Posts published in “Hashtable”

花花酱 LeetCode 49. Group Anagrams

By zxi on October 2, 2019

Given an array of strings, group anagrams together.

Example:

Input: ["eat", "tea", "tan", "ate", "nat", "bat"],
Output:
[
  ["ate","eat","tea"],
  ["nat","tan"],
  ["bat"]
]

Note:

All inputs will be in lowercase.
The order of your output does not matter.

Solution: HashTable

The sorted word will be the key of each group

Time complexity: O(sum(l*log(l)))
Space complexity: O(sum(l))

C++

// Author: Huahua

class Solution {

public:

vector<vector<string>> groupAnagrams(vector<string>& strs) {

vector<vector<string>> ans;

unordered_map<string, vector<int>> m;

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

string c = strs[i];

sort(begin(c), end(c));

m[c].push_back(i);

}

for (const auto& kv : m) {

ans.push_back({});

for (int i : kv.second)

ans.back().push_back(strs[i]);

}

return ans;

}

};

花花酱 LeeCode 1207. Unique Number of Occurrences

By zxi on September 29, 2019

Given an array of integers arr, write a function that returns true if and only if the number of occurrences of each value in the array is unique.

Example 1:

Input: arr = [1,2,2,1,1,3]
Output: true
Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.

Example 2:

Input: arr = [1,2]
Output: false

Example 3:

Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
Output: true

Constraints:

1 <= arr.length <= 1000
-1000 <= arr[i] <= 1000

Solution: HashTable

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

C++

// Author: Huahua

class Solution {

public:

bool uniqueOccurrences(vector<int>& arr) {

unordered_map<int, int> c;

for (int a : arr) ++c[a];

unordered_set<int> s;

for (const auto& kv : c)

if (!s.insert(kv.second).second)

return false;

return true;

}

};

花花酱 LeetCode 1189. Maximum Number of Balloons

By zxi on September 15, 2019

Given a string text, you want to use the characters of text to form as many instances of the word “balloon” as possible.

You can use each character in text at most once. Return the maximum number of instances that can be formed.

Example 1:

Input: text = "nlaebolko"
Output: 1

Example 2:

Input: text = "loonbalxballpoon"
Output: 2

Example 3:

Input: text = "leetcode"
Output: 0

Constraints:

1 <= text.length <= 10^4
text consists of lower case English letters only.

Solution: HashTable

Use a hashtable to count the occurrence of each letter and find the bottleneck one.

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

C++

// Author: Huahua

class Solution {

public:

int maxNumberOfBalloons(string text) {

map<char, int> c;

for (char t : text) ++c[t];

return min({c['b'], c['a'], c['l'] / 2, c['o'] / 2, c['n']});

}

};

花花酱 LeetCode 1178. Number of Valid Words for Each Puzzle

By zxi on September 2, 2019

With respect to a given puzzle string, a word is valid if both the following conditions are satisfied:

word contains the first letter of puzzle.
For each letter in word, that letter is in puzzle.
For example, if the puzzle is “abcdefg”, then valid words are “faced”, “cabbage”, and “baggage”; while invalid words are “beefed” (doesn’t include “a”) and “based” (includes “s” which isn’t in the puzzle).

Return an array answer, where answer[i] is the number of words in the given word list words that are valid with respect to the puzzle puzzles[i].

Example :

Input: 
words = ["aaaa","asas","able","ability","actt","actor","access"], 
puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
Output: [1,1,3,2,4,0]
Explanation:
1 valid word for "aboveyz" : "aaaa" 
1 valid word for "abrodyz" : "aaaa"
3 valid words for "abslute" : "aaaa", "asas", "able"
2 valid words for "absoryz" : "aaaa", "asas"
4 valid words for "actresz" : "aaaa", "asas", "actt", "access"
There're no valid words for "gaswxyz" cause none of the words in the list contains letter 'g'.

