all the next greater numbers for nums1‘s elements in the corresponding places of nums2.

The Next Greater Number of a number x in nums1 is the first greater number to its right in nums2. If it does not exist, output -1 for this number.

Example 1:

Input: nums1 = [4,1,2], nums2 = [1,3,4,2].

Output: [-1,3,-1]

Explanation:

For number 4 in the first array, you cannot find the next greater number for it in the second array, so output -1.

For number 1 in the first array, the next greater number for it in the second array is 3.

For number 2 in the first array, there is no next greater number for it in the second array, so output -1.

Example 2:

Input: nums1 = [2,4], nums2 = [1,2,3,4].

Output: [3,-1]

Explanation:

For number 2 in the first array, the next greater number for it in the second array is 3.

For number 4 in the first array, there is no next greater number for it in the second array, so output -1.

Note:

All elements in nums1 and nums2 are unique.
The length of both nums1 and nums2 would not exceed 1000.

Solution 1: Brute Force

Time complexity: O(n^2)

Space complexity: O(1)

// Author: Huahua

// Running time: 13 ms

class Solution {

public:

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

vector<int> ans;

for (int num : findNums) {

ans.push_back(-1);

int index = std::find(nums.begin(), nums.end(), num) - nums.begin();

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

if (nums[i] > num) {

ans.back() = nums[i];

break;

}

return ans;

}

};

Solution 2: HashTable + Brute Force

Time complexity: O(n^2)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 15 ms

class Solution {

public:

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

unordered_map<int, int> indices;

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

indices[nums[i]] = i;

vector<int> ans;

for (int num : findNums) {

ans.push_back(-1);

for (int i = indices[num]; i < nums.size(); ++i)

if (nums[i] > num) {

ans.back() = nums[i];

break;

}

return ans;

}

};

Solution 3: Stack + HashTable

Using a stack to store the nums whose next greater isn’t found yet.

// Author: Huahua

// Running time: 11 ms

class Solution {

public:

vector<int> nextGreaterElement(vector<int>& findNums, vector<int>& nums) {

stack<int> s;

unordered_map<int, int> next;

for (int num : nums) {

while (!s.empty() && num > s.top()) {

next[s.top()] = num;

s.pop();

}

s.push(num);

}

vector<int> ans;

for (int num : findNums)

ans.push_back(next.count(num) ? next[num] : -1);

return ans;

}

};

花花酱 LeetCode 793. Preimage Size of Factorial Zeroes Function

By zxi on March 4, 2018

Let f(x) be the number of zeroes at the end of x!. (Recall that x! = 1 * 2 * 3 * ... * x, and by convention, 0! = 1.)

For example, f(3) = 0 because 3! = 6 has no zeroes at the end, while f(11) = 2 because 11! = 39916800 has 2 zeroes at the end. Given K, find how many non-negative integers x have the property that f(x) = K.

Example 1:

Input: K = 0

Output: 5

Explanation: 0!, 1!, 2!, 3!, and 4! end with K = 0 zeroes.

Example 2:

Input: K = 5

Output: 0

Explanation: There is no x such that x! ends in K = 5 zeroes.

Note:

K will be an integer in the range [0, 10^9].

Idea:

First we need to compute how many trailing zeros n! has.

See 花花酱 LeetCode 172. Factorial Trailing Zeroes for details

It’s hard to say how many numbers have trailing zeros equals to K, but we can find the largest number p whose trailing zeros is K using binary search. (p+1)! has more than K trailing zeros. And do the same thing to find the largest number q whose trailing zeros is K – 1 using binary search.

Then we know that are exact p numbers 1,2,…,p whose trailing zeros are less or equal to K.

And exact q numbers 1, 2, …, q whose trailing zeros are less or equal to K – 1.

q + 1, q + 2, …, m (m – q numbers in total) the numbers with trailing zeros equal to K.

Solution 1: Math + Binary Search

Time complexity: O(log2(INT_MAX)*log5(INT_MAX))

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

int preimageSizeFZF(int K) {

return g(K) - g(K - 1);

}

private:

// Find the largest number l, that numZeros(l!) == K and numZeros((l+1)!) > K

int g(int K) {

int l = 0;

int r = INT_MAX;

while (l < r) {

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

int zeros = numZeros(m);

if (zeros <= K)

l = m + 1;

else

r = m;

}

return l;

}

int numZeros(int n) {

return n < 5 ? 0 : n / 5 + numZeros(n / 5);

}

};

花花酱 LeetCode 792. Number of Matching Subsequences

By zxi on March 3, 2018

题目大意：给你一些单词，问有多少单词出现在字符串S的子序列中。

https://leetcode.com/problems/number-of-matching-subsequences/description/

Given string S and a dictionary of words words, find the number of words[i] that is a subsequence of S.

Example :

Input:

S = "abcde"

words = ["a", "bb", "acd", "ace"]

Output: 3

Explanation: There are three words in <code>words</code> that are a subsequence of <code>S</code>: "a", "acd", "ace".

Note:

All words in words and S will only consists of lowercase letters.
The length of S will be in the range of [1, 50000].
The length of words will be in the range of [1, 5000].
The length of words[i] will be in the range of [1, 50].

