Posts tagged as “hashtable”

花花酱 LeetCode 383. Ransom Note

By zxi on March 15, 2018

题目大意：给你一个字符串，问能否用它其中的字符组成另外一个字符串，每个字符只能使用一次。

Problem:

Given an arbitrary ransom note string and another string containing letters from all the magazines, write a function that will return true if the ransom note can be constructed from the magazines ; otherwise, it will return false.

Each letter in the magazine string can only be used once in your ransom note.

Note:
You may assume that both strings contain only lowercase letters.

canConstruct("a", "b") -> false
canConstruct("aa", "ab") -> false
canConstruct("aa", "aab") -> true

Solution: HashTable

Time complexity: O(n + m)

Space complexity: O(128)

C++

// Author: Huahua

// Running time: 24 ms

class Solution {

public:

bool canConstruct(string ransomNote, string magazine) {

vector<int> counts(128, 0);

for (const char c : magazine)

++counts[c];

for (const char c : ransomNote)

if (--counts[c] < 0) return false;

return true;

}

};

// Author: Huahua

// Running time: 9 ms (beats 100%)

static int x=[](){

std::ios::sync_with_stdio(false);

cin.tie(NULL);

return 0;

}();

class Solution {

public:

bool canConstruct(const string& ransomNote, const string& magazine) {

vector<int> counts(128, 0);

for (const char c : magazine)

++counts[c];

for (const char c : ransomNote)

if (--counts[c] < 0) return false;

return true;

}

};

Python3

"""

Author: Huahua

Running time: 64 ms

"""

class Solution:

def canConstruct(self, ransomNote, magazine):

return not collections.Counter(ransomNote) - collections.Counter(magazine);

花花酱 LeetCode 697. Degree of an Array

By zxi on March 8, 2018

题目大意：求”度”相同的最短子数组的长度。

Problem:

https://leetcode.com/problems/degree-of-an-array/description/

Given a non-empty array of non-negative integers nums, the degree of this array is defined as the maximum frequency of any one of its elements.

Your task is to find the smallest possible length of a (contiguous) subarray of nums, that has the same degree as nums.

Example 1:

Input: [1, 2, 2, 3, 1]

Output: 2

Explanation:

The input array has a degree of 2 because both elements 1 and 2 appear twice.

Of the subarrays that have the same degree:

[1, 2, 2, 3, 1], [1, 2, 2, 3], [2, 2, 3, 1], [1, 2, 2], [2, 2, 3], [2, 2]

The shortest length is 2. So return 2.

Example 2:

1 2	Input: [1,2,2,3,1,4,2] Output: 6

Solution 1: Hashtable

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 33 ms (beats 90.96%)

class Solution {

public:

int findShortestSubArray(vector<int>& nums) {

unordered_map<int, vector<int>> indices;

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

indices[nums[i]].push_back(i);

size_t degree = 0;

for (const auto& p : indices)

degree = max(degree, p.second.size());

int ans = nums.size();

for (const auto& p : indices) {

if (p.second.size() != degree) continue;

ans = min(ans, p.second.back() - p.second.front() + 1);

}

return ans;

}

};

花花酱 LeetCode 454. 4Sum II

By zxi on March 5, 2018

题目大意：给你4个组数：A, B, C, D。问有多少组组合的A[i] + B[j] + C[k] + D[l] 的和为0。

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l]is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -2²⁸ to 2²⁸ – 1 and the result is guaranteed to be at most 2³¹ – 1.

Example:

Input:

A = [ 1, 2]

B = [-2,-1]

C = [-1, 2]

D = [ 0, 2]

Output:

Explanation:

The two tuples are:

1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0

2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

Solution 1: HashTable

Split the arrays into two groups: (A, B), (C, D)

Create the Minkowski sum of two groups and count the occurrence of each element.

e.g. A = [1, 2, 3], B = [0, -1, 1]

Minkowski sum(A, B) SAB= [0, 1, 2, 3, 4] => [0:1, 1:2, 2:3, 3:2, 4:1]

for each element s in SAB, check whether -s is in Minkowski sum(C, D) SCD

ans = sum(SAB[s] * SCD[-s]), for all s, that s in SAB and -s in SCD

Time complexity: O(n^2)

Space complexity: O(n^2)

C++

// Author: Huahua

// Running time: 217 ms

class Solution {

public:

int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {

unordered_map<int, int> sums;

for (int a : A)

for (int b : B)

++sums[a + b];

int ans = 0;

for (int c : C)

for (int d : D) {

auto it = sums.find(- c - d);

if (it != sums.end()) ans += it->second;

}

return ans;

}

};

Python3

"""

Author: Huahua

Running time: 672 ms

"""

class Solution:

def fourSumCount(self, A, B, C, D):

s = collections.Counter([a + b for a in A for b in B])

return sum(s[-c - d] for c in C for d in D)

Related Problems:

[解题报告] LeetCode 1. Two Sum – 花花酱

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 575. Distribute Candies

By zxi on March 5, 2018

题目大意：有一些不同种类的糖果，男生女生各得一半数量的糖果，问女生最多可能得到多少种不同种类的糖果。

Given an integer array with even length, where different numbers in this array represent different kinds of candies. Each number means one candy of the corresponding kind. You need to distribute these candies equally in number to brother and sister. Return the maximum number of kinds of candies the sister could gain.

Example 1:

Input: candies = [1,1,2,2,3,3]

Output: 3

Explanation:

There are three different kinds of candies (1, 2 and 3), and two candies for each kind.

Optimal distribution: The sister has candies [1,2,3] and the brother has candies [1,2,3], too.

The sister has three different kinds of candies.

Example 2:

Input: candies = [1,1,2,3]

Output: 2

Explanation: For example, the sister has candies [2,3] and the brother has candies [1,1].

The sister has two different kinds of candies, the brother has only one kind of candies.

Note:

The length of the given array is in range [2, 10,000], and will be even.
The number in given array is in range [-100,000, 100,000].

Solution 1: Greedy

Give all unique candies to sisters until they have n/2 candies.

Time complexity: O(n)

Space complexity: O(n)

C++ Hashset

// Author: Huahua

// Running time: 258 ms

class Solution {

public:

int distributeCandies(vector<int>& candies) {

unordered_set<int> kinds(candies.begin(), candies.end());

return min(kinds.size(), candies.size() / 2);

}

};

C++ Bitset

// Author: Huahua

// Running time: 195 ms (beats 97.61%)

class Solution {

public:

int distributeCandies(vector<int>& candies) {

constexpr int kMaxKinds = 200001;

constexpr int kOffset = 100000;

bitset<kMaxKinds> s;

for (int candy : candies)

if (!s.test(candy + kOffset))

s.set(candy + kOffset);

return std::min(s.count(), candies.size() / 2);

}

};