花花酱 LeetCode 736. Parse Lisp Expression

By zxi on December 1, 2017

Problem:

You are given a string expression representing a Lisp-like expression to return the integer value of.

The syntax for these expressions is given as follows.

An expression is either an integer, a let-expression, an add-expression, a mult-expression, or an assigned variable. Expressions always evaluate to a single integer.
(An integer could be positive or negative.)
A let-expression takes the form (let v1 e1 v2 e2 ... vn en expr), where let is always the string "let", then there are 1 or more pairs of alternating variables and expressions, meaning that the first variable v1is assigned the value of the expression e1, the second variable v2 is assigned the value of the expression e2, and so on sequentially; and then the value of this let-expression is the value of the expression expr.
An add-expression takes the form (add e1 e2) where add is always the string "add", there are always two expressions e1, e2, and this expression evaluates to the addition of the evaluation of e1 and the evaluation of e2.
A mult-expression takes the form (mult e1 e2) where mult is always the string "mult", there are always two expressions e1, e2, and this expression evaluates to the multiplication of the evaluation of e1 and the evaluation of e2.
For the purposes of this question, we will use a smaller subset of variable names. A variable starts with a lowercase letter, then zero or more lowercase letters or digits. Additionally for your convenience, the names “add”, “let”, or “mult” are protected and will never be used as variable names.
Finally, there is the concept of scope. When an expression of a variable name is evaluated, within the context of that evaluation, the innermost scope (in terms of parentheses) is checked first for the value of that variable, and then outer scopes are checked sequentially. It is guaranteed that every expression is legal. Please see the examples for more details on scope.

Evaluation Examples:

Input: (add 1 2)

Output: 3

Input: (mult 3 (add 2 3))

Output: 15

Input: (let x 2 (mult x 5))

Output: 10

Input: (let x 2 (mult x (let x 3 y 4 (add x y))))

Output: 14

Explanation: In the expression (add x y), when checking for the value of the variable x,

we check from the innermost scope to the outermost in the context of the variable we are trying to evaluate.

Since x = 3 is found first, the value of x is 3.

Input: (let x 3 x 2 x)

Output: 2

Explanation: Assignment in let statements is processed sequentially.

Input: (let x 1 y 2 x (add x y) (add x y))

Output: 5

Explanation: The first (add x y) evaluates as 3, and is assigned to x.

The second (add x y) evaluates as 3+2 = 5.

Input: (let x 2 (add (let x 3 (let x 4 x)) x))

Output: 6

Explanation: Even though (let x 4 x) has a deeper scope, it is outside the context

of the final x in the add-expression. That final x will equal 2.

Input: (let a1 3 b2 (add a1 1) b2)

Output 4

Explanation: Variable names can contain digits after the first character.

Note:

The given string expression is well formatted: There are no leading or trailing spaces, there is only a single space separating different components of the string, and no space between adjacent parentheses. The expression is guaranteed to be legal and evaluate to an integer.
The length of expression is at most 2000. (It is also non-empty, as that would not be a legal expression.)
The answer and all intermediate calculations of that answer are guaranteed to fit in a 32-bit integer.

Idea:

Recursive parsing

Time complexity: O(n^2) in worst case O(n) in practice

Space complexity: O(n)

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

int evaluate(string expression) {

scopes_.clear();

int pos = 0;

return eval(expression, pos);

}

private:

int eval(const string& s, int& pos) {

scopes_.push_front(unordered_map<string, int>());

int value = 0; // The return value of current expr

if (s[pos] == '(') ++pos;

// command, variable or number

const string token = getToken(s, pos);

if (token == "add") {

int v1 = eval(s, ++pos);

int v2 = eval(s, ++pos);

value = v1 + v2;

} else if (token == "mult") {

int v1 = eval(s, ++pos);

int v2 = eval(s, ++pos);

value = v1 * v2;

} else if (token == "let") {

string var;

// expecting " var1 exp1 var2 exp2 ... last_expr)"

while (s[pos] != ')') {

++pos;

// Must be last_expr

if (s[pos] == '(') {

value = eval(s, ++pos);

break;

