Leetcode #990: Satisfiability of Equality Equations
In this guide, we solve Leetcode #990 Satisfiability of Equality Equations in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Union Find, Graph, Array, String
Intuition
The data forms a graph, so we should explore nodes and edges systematically.
A traversal ensures we visit each node once while maintaining the needed state.
Approach
Build an adjacency list and traverse with BFS or DFS.
Aggregate results as you visit nodes.
Steps:
- Build the graph.
- Traverse with BFS/DFS.
- Accumulate the required output.
Example
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.
Python Solution
class Solution:
def equationsPossible(self, equations: List[str]) -> bool:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]
p = list(range(26))
for e in equations:
a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
if e[1] == '=':
p[find(a)] = find(b)
for e in equations:
a, b = ord(e[0]) - ord('a'), ord(e[-1]) - ord('a')
if e[1] == '!' and find(a) == find(b):
return False
return True
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.