花花酱 LeetCode 11. Container With Most Water

By zxi on December 3, 2017

Problem:

Given n non-negative integers a₁, a₂, …, a_n, where each represents a point at coordinate (i, a_i). n vertical lines are drawn such that the two endpoints of line i is at (i, a_i) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

Note: You may not slant the container and n is at least 2.

Examples:

input: [1 3 2 4]
output: 6
explanation: use 3, 4, we have the following, which contains (4th-2nd) * min(3, 4) = 2 * 3 = 6 unit of water.

|~~|

Idea:

Two pointers

Time complexity: O(n)

Space complexity: O(1)

Solution:

C++ two pointers

// Author: Huahua

// Runtime: 19 ms

class Solution {

public:

int maxArea(const vector<int>& height) {

int ans = 0;

int l = 0;

int r = height.size() - 1;

while (l < r) {

int h = min(height[l], height[r]);

ans = max(ans, h * (r - l));

if (height[l] < height[r])

++l;

else

--r;

}

return ans;

}

};

Java

// Author: Huahua

// Runtime: 13 ms

class Solution {

public int maxArea(int[] height) {

int l = 0;

int r = height.length - 1;

int ans = 0;

while (l < r) {

int h = Math.min(height[l], height[r]);

ans = Math.max(ans, h * (r - l));

if (height[l] < height[r])

++l;

else

--r;

}

return ans;

}

花花酱 LeetCode 399. Evaluate Division

By zxi on December 1, 2017

题目大意：给你一些含有变量名的分式的值，让你计算另外一些分式的值，如果不能计算返回-1。

Problem:

Equations are given in the format A / B = k, where A and B are variables represented as strings, and k is a real number (floating point number). Given some queries, return the answers. If the answer does not exist, return -1.0.

Example:
Given a / b = 2.0, b / c = 3.0.
queries are: a / c = ?, b / a = ?, a / e = ?, a / a = ?, x / x = ? .
return [6.0, 0.5, -1.0, 1.0, -1.0 ].

The input is: vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries , where equations.size() == values.size(), and the values are positive. This represents the equations. Return vector<double>.

According to the example above:

equations = [ ["a", "b"], ["b", "c"] ],

values = [2.0, 3.0],

queries = [ ["a", "c"], ["b", "a"], ["a", "e"], ["a", "a"], ["x", "x"] ].

The input is always valid. You may assume that evaluating the queries will result in no division by zero and there is no contradiction.

Solution 1: DFS

C++

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

vector<double> calcEquation(vector<pair<string, string>> equations, vector<double>& values, vector<pair<string, string>> queries) {

// g[A][B] = k -> A / B = k

unordered_map<string, unordered_map<string, double>> g;

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

const string& A = equations[i].first;

const string& B = equations[i].second;

const double k = values[i];

g[A][B] = k;

g[B][A] = 1.0 / k;

}

vector<double> ans;

for (const auto& pair : queries) {

const string& X = pair.first;

const string& Y = pair.second;

if (!g.count(X) || !g.count(Y)) {

ans.push_back(-1.0);

continue;

}

unordered_set<string> visited;

ans.push_back(divide(X, Y, g, visited));

}

return ans;

}

private:

// get result of A / B

double divide(const string& A, const string& B,

unordered_map<string, unordered_map<string, double>>& g,

unordered_set<string>& visited) {

if (A == B) return 1.0;

visited.insert(A);

for (const auto& pair : g[A]) {

const string& C = pair.first;

if (visited.count(C)) continue;

double d = divide(C, B, g, visited); // d = C / B

// A / B = C / B * A / C

if (d > 0) return d * g[A][C];

}

return -1.0;

}

};

Java

// Updated: 2020/9/27

// Author: Huahua

// Running time: 1 ms

class Solution {

private Map<String, HashMap<String, Double>> g = new HashMap<>();

public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {

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

String x = equations.get(i).get(0);

String y = equations.get(i).get(1);

double k = values[i];

g.computeIfAbsent(x, l -> new HashMap<String, Double>()).put(y, k);

g.computeIfAbsent(y, l -> new HashMap<String, Double>()).put(x, 1.0 / k);

}

double[] ans = new double[queries.size()];

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

String x = queries.get(i).get(0);

String y = queries.get(i).get(1);

if (!g.containsKey(x) || !g.containsKey(y)) {

ans[i] = -1.0;

} else {

ans[i] = divide(x, y, new HashSet<String>());

}

return ans;

}

private double divide(String x, String y, Set<String> visited) {

if (x.equals(y)) return 1.0;

visited.add(x);

if (!g.containsKey(x)) return -1.0;

for (String n : g.get(x).keySet()) {

if (visited.contains(n)) continue;

visited.add(n);

double d = divide(n, y, visited);

if (d > 0) return d * g.get(x).get(n);

}

return -1.0;

}

Python3

"""

Author: Huahua

Running time: 32 ms (beats 100%)