}

// Get a token, could be "x" or "-12" for last_expr

var = getToken(s, pos);

// End of let, var is last_expr

if (s[pos] == ')') {

if (isalpha(var[0]))

value = getValue(var);

else

value = stoi(var);

break;

}

// x -12 -> set x to -12 and store in the current scope and take it as the current return value

value = scopes_.front()[var] = eval(s, ++pos);

}

} else if (isalpha(token[0])) {

value = getValue(token); // symbol

} else {

value = std::stoi(token); // number

}

if (s[pos] == ')') ++pos;

scopes_.pop_front();

return value;

}

int getValue(const string& symbol) {

for (const auto& scope : scopes_)

if (scope.count(symbol)) return scope.at(symbol);

return 0;

}

// Get a token from current pos.

// "let x" -> "let"

// "-12 (add x y)" -> "-12"

string getToken(const string& s, int& pos) {

string token;

while (pos < s.length()) {

if (s[pos] == ')' || s[pos] == ' ') break;

token += s[pos++];

}

return token;

}

deque<unordered_map<string, int>> scopes_;

};

花花酱 LeetCode Disjoint set / Union Find Forest SP1

By zxi on November 29, 2017

Disjoint-set/Union-find Forest

Find(x): find the root/cluster-id of x

Union(x, y): merge two clusters

Check whether two elements are in the same set or not in O(1)*.

Find: O(ɑ(n))* ≈ O(1)

Union: O(ɑ(n))* ≈ O(1)

Space: O(n)

Without optimization: Find: O(n)

Two key optimizations:

Path compression: make tree flat
Union by rank: merge low rank tree to high rank one

*: amortized

ɑ(.): inverse Ackermann function

Implementations:

C++

// Author: Huahua

class UnionFindSet {

public:

UnionFindSet(int n) {

ranks_ = vector<int>(n + 1, 0);

parents_ = vector<int>(n + 1, 0);

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

parents_[i] = i;

}

// Merge sets that contains u and v.

// Return true if merged, false if u and v are already in one set.

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

// Meger low rank tree into high rank tree

if (ranks_[pv] < ranks_[pu])

parents_[pv] = pu;

else if (ranks_[pu] < ranks_[pv])

parents_[pu] = pv;

else {

parents_[pv] = pu;

ranks_[pu] += 1;

}

return true;

}

// Get the root of u.

int Find(int u) {

// Compress the path during traversal

if (u != parents_[u])

parents_[u] = Find(parents_[u]);

return parents_[u];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

Java

// Author: Huahua

class UnionFindSet {

private int[] parents_;

private int[] ranks_;

public UnionFindSet(int n) {

parents_ = new int[n + 1];

ranks_ = new int[n + 1];

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

parents_[i] = i;

ranks_[i] = 1;

}

public boolean Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

if (ranks_[pv] > ranks_[pu])

parents_[pu] = pv;

else if (ranks_[pu] > ranks_[pv])

parents_[pv] = pu;

else {

parents_[pv] = pu;

ranks_[pu] += 1;

}

return true;

}

public int Find(int u) {

while (parents_[u] != u) {

parents_[u] = parents_[parents_[u]];

u = parents_[u];

}

return u;

}

Python

"""

Author: Huahua

"""

class UnionFindSet:

def __init__(self, n):

self._parents = [i for i in range(n + 1)]

self._ranks = [1 for i in range(n + 1)]

def find(self, u):

while u != self._parents[u]:

self._parents[u] = self._parents[self._parents[u]]

u = self._parents[u]

return u

def union(self, u, v):

pu, pv = self.find(u), self.find(v)

if pu == pv: return False

if self._ranks[pu] < self._ranks[pv]:

self._parents[pu] = pv

elif self._ranks[pu] > self._ranks[pv]:

self._parents[pv] = pu

else:

self._parents[pv] = pu

self._ranks[pu] += 1

return True

Union-Find Problems

References

花花酱 LeetCode 737. Sentence Similarity II

By zxi on November 28, 2017

Problem:

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.