Constraints:

1 <= words.length <= 10^5
4 <= words[i].length <= 50
1 <= puzzles.length <= 10^4
puzzles[i].length == 7
words[i][j], puzzles[i][j] are English lowercase letters.
Each puzzles[i] doesn’t contain repeated characters.

Solution: Subsets

Preprocessing:
Compress each word to a bit map, and compute the frequency of each bit map.
Since there are at most |words| bitmaps while its value ranging from 0 to 2^26, thus it’s better to use a hashtable instead of an array.

Query:
Use the same way to compress a puzzle into a bit map.
Try all subsets (at most 128) of the puzzle (the bit of the first character is be must), and check how many words match each subset.

words = [“aaaa”,”asas”,”able”,”ability”,”actt”,”actor”,”access”],
puzzle = “abslute”
bitmap(“aaaa”) = {0}
bitmap(“asas”) = {0, 18}
bitmap(“able”) = {0,1,4,11}
bitmap(“actt”) = {0, 2, 19}
bitmap(“actor”) = {0, 2, 14, 17, 19}
bitmap(“access”) = {0, 2, 4, 18}

bitmap(“abslute”) = {0, 1, 4, 11, 18, 19, 20}

Time complexity: O(sum(len(w_i)) + |puzzles|)
Space complexity: O(|words|)

C++

// Author: Huahua, 136 ms, 29.7 MB

class Solution {

public:

vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) {

vector<int> ans;

unordered_map<int, int> freq;

for (const string& word : words) {

int mask = 0;

for (char c : word)

mask |= 1 << (c - 'a');

++freq[mask];

}

for (const string& p : puzzles) {

int mask = 0;

for (char c : p)

mask |= 1 << (c - 'a');

int first = p[0] - 'a';

int curr = mask;

int total = 0;

while (curr) {

if ((curr >> first) & 1) {

auto it = freq.find(curr);

if (it != freq.end()) total += it->second;

}

curr = (curr - 1) & mask;

}

ans.push_back(total);

}

return ans;

}

};

花花酱 LeetCode 1177. Can Make Palindrome from Substring

By zxi on September 2, 2019

Given a string s, we make queries on substrings of s.

For each query queries[i] = [left, right, k], we may rearrange the substring s[left], ..., s[right], and then choose up to k of them to replace with any lowercase English letter.

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return an array answer[], where answer[i] is the result of the i-th query queries[i].

Note that: Each letter is counted individually for replacement so if for example s[left..right] = "aaa", and k = 2, we can only replace two of the letters. (Also, note that the initial string s is never modified by any query.)

Example :

Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
Explanation:
queries[0] : substring = "d", is palidrome.
queries[1] : substring = "bc", is not palidrome.
queries[2] : substring = "abcd", is not palidrome after replacing only 1 character.
queries[3] : substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries[4] : substring = "abcda", could be changed to "abcba" which is palidrome.

Constraints:

1 <= s.length, queries.length <= 10^5
0 <= queries[i][0] <= queries[i][1] < s.length
0 <= queries[i][2] <= s.length
s only contains lowercase English letters.

Solution: Prefix frequency

Compute the prefix frequency of each characters, then we can efficiently compute the frequency of each characters in the substring in O(1) time. Count the number odd frequency characters o, we can convert it to a palindrome if o / 2 <= k.

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

C++

// Author: Huahua

class Solution {

public:

vector<bool> canMakePaliQueries(string s, vector<vector<int>>& queries) {

vector<bool> ans;

int n = s.length();

vector<vector<int>> f(n + 1, vector<int>(26));

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

++f[i][s[i] - 'a'];

f[i + 1] = f[i];

}

for (const auto& q : queries) {

int l = q[0];

int r = q[1];

int k = q[2];

int o = 0;

for (int i = 0; i < 26; ++i) {

int c = f[r][i] - (l == 0 ? 0 : f[l - 1][i]);

o += c % 2;

}

ans.push_back(o / 2 <= k);

}

return ans;

}

};