"""

class Solution:

def calcEquation(self, equations, values, queries):

def divide(x, y, visited):

if x == y: return 1.0

visited.add(x)

for n in g[x]:

if n in visited: continue

visited.add(n)

d = divide(n, y, visited)

if d > 0: return d * g[x][n]

return -1.0

g = collections.defaultdict(dict)

for (x, y), v in zip(equations, values):

g[x][y] = v

g[y][x] = 1.0 / v

ans = [divide(x, y, set()) if x in g and y in g else -1 for x, y in queries]

return ans

Solution 2: Union Find

C++

// Author: Huahua

// Runtime: 3 ms

class Solution {

public:

vector<double> calcEquation(const vector<pair<string, string>>& equations, vector<double>& values, const vector<pair<string, string>>& queries) {

// parents["A"] = {"B", 2.0} -> A = 2.0 * B

// parents["B"] = {"C", 3.0} -> B = 3.0 * C

unordered_map<string, pair<string, double>> parents;

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

const string& A = equations[i].first;

const string& B = equations[i].second;

const double k = values[i];

// Neighter is in the forrest

if (!parents.count(A) && !parents.count(B)) {

parents[A] = {B, k};

parents[B] = {B, 1.0};

} else if (!parents.count(A)) {

parents[A] = {B, k};

} else if (!parents.count(B)) {

parents[B] = {A, 1.0 / k};

} else {

auto& rA = find(A, parents);

auto& rB = find(B, parents);

parents[rA.first] = {rB.first, k / rA.second * rB.second};

}

vector<double> ans;

for (const auto& pair : queries) {

const string& X = pair.first;

const string& Y = pair.second;

if (!parents.count(X) || !parents.count(Y)) {

ans.push_back(-1.0);

continue;

}

auto& rX = find(X, parents); // {rX, X / rX}

auto& rY = find(Y, parents); // {rY, Y / rY}

if (rX.first != rY.first)

ans.push_back(-1.0);

else // X / Y = (X / rX / (Y / rY))

ans.push_back(rX.second / rY.second);

}

return ans;

}

private:

pair<string, double>& find(const string& C, unordered_map<string, pair<string, double>>& parents) {

if (C != parents[C].first) {

const auto& p = find(parents[C].first, parents);

parents[C].first = p.first;

parents[C].second *= p.second;

}

return parents[C];

}

};

Java

// Updated: 9/27/2020

// Author: Huahua

// Running time: 3 ms

class Solution {

class Node {

public String parent;

public double ratio;

public Node(String parent, double ratio) {

this.parent = parent;

this.ratio = ratio;

}

class UnionFindSet {

private Map<String, Node> parents = new HashMap<>();

public Node find(String s) {

if (!parents.containsKey(s)) return null;

Node n = parents.get(s);

if (!n.parent.equals(s)) {

Node p = find(n.parent);

n.parent = p.parent;

n.ratio *= p.ratio;

}

return n;

}

public void union(String A, String B, double ratio) {

boolean hasA = parents.containsKey(A);

boolean hasB = parents.containsKey(B);

if (!hasA && !hasB) {

parents.put(A, new Node(B, ratio));

parents.put(B, new Node(B, 1.0));

} else if (!hasA) {

parents.put(A, new Node(B, ratio));

} else if (!hasB) {

parents.put(B, new Node(A, 1.0 / ratio));

} else {

Node rA = find(A);

Node rB = find(B);

parents.put(rA.parent,

new Node(rB.parent, ratio / rA.ratio * rB.ratio));

}

public double[] calcEquation(List<List<String>> equations, double[] values, List<List<String>> queries) {

UnionFindSet u = new UnionFindSet();

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

u.union(equations.get(i).get(0), equations.get(i).get(1), values[i]);

double[] ans = new double[queries.size()];

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

Node rx = u.find(queries.get(i).get(0));

Node ry = u.find(queries.get(i).get(1));

if (rx == null || ry == null || !rx.parent.equals(ry.parent))

ans[i] = -1.0;

else

ans[i] = rx.ratio / ry.ratio;

}

return ans;

}

Python3

"""

Author: Huahua

Running time: 32 ms (beats 100%)

"""

class Solution:

def calcEquation(self, equations, values, queries):

def find(x):

if x != U[x][0]:

px, pv = find(U[x][0])

U[x] = (px, U[x][1] * pv)

return U[x]

def divide(x, y):

rx, vx = find(x)

ry, vy = find(y)

if rx != ry: return -1.0

return vx / vy

U = {}

for (x, y), v in zip(equations, values):

if x not in U and y not in U:

U[x] = (y, v)

U[y] = (y, 1.0)

elif x not in U:

U[x] = (y, v)

elif y not in U:

U[y] = (x, 1.0 / v)

else:

rx, vx = find(x)

ry, vy = find(y)

U[rx] = (ry, v / vx * vy)

ans = [divide(x, y) if x in U and y in U else -1 for x, y in queries]

return ans

Related Problems:

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;

}

};

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References

花花酱 LeetCode 737. Sentence Similarity II