For example, words1 = ["great", "acting", "skills"] and words2 = ["fine", "drama", "talent"] are similar, if the similar word pairs are pairs = [["great", "good"], ["fine", "good"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is transitive. For example, if “great” and “good” are similar, and “fine” and “good” are similar, then “great” and “fine” are similar.

Similarity is also symmetric. For example, “great” and “fine” being similar is the same as “fine” and “great” being similar.

Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.

Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].

Note:

The length of words1 and words2 will not exceed 1000.
The length of pairs will not exceed 2000.
The length of each pairs[i] will be 2.
The length of each words[i] and pairs[i][j] will be in the range [1, 20].

题目大意：

给你两个句子（由单词数组表示）和一些近义词对，问你这两个句子是否相似，即每组相对应的单词都要相似。

注意相似性可以传递，比如只给你”great”和”fine”相似、”fine”和”good”相似，能推断”great”和”good”也相似。

Idea:

DFS / Union Find

Solution1:

Time complexity: O(|Pairs| * |words1|)

Space complexity: O(|Pairs|)

C++ / DFS

// Author: Huahua

// Runtime: 346 ms

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

g_.clear();

for (const auto& p : pairs) {

g_[p.first].insert(p.second);

g_[p.second].insert(p.first);

}

unordered_set<string> visited;

for (int i = 0; i < words1.size(); ++i) {

visited.clear();

if (!dfs(words1[i], words2[i], visited)) return false;

}

return true;

}

private:

bool dfs(const string& src, const string& dst, unordered_set<string>& visited) {

if (src == dst) return true;

visited.insert(src);

for (const auto& next : g_[src]) {

if (visited.count(next)) continue;

if (dfs(next, dst, visited)) return true;

}

return false;

}

unordered_map<string, unordered_set<string>> g_;

};

Time complexity: O(|Pairs| + |words1|)

Space complexity: O(|Pairs|)

C++ / DFS Optimized

// Author: Huahua

// Runtime 229 ms

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

g_.clear();

ids_.clear();

for (const auto& p : pairs) {

g_[p.first].insert(p.second);

g_[p.second].insert(p.first);

}

int id = 0;

for (const auto& p : pairs) {

if(!ids_.count(p.first)) dfs(p.first, ++id);

if(!ids_.count(p.second)) dfs(p.second, ++id);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

auto it1 = ids_.find(words1[i]);

auto it2 = ids_.find(words2[i]);

if (it1 == ids_.end() || it2 == ids_.end()) return false;

if (it1->second != it2->second) return false;

}

return true;

}

private:

bool dfs(const string& curr, int id) {

ids_[curr] = id;

for (const auto& next : g_[curr]) {

if (ids_.count(next)) continue;

if (dfs(next, id)) return true;

}

return false;

}

unordered_map<string, int> ids_;

unordered_map<string, unordered_set<string>> g_;

};

Solution2:

Time complexity: O(|Pairs| + |words1|)

Space complexity: O(|Pairs|)

C++ / Union Find

// Author: Huahua

// Runtime: 219 ms

class UnionFindSet {

public:

bool Union(const string& word1, const string& word2) {

const string& p1 = Find(word1, true);

const string& p2 = Find(word2, true);

if (p1 == p2) return false;

parents_[p1] = p2;

return true;

}

const string& Find(const string& word, bool create = false) {

if (!parents_.count(word)) {

if (!create) return word;

return parents_[word] = word;

}

string w = word;

while (w != parents_[w]) {

parents_[w] = parents_[parents_[w]];

w = parents_[w];

}

return parents_[w];

}

private:

unordered_map<string, string> parents_;

};

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

UnionFindSet s;

for (const auto& pair : pairs)

s.Union(pair.first, pair.second);

for (int i = 0; i < words1.size(); ++i)

if (s.Find(words1[i]) != s.Find(words2[i])) return false;

return true;

}

};

C++ / Union Find, Optimized

// Author: Huahua

// Runtime: 175 ms (< 98.45%)

class UnionFindSet {

public:

