Merge pull request #16 from mrudul0026/main

LC-990 Solution

Author: Jayanti2919

LeetCode Questions/C++/1. Problems/990-SatisfiabilityOfEqualityEquations.cpp

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+/*
+Satisfiability of Equality Equations
+
+You are given an array of strings equations that represent relationships between 
+variables where each string equations[i] is of length 4 and takes one of 
+two different forms: "xi==yi" or "xi!=yi".Here, xi and yi are lowercase letters 
+(not necessarily different) that represent one-letter variable names.
+
+Return true if it is possible to assign integers to variable names so as 
+to satisfy all the given equations, or false otherwise.
+
+Input: equations = ["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.
+
+Input: equations = ["b==a","a==b"]
+Output: true
+Explanation: We could assign a = 1 and b = 1 to satisfy both equations.
+
+Input: equations = ["a==b","b==c","a==c"]
+Output: true
+
+Input: equations = ["a==b","b!=c","c==a"]
+Output: false
+
+Input: equations = ["c==c","b==d","x!=z"]
+Output: true
+
+Constraints:
+    1 <= equations.length <= 500
+    equations[i].length == 4
+    equations[i][0] is a lowercase letter.
+    equations[i][1] is either '=' or '!'.
+    equations[i][2] is '='.
+    equations[i][3] is a lowercase letter.
+
+*/
+
+/*
+Idea: 
+Since only atmost 2 variables are involved in an equation and the equation can only
+be == or !=, the only possible scenerio where unsatisfiability takes place is when
+two variables which were declared directly or indirectly equal is declared unequal.
+
+So we will first process all the equalities first. for every equality we will indicate it
+by an edge and any two variables directly/indirectly connected by edge means they are equal
+this can by captured using union operation because if 2 nodes have indirect/direct connection
+they will yield the same parent/component when doing find.
+
+Now we will process the inequalities. Here we will just check if find is same then return false
+because same component variables are equal but now contradiction is happening and hence unsatisfiable.
+
+If no contradiction after processing everything, then return true
+
+Time Complexity: O(N) where N is equations.length
+Space Complexity: O(26) = O(1)
+*/
+
+
+class Solution {
+public:
+    vector<int> p,sz;
+    int find(int a){
+        if(p[a]!=a)p[a]=find(p[a]);
+        return p[a];
+    }
+    void unite(int a ,int b){
+        a=find(a);
+        b=find(b);
+        p[a]=p[b];
+
+    }
+    bool equationsPossible(vector<string>& equations) {
+        p.resize(26),sz.resize(26,1);
+        for(int i=0;i<26;i++)p[i]=i;
+        for(auto &s: equations){
+            if(s[1]=='=')unite(s[0]-'a',s[3]-'a');
+        }
+        for(auto &s: equations){
+            if(s[1]!='='){
+                if(find(s[0]-'a')==find(s[3]-'a'))return false; 
+            }
+        }
+        return true;
+    }
+};