Question

2307. Check for Contradictions in Equations

中文文档

Description

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. Return true_ if there is a contradiction, or _false_ otherwise_.

Note:

• When checking if two numbers are equal, check that their absolute difference is less than 10-5.
• The testcases are generated such that there are no cases targeting precision, i.e. using double is enough to solve the problem.

 

Example 1:

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.

Example 2:

Input: equations = [["le","et"],["le","code"],["code","et"]], values = [2,5,0.5]
Output: true
Explanation:
The given equations are: le / et = 2, le / code = 5, code / et = 0.5
Based on the first two equations, we get code / et = 0.4.
Since the third equation is code / et = 0.5, we get a contradiction.

 

Constraints:

• 1 <= equations.length <= 100
• equations[i].length == 2
• 1 <= Ai.length, Bi.length <= 5
• Ai, Bi consist of lowercase English letters.
• equations.length == values.length
• 0.0 < values[i] <= 10.0
• values[i] has a maximum of 2 decimal places.

Solutions

Solution 1: Weighted Union-Find

First, we convert the strings into integers starting from $0$. Then, we traverse all the equations, map the two strings in each equation to the corresponding integers $a$ and $b$. If these two integers are not in the same set, we merge them into the same set and record the weights of the two integers, which is the ratio of $a$ to $b$. If these two integers are in the same set, we check whether their weights satisfy the equation. If not, we return true.

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

Similar problems:

• 399. Evaluate Division

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

class Solution {
  private int[] p;
  private double[] w;

  public boolean checkContradictions(List<List<String>> equations, double[] values) {
    Map<String, Integer> d = new HashMap<>();
    int n = 0;
    for (var e : equations) {
      for (var s : e) {
        if (!d.containsKey(s)) {
          d.put(s, n++);
        }
      }
    }
    p = new int[n];
    w = new double[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
      w[i] = 1.0;
    }
    final double eps = 1e-5;
    for (int i = 0; i < equations.size(); ++i) {
      int a = d.get(equations.get(i).get(0)), b = d.get(equations.get(i).get(1));
      int pa = find(a), pb = find(b);
      double v = values[i];
      if (pa != pb) {
        p[pb] = pa;
        w[pb] = v * w[a] / w[b];
      } else if (Math.abs(v * w[a] - w[b]) >= eps) {
        return true;
      }
    }
    return false;
  }

  private int find(int x) {
    if (p[x] != x) {
      int root = find(p[x]);
      w[x] *= w[p[x]];
      p[x] = root;
    }
    return p[x];
  }
}

class Solution {
public:
  bool checkContradictions(vector<vector<string>>& equations, vector<double>& values) {
    unordered_map<string, int> d;
    int n = 0;
    for (auto& e : equations) {
      for (auto& s : e) {
        if (!d.count(s)) {
          d[s] = n++;
        }
      }
    }
    vector<int> p(n);
    iota(p.begin(), p.end(), 0);
    vector<double> w(n, 1.0);
    function<int(int)> find = [&](int x) -> int {
      if (p[x] != x) {
        int root = find(p[x]);
        w[x] *= w[p[x]];
        p[x] = root;
      }
      return p[x];
    };
    for (int i = 0; i < equations.size(); ++i) {
      int a = d[equations[i][0]], b = d[equations[i][1]];
      double v = values[i];
      int pa = find(a), pb = find(b);
      if (pa != pb) {
        p[pb] = pa;
        w[pb] = v * w[a] / w[b];
      } else if (fabs(v * w[a] - w[b]) >= 1e-5) {
        return true;
      }
    }
    return false;
  }
};

func checkContradictions(equations [][]string, values []float64) bool {
  d := make(map[string]int)
  n := 0

  for _, e := range equations {
    for _, s := range e {
      if _, ok := d[s]; !ok {
        d[s] = n
        n++
      }
    }
  }

  p := make([]int, n)
  for i := range p {
    p[i] = i
  }

  w := make([]float64, n)
  for i := range w {
    w[i] = 1.0
  }

  var find func(int) int
  find = func(x int) int {
    if p[x] != x {
      root := find(p[x])
      w[x] *= w[p[x]]
      p[x] = root
    }
    return p[x]
  }
  for i, e := range equations {
    a, b := d[e[0]], d[e[1]]
    v := values[i]

    pa, pb := find(a), find(b)
    if pa != pb {
      p[pb] = pa
      w[pb] = v * w[a] / w[b]
    } else if v*w[a]-w[b] >= 1e-5 || w[b]-v*w[a] >= 1e-5 {
      return true
    }
  }

  return false
}

function checkContradictions(equations: string[][], values: number[]): boolean {
  const d: { [key: string]: number } = {};
  let n = 0;

  for (const e of equations) {
    for (const s of e) {
      if (!(s in d)) {
        d[s] = n;
        n++;
      }
    }
  }

  const p: number[] = Array.from({ length: n }, (_, i) => i);
  const w: number[] = Array.from({ length: n }, () => 1.0);

  const find = (x: number): number => {
    if (p[x] !== x) {
      const root = find(p[x]);
      w[x] *= w[p[x]];
      p[x] = root;
    }
    return p[x];
  };

  for (let i = 0; i < equations.length; i++) {
    const a = d[equations[i][0]];
    const b = d[equations[i][1]];
    const v = values[i];

    const pa = find(a);
    const pb = find(b);

    if (pa !== pb) {
      p[pb] = pa;
      w[pb] = (v * w[a]) / w[b];
    } else if (Math.abs(v * w[a] - w[b]) >= 1e-5) {
      return true;
    }
  }

  return false;
}