Solution 1: Brute Force

Time complexity: O((S + L) * W)

C++ w/o cache TLE

Space complexity: O(1)

C++ w/ cache 155 ms

Space complexity: O(W * L)

// Author: Huahua

// Running time: 155 ms

class Solution {

public:

int numMatchingSubseq(const string& S, vector<string>& words) {

int ans = 0;

unordered_map<string, bool> m;

for (const string& word : words) {

auto it = m.find(word);

if (it == m.end()) {

bool match = isMatch(word, S);

ans += m[word] = match;

} else {

ans += it->second;

}

return ans;

}

private:

bool isMatch(const string& word, const string& S) {

int start = 0;

for (const char c : word) {

bool found = false;

for (int i = start; i < S.length(); ++i)

if (S[i] == c) {

found = true;

start = i + 1;

break;

}

if (!found) return false;

}

return true;

}

};

Solution 2: Indexing+ Binary Search

Time complexity: O(S + W * L * log(S))

Space complexity: O(S)

S: length of S

W: number of words

L: length of a word

C++

// Author: Huahua

// Running time: 183 ms

class Solution {

public:

int numMatchingSubseq(const string& S, vector<string>& words) {

vector<vector<int>> pos(128);

for (int i = 0; i < S.length(); ++i)

pos[S[i]].push_back(i);

int ans = 0;

for (const string& word : words)

ans += isMatch(word, pos);

return ans;

}

private:

unordered_map<string, bool> m_;

bool isMatch(const string& word, const vector<vector<int>>& pos) {

if (m_.count(word)) return m_[word];

int last_index = -1;

for (const char c : word) {

const vector<int>& p = pos[c];

const auto it = std::lower_bound(p.begin(), p.end(), last_index + 1);

if (it == p.end()) return m_[word] = false;

last_index = *it;

}

return m_[word] = true;

}

};

Java

// Author: Huahua

// Running time: 135 ms

class Solution {

private List<List<Integer>> pos;

private boolean isMatch(String word) {

int l = -1;

for (char c : word.toCharArray()) {

List<Integer> p = pos.get(c);

int index = Collections.binarySearch(p, l + 1);

if (index < 0) index = -index - 1;

if (index >= p.size()) return false;

l = p.get(index);

}

return true;

}

public int numMatchingSubseq(String S, String[] words) {

pos = new ArrayList<>();

for (int i = 0; i < 128; ++i)

pos.add(new ArrayList<Integer>());

char[] s = S.toCharArray();

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

pos.get(s[i]).add(i);

int ans = 0;

for (String word : words)

if (isMatch(word)) ++ans;

return ans;

}

Python 3:

w/o cache

"""

Author: Huahua

Running time: 1120 ms

"""

class Solution:

def numMatchingSubseq(self, S, words):

import bisect

def isMatch(word):

l = -1

for c in word:

index = bisect.bisect_left(pos[c], l + 1)

if index == len(pos[c]): return 0

l = pos[c][index]

return 1

pos = {chr(i + ord('a')) : [] for i in range(26)}

for i, c in enumerate(S):

pos[c].append(i)

return sum(map(isMatch, words))

w/ cache

"""

Author: Huahua

Running time: 632 ms

"""

class Solution:

def numMatchingSubseq(self, S, words):

import bisect

def isMatch(word):

if word in m: return m[word]

l = -1

for c in word:

index = bisect.bisect_left(pos[c], l + 1)

if index == len(pos[c]):

m[word] = 0

return 0

l = pos[c][index]

m[word] = 1

return 1

m = {}

pos = {chr(i + ord('a')) : [] for i in range(26)}

for i, c in enumerate(S):

pos[c].append(i)

return sum(map(isMatch, words))

花花酱 LeetCode 795. Number of Subarrays with Bounded Maximum

By zxi on March 3, 2018

题目大意：问一个数组中有多少个子数组的最大元素值在[L, R]的范围里。

We are given an array A of positive integers, and two positive integers L and R (L <= R).

Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least L and at most R.

Example :

Input:

A = [2, 1, 4, 3]

L = 2

R = 3

Output: 3

Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].

Solution 1:

C++

// Author: Huahua

// Running time: 42 ms

class Solution {

public:

int numSubarrayBoundedMax(vector<int>& A, int L, int R) {

return count(A, R) - count(A, L - 1);

}

private:

// Count # of sub arrays whose max element is <= N

int count(const vector<int>& A, int N) {

int ans = 0;

int cur = 0;

for (int a: A) {

if (a <= N)

ans += ++cur;

else

cur = 0;

}

return ans;

}

};

Solution 2: One pass

C++

// Author: Huahua

// Running time: 38 ms

class Solution {

public:

int numSubarrayBoundedMax(vector<int>& A, int L, int R) {

int ans = 0;

int left = -1;

int right = -1;

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

if (A[i] >= L) right = i;

if (A[i] > R) left = i;

ans += (right - left);

}

return ans;

}

};

花花酱 LeetCode 378. Kth Smallest Element in a Sorted Matrix

By zxi on February 20, 2018

题目大意：给你一个矩阵，每行每列各自排序。找出矩阵中第K小的元素。

Given a n x n matrix where each of the rows and columns are sorted in ascending order, find the kth smallest element in the matrix.

Note that it is the kth smallest element in the sorted order, not the kth distinct element.

Example:

matrix = [

[ 1, 5, 9],

[10, 11, 13],

[12, 13, 15]

k = 8,

return 13.

Note:
You may assume k is always valid, 1 ≤ k ≤ n².

Solution 1: Binary Search

Find the smallest x, such that there are k elements that are smaller or equal to x.

Time complexity: O(nlogn*log(max – min))

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 24 ms (beats 97.12%)

class Solution {

public:

int kthSmallest(vector<vector<int>>& matrix, int k) {

const int n = matrix.size();

long l = matrix[0][0];

long r = matrix[n - 1][n - 1] + 1;

while (l < r) {

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

int total = 0;

int s = n;

for (const auto& row : matrix)

total += (s = distance(begin(row), upper_bound(begin(row), begin(row) + s, m)));

if (total >= k)

r = m;

else

l = m + 1;

}

return l;

}

};