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Check if it is possible to assign values such that all the given relations are satisfied
  • Difficulty Level : Medium
  • Last Updated : 18 Mar, 2021
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Given an array of strings arr[], where each arr[i] is of the form “i==j” or “i!=j”, where i and j are variables representing relationships among them, the task is to check if it is possible to assign values to the variables that satisfy all the relations. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: arr[] = {“i==j”, ” j!=i”}
Output: No
Explanation:
First relation holds true for values i = 1 and j = 1, but the second relation fails for the same set of values. Therefore, print No.

Input: arr[] = {“p==q”, “q==r”, “p==r”]
Output: Yes
Explanation:
The assignment of the value 1 in p, q and r satisfies all 3 relations. Therefore, print “Yes”.

Approach: The approach for solving the given problem is to use Union-Find Algorithm. The idea is to traverse the array of strings and go to every equality relation and perform union operation on the two variables. After the traversal, go to every non-equality relation in each string and check if the parent of the two variables is the same or not. Follow the steps below to solve the problem:



  1. Initialize an array, parent[] of size 26 with 0 and a variable answer as true to store the required result.
  2. Traverse the array parent[] using the variable i, and set parent[i] as i.
  3. Now, traverse each string S in the array of strings and if the value of S[i][1] is ‘=’, then perform union operation on S[i][0 – ‘a’] and S[i][3 – ‘a’].
  4. Again, traverse each the string S in the array of string and do the following:
    1. If the value of S[i][1] is equal to ‘!’, store the parent of S[i][0 – ‘a’] and S[i][3 – ‘a’] in X and Y respectively.
    2. If the value of X is equal to Y, set the answer as false.
  5. After the above steps, if the value of the answer is true then print “Yes” else print “No”.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find parent of each node
int find(int v, vector<int>& parent)
{
    // If parent[v] is not v, then
    // recursively call to find the
    // parent of parent[v]
    if (parent[v] != v)
        return parent[v] = find(parent[v],
                                parent);
 
    // Otherwise, return v
    return v;
}
 
// Function to perform union of the
// two variables
void unions(int a, int b,
            vector<int>& parent)
{
    // Stores the parent of a and b
    int x = find(a, parent);
    int y = find(b, parent);
 
    // If they are not equal, make
    // parent[x] = parent[y]
    if (x != y)
        parent[x] = parent[y];
}
 
// Funtion to check whether it is
// possible to assign values to
// variables to satisfy relations
bool equationsPossible(
    vector<string>& relations)
{
    // Initialize parent array as 0s
    vector<int> parent(26, 0);
 
    // Iterate in range [0, 25]
    // and make parent[i] = i
    for (int i = 0; i < 26; i++) {
        parent[i] = i;
    }
 
    // Store the size of the string
    int n = relations.size();
 
    // Traverse the string
    for (auto string : relations) {
 
        // Check if it is of the
        // form "i==j" or not
        if (string[1] == '=')
 
            // Take union of both
            // the variables
            unions(string[0] - 'a',
                   string[3] - 'a',
                   parent);
    }
 
    // Traverse the string
    for (auto string : relations) {
 
        // Check if it is of the
        // form "i!=j" or not
        if (string[1] == '!') {
 
            // Store the parent of
            // i and j
            int x = find(string[0] - 'a',
                         parent);
            int y = find(string[3] - 'a',
                         parent);
 
            // If they are equal,
            // then return false
            if (x == y)
                return false;
        }
    }
 
    return true;
}
 
// Driver Code
int main()
{
    vector<string> relations
        = { "i==j", "j!=i" };
 
    if (equationsPossible(relations)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to find parent of each node
static int find(int v, ArrayList<Integer> parent)
{
     
    // If parent[v] is not v, then
    // recursively call to find the
    // parent of parent[v]
    if (parent.get(v) != v)
    {
        parent.set(v, find(parent.get(v), parent));
        return parent.get(v);
    }
     
    // Otherwise, return v
    return v;
}
 
// Function to perform union of the
// two variables
static void unions(int a, int b,
                   ArrayList<Integer> parent)
{
    // Stores the parent of a and b
    int x = find(a, parent);
    int y = find(b, parent);
     
    // If they are not equal, make
    // parent[x] = parent[y]
    if (x != y)
        parent.set(x, parent.get(y));
}
 
// Funtion to check whether it is
// possible to assign values to
// variables to satisfy relations
static boolean equationsPossible(ArrayList<String> relations)
{
     
    // Initialize parent array as 0s
    ArrayList<Integer> parent = new ArrayList<Integer>();
    for(int i = 0; i < 26; i++)
        parent.add(0);
     
    // Iterate in range [0, 25]
    // and make parent[i] = i
    for(int i = 0; i < 26; i++)
    {
        parent.set(i, i);
    }
     
    // Store the size of the string
    int n = relations.size();
     
    // Traverse the string
    for(String str : relations)
    {
     
        // Check if it is of the
        // form "i==j" or not
        if (str.charAt(1) == '=')
         