UnionFindSet(int n) {

parents_ = vector<int>(n + 1, 0);

ranks_ = vector<int>(n + 1, 0);

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

parents_[i] = i;

}

bool Union(int u, int v) {

int pu = Find(u);

int pv = Find(v);

if (pu == pv) return false;

if (ranks_[pu] > ranks_[pv]) {

parents_[pv] = pu;

} else if (ranks_[pv] > ranks_[pu]) {

parents_[pu] = pv;

} else {

parents_[pu] = pv;

++ranks_[pv];

}

return true;

}

int Find(int id) {

if (id != parents_[id])

parents_[id] = Find(parents_[id]);

return parents_[id];

}

private:

vector<int> parents_;

vector<int> ranks_;

};

class Solution {

public:

bool areSentencesSimilarTwo(vector<string>& words1, vector<string>& words2, vector<pair<string, string>>& pairs) {

if (words1.size() != words2.size()) return false;

UnionFindSet s(pairs.size() * 2);

unordered_map<string, int> indies; // word to index

for (const auto& pair : pairs) {

int u = getIndex(pair.first, indies, true);

int v = getIndex(pair.second, indies, true);

s.Union(u, v);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

int u = getIndex(words1[i], indies);

int v = getIndex(words2[i], indies);

if (u < 0 || v < 0) return false;

if (s.Find(u) != s.Find(v)) return false;

}

return true;

}

private:

int getIndex(const string& word, unordered_map<string, int>& indies, bool create = false) {

auto it = indies.find(word);

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

if (!create) return INT_MIN;

int index = indies.size();

indies.emplace(word, index);

return index;

}

return it->second;

}

};

Related Problems:

花花酱 LeetCode 734. Sentence Similarity

By zxi on November 28, 2017

Problem:

Given two sentences words1, words2 (each represented as an array of strings), and a list of similar word pairs pairs, determine if two sentences are similar.

For example, “great acting skills” and “fine drama talent” are similar, if the similar word pairs are pairs = [["great", "fine"], ["acting","drama"], ["skills","talent"]].

Note that the similarity relation is not transitive. For example, if “great” and “fine” are similar, and “fine” and “good” are similar, “great” and “good” are not necessarily similar.

However, similarity is symmetric. For example, “great” and “fine” being similar is the same as “fine” and “great” being similar.

Also, a word is always similar with itself. For example, the sentences words1 = ["great"], words2 = ["great"], pairs = [] are similar, even though there are no specified similar word pairs.

Finally, sentences can only be similar if they have the same number of words. So a sentence like words1 = ["great"] can never be similar to words2 = ["doubleplus","good"].

Note:

The length of words1 and words2 will not exceed 1000.
The length of pairs will not exceed 2000.
The length of each pairs[i] will be 2.
The length of each words[i] and pairs[i][j] will be in the range [1, 20].

题目大意：

给你两个句子（由单词数组表示）和一些近义词对，问你这两个句子是否相似，即每组相对应的单词都要相似。

注意相似性不能传递，比如给只你”great”和”fine”相似、”fine”和”good”相似，这种情况下”great”和”good”不相似。

Idea:

Use hashtable to store mapping from word to its similar words.

Solution: HashTable

Time complexity: O(|pairs| + |words1|)

Space complexity: O(|pairs|)

C++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

bool areSentencesSimilar(vector<string>& words1, vector<string>& words2, vector<pair<string, string>> pairs) {

if (words1.size() != words2.size()) return false;

unordered_map<string, unordered_set<string>> similar_words;

for (const auto& pair : pairs) {

similar_words[pair.first].insert(pair.second);

similar_words[pair.second].insert(pair.first);

}

for (int i = 0; i < words1.size(); ++i) {

if (words1[i] == words2[i]) continue;

if (!similar_words[words1[i]].count(words2[i])) return false;

}

return true;

}

};

Java

// Author: Huahua

// Runtime: 10 ms

class Solution {

public boolean areSentencesSimilar(String[] words1, String[] words2, String[][] pairs) {

if (words1.length != words2.length) return false;

