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SatisfiabilityOfEqualityEquations.cpp
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SatisfiabilityOfEqualityEquations.cpp
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// Source : https://leetcode.com/problems/satisfiability-of-equality-equations/
// Author : Hao Chen
// Date : 2019-03-25
/*****************************************************************************************************
*
* Given an array equations of strings that represent relationships between variables, each string
* equations[i] has length 4 and takes one of two different forms: "a==b" or "a!=b". Here, a and b
* are lowercase letters (not necessarily different) that represent one-letter variable names.
*
* Return true if and only if it is possible to assign integers to variable names so as to satisfy all
* the given equations.
*
* Example 1:
*
* Input: ["a==b","b!=a"]
* Output: false
* Explanation: If we assign say, a = 1 and b = 1, then the first equation is satisfied, but not the
* second. There is no way to assign the variables to satisfy both equations.
*
* Example 2:
*
* Input: ["b==a","a==b"]
* Output: true
* Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
*
* Example 3:
*
* Input: ["a==b","b==c","a==c"]
* Output: true
*
* Example 4:
*
* Input: ["a==b","b!=c","c==a"]
* Output: false
*
* Example 5:
*
* Input: ["c==c","b==d","x!=z"]
* Output: true
*
* Note:
*
* 1 <= equations.length <= 500
* equations[i].length == 4
* equations[i][0] and equations[i][3] are lowercase letters
* equations[i][1] is either '=' or '!'
* equations[i][2] is '='
*
******************************************************************************************************/
class Solution {
private:
map<char, char> eq_map; // the table store the equivalent relationship
void init() {
for (char ch='a'; ch<='z'; ch++) {
eq_map[ch] = ch;
}
}
public:
Solution() {
init();
}
// find the tail node,
// if a == b == c == d, find(a) would return d;
// find(b) would return d;
char find(char ch) {
while(ch != eq_map[ch]) {
ch = eq_map[ch];
}
return ch;
}
//join x equivalent chain with y equivalent chain
void join(char x, char y) {
char tx = find(x);
char ty = find(y);
if (tx != ty) {
eq_map[tx] = ty;
}
}
bool equationsPossible(vector<string>& equations) {
for (auto e : equations ) {
if (e[1] == '=' && e[0] != e[3]) { // equal
join(e[0], e[3]);
}
}
for (auto e : equations) {
if (e[1] == '!' && (e[0] == e[3] || find(e[0]) == find(e[3]) ) ) {
return false;
}
}
return true;
}
};