        // Take union of both
        // the variables
        unions((int)str.charAt(0) - 97,
               (int)str.charAt(3) - 97,
               parent);
    }
     
    // Traverse the string
    for(String str : relations)
    {
     
        // Check if it is of the
        // form "i!=j" or not
        if (str.charAt(1) == '!')
        {
         
            // Store the parent of
            // i and j
            int x = find((int)str.charAt(0) - 97,
                         parent);
            int y = find((int)str.charAt(3) - 97,
                         parent);
             
            // If they are equal,
            // then return false
            if (x == y)
                return false;
        }
    }
    return true;
}
 
// Driver Code
public static void main (String[] args)
{
    ArrayList<String> relations = new ArrayList<String>(
        Arrays.asList("i==j", "j!=i"));
    if (equationsPossible(relations))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}
 
// This code is contributed by rag2127

Python3




# Python3 program for the above approach
 
# Function to find parent of each node
def find(v, parent):
   
    # If parent[v] is not v, then
    # recursively call to find the
    # parent of parent[v]
    if (parent[v] != v):
        parent[v] = find(parent[v], parent)
        return parent[v]
 
    # Otherwise, return v
    return v
 
# Function to perform union of the
# two variables
def unions(a, b, parent):
     
    # Stores the parent of a and b
    x = find(a, parent)
    y = find(b, parent)
 
    # If they are not equal, make
    # parent[x] = parent[y]
    if (x != y):
        parent[x] = parent[y]
 
# Funtion to check whether it is
# possible to assign values to
# variables to satisfy relations
def equationsPossible(relations):
     
    # Initialize parent array as 0s
    parent = [0]*(26)
 
    # Iterate in range [0, 25]
    # and make parent[i] = i
    for i in range(26):
        parent[i] = i
 
    # Store the size of the string
    n = len(relations)
 
    # Traverse the string
    for string in relations:
 
        # Check if it is of the
        # form "i==j" or not
        if (string[1] == '='):
 
            # Take union of both
            # the variables
            unions(ord(string[0]) - ord('a'),ord(string[3]) - ord('a'), parent)
 
    # Traverse the string
    for string in relations:
 
        # Check if it is of the
        # form "i!=j" or not
        if (string[1] == '!'):
 
            # Store the parent of
            # i and j
            x = find(ord(string[0]) - ord('a'), parent)
            y = find(ord(string[3]) - ord('a'), parent)
 
            # If they are equal,
            # then return false
            if (x == y):
                return False
    return True
 
# Driver Code
if __name__ == '__main__':
    relations = ["i==j", "j!=i"]
    if (equationsPossible(relations)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by mohit kumar 29.

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
  // Function to find parent of each node
  static int find(int v, List<int> parent)
  {
    // If parent[v] is not v, then
    // recursively call to find the
    // parent of parent[v]
    if (parent[v] != v)
      return parent[v] = find(parent[v],
                              parent);
 
    // Otherwise, return v
    return v;
  }
 
  // Function to perform union of the
  // two variables
  static void unions(int a, int b,
                     List<int> parent)
  {
    // Stores the parent of a and b
    int x = find(a, parent);
    int y = find(b, parent);
 
    // If they are not equal, make
    // parent[x] = parent[y]
    if (x != y)
      parent[x] = parent[y];
  }
 
  // Funtion to check whether it is
  // possible to assign values to
  // variables to satisfy relations
  static bool equationsPossible(List<string> relations)
  {
    // Initialize parent array as 0s
    List<int> parent = new List<int>();
    for(int i=0;i<26;i++)
      parent.Add(0);
 
    // Iterate in range [0, 25]
    // and make parent[i] = i
    for (int i = 0; i < 26; i++) {
      parent[i] = i;
    }
 
    // Store the size of the string
    int n = relations.Count;
 
    // Traverse the string
    foreach( string str in relations) {
 
      // Check if it is of the
      // form "i==j" or not
      if (str[1] == '=')
 
        // Take union of both
        // the variables
        unions((int)str[0] - 97,
               (int)str[3] - 97,
               parent);
    }
 
    // Traverse the string
    foreach (string str in relations) {
 
      // Check if it is of the
      // form "i!=j" or not
      if (str[1] == '!') {
 
        // Store the parent of
        // i and j
        int x = find((int)str[0] - 97,
                     parent);
        int y = find((int)str[3] - 97,
                     parent);
 
        // If they are equal,
        // then return false
        if (x == y)
          return false;
      }
    }
 
    return true;
  }
 
  // Driver Code
  public static void Main()
  {
    List<string> relations = new List<string>{ "i==j", "j!=i" };
 
    if (equationsPossible(relations)) {
      Console.WriteLine("Yes");
    }
    else {
      Console.WriteLine("No");
    }
 
  }
}
 
// This code is contributed by bgangwar59.
Output: 
No

 

Time Complexity: O(N*log M), where M is the number of unique variables in arr[]. 
Auxiliary Space: O(1) 
 

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