Map<String, Set<String>> similar_words = new HashMap<>();

for (String[] pair : pairs) {

if (!similar_words.containsKey(pair[0]))

similar_words.put(pair[0], new HashSet<>());

if (!similar_words.containsKey(pair[1]))

similar_words.put(pair[1], new HashSet<>());

similar_words.get(pair[0]).add(pair[1]);

similar_words.get(pair[1]).add(pair[0]);

}

for (int i = 0; i < words1.length; ++i) {

if (words1[i].equals(words2[i])) continue;

if (!similar_words.containsKey(words1[i])) return false;

if (!similar_words.get(words1[i]).contains(words2[i])) return false;

}

return true;

}

Python

"""

Author: Huahua

Runtime: 62 ms

"""

class Solution:

def areSentencesSimilar(self, words1, words2, pairs):

if len(words1) != len(words2): return False

similar_words = {}

for w1, w2 in pairs:

if not w1 in similar_words: similar_words[w1] = set()

if not w2 in similar_words: similar_words[w2] = set()

similar_words[w1].add(w2)

similar_words[w2].add(w1)

for w1, w2 in zip(words1, words2):

if w1 == w2: continue

if w1 not in similar_words: return False

if w2 not in similar_words[w1]: return False

return True

Related Problems:

[解题报告] LeetCode 737. Sentence Similarity II

花花酱 LeetCode 733. Flood Fill

By zxi on November 28, 2017

Problem:

An image is represented by a 2-D array of integers, each integer representing the pixel value of the image (from 0 to 65535).

Given a coordinate (sr, sc) representing the starting pixel (row and column) of the flood fill, and a pixel value newColor, “flood fill” the image.

To perform a “flood fill”, consider the starting pixel, plus any pixels connected 4-directionally to the starting pixel of the same color as the starting pixel, plus any pixels connected 4-directionally to those pixels (also with the same color as the starting pixel), and so on. Replace the color of all of the aforementioned pixels with the newColor.

At the end, return the modified image.

Example 1:

Input:

image = [[1,1,1],[1,1,0],[1,0,1]]

sr = 1, sc = 1, newColor = 2

Output: [[2,2,2],[2,2,0],[2,0,1]]

Explanation:

From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected

by a path of the same color as the starting pixel are colored with the new color.

Note the bottom corner is not colored 2, because it is not 4-directionally connected

to the starting pixel.

Note:

The length of image and image[0] will be in the range [1, 50].
The given starting pixel will satisfy 0 <= sr < image.length and 0 <= sc < image[0].length.
The value of each color in image[i][j] and newColor will be an integer in [0, 65535].

Idea

DFS

Solution:

C++ / DFS

Time complexity: O(mn)

Space complexity: O(1)

// Author: Huahua

// Runtime: 33 ms

class Solution {

public:

vector<vector<int>> floodFill(vector<vector<int>>& image,

int sr, int sc, int newColor) {

if (image[sr][sc] == newColor) return image;

int m = image.size();

int n = image[0].size();

floodFill(image, sc, sr, n, m, image[sr][sc], newColor);

return image;

}

private:

void floodFill(vector<vector<int>>& image,

int x, int y, int n, int m, int orgColor, int newColor) {

if (x < 0 || x >= n || y < 0 || y >= m) return;

if (image[y][x] != orgColor) return;

image[y][x] = newColor;

floodFill(image, x + 1, y, n, m, orgColor, newColor);

floodFill(image, x - 1, y, n, m, orgColor, newColor);

floodFill(image, x, y + 1, n, m, orgColor, newColor);

floodFill(image, x, y - 1, n, m, orgColor, newColor);

}

};

Related Problems:

Huahua's Tech Road

花花酱 LeetCode 736. Parse Lisp Expression

花花酱 LeetCode Disjoint set / Union Find Forest SP1

Implementations:

C++

Java

Python

Union-Find Problems

References

花花酱 LeetCode 737. Sentence Similarity II

花花酱 LeetCode 734. Sentence Similarity

Solution: HashTable

C++

Java

Python

花花酱 LeetCode 733. Flood Fill