Leetcode #2307: Check for Contradictions in Equations

In this guide, we solve Leetcode #2307 Check for Contradictions in 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.

Leetcode

Problem Statement

You are given a 2D array of strings equations and an array of real numbers values, where equations[i] = [Ai, Bi] and values[i] means that Ai / Bi = values[i]. Determine if there exists a contradiction in the equations.

Quick Facts

Difficulty: Hard
Premium: Yes
Tags: Depth-First Search, Union Find, Graph, Array

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","c"],["a","c"]], values = [3,0.5,1.5]
Output: false
Explanation:
The given equations are: a / b = 3, b / c = 0.5, a / c = 1.5
There are no contradictions in the equations. One possible assignment to satisfy all equations is:
a = 3, b = 1 and c = 2.

Python Solution

class Solution:
    def checkContradictions(
        self, equations: List[List[str]], values: List[float]
    ) -> bool:
        def find(x: int) -> int:
            if p[x] != x:
                root = find(p[x])
                w[x] *= w[p[x]]
                p[x] = root
            return p[x]

        d = defaultdict(int)
        n = 0
        for e in equations:
            for s in e:
                if s not in d:
                    d[s] = n
                    n += 1
        p = list(range(n))
        w = [1.0] * n
        eps = 1e-5
        for (a, b), v in zip(equations, values):
            a, b = d[a], d[b]
            pa, pb = find(a), find(b)
            if pa != pb:
                p[pb] = pa
                w[pb] = v * w[a] / w[b]
            elif abs(v * w[a] - w[b]) >= eps:
                return True
        return False

Complexity

The time complexity is $O(n \times \log n)$ or $O(n \times \alpha(n))$ , and the space complexity is $O(n)$ . The space complexity is $O(n)$ .

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.

Python Solution

class Solution:
    def checkContradictions(
        self, equations: List[List[str]], values: List[float]
    ) -> bool:
        def find(x: int) -> int:
            if p[x] != x:
                root = find(p[x])
                w[x] *= w[p[x]]
                p[x] = root
            return p[x]

        d = defaultdict(int)
        n = 0
        for e in equations:
            for s in e:
                if s not in d:
                    d[s] = n
                    n += 1
        p = list(range(n))
        w = [1.0] * n
        eps = 1e-5
        for (a, b), v in zip(equations, values):
            a, b = d[a], d[b]
            pa, pb = find(a), find(b)
            if pa != pb:
                p[pb] = pa
                w[pb] = v * w[a] / w[b]
            elif abs(v * w[a] - w[b]) >= eps:
                return True
        return False

Leetcode #2307: Check for Contradictions in Equations

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview

Leetcode #2307: Check for Contradictions in Equations

Problem Statement

Quick Facts

Intuition

Approach

Example

Python Solution

Complexity

Edge Cases and Pitfalls

Summary

Ace your next coding interview