Check if the given Binary Expressions are valid

0 0 8 minutes read

Given n expressions of the type x = y and x != y where 1 ? x, y ? n, the task is to check whether the integers from 1 to n can be assigned to x and y such that all the equations are satisfied.

Examples:

Input: x[] = {1, 2, 3}, op[] = {“=”, “=”, “!=”}, y[] = {2, 3, 1}
Output: Invalid
If 1 = 2 and 2 = 3 then 3 must be equal to 1.

Input: x[] = {1, 2}, op[] = {“=”, “=”}, y[] = {2, 3}
Output: Valid

Approach: The idea is to use union-find. For each statement check if there exists “=” sign then find the parent values for the two variables and union them using the rank method. Now once the union has been done for all the variables having “=” operator between them, start checking for “!=” operator, if this operator exists between any two variables whose parents are the same then the expressions are invalid, else they are valid.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the parent of an integer
int findParent(int i, vector<int> parent)
{
    if (parent[i] == i)
        return i;
    return findParent(parent[i], parent);
}
 
// Find union for both the integers x and y
// using rank method
void findUnion(int x, int y, vector<int>& parent,
               vector<int>& rank)
{
    int xroot = findParent(x, parent);
    int yroot = findParent(y, parent);
 
    // Union using rank method
    if (xroot != yroot) {
        if (rank[xroot] > rank[yroot]) {
            parent[y] = x;
            rank[xroot]++;
        }
        else if (rank[x] < rank[y]) {
            parent[x] = y;
            rank[yroot]++;
        }
        else {
            parent[y] = x;
            rank[xroot]++;
        }
    }
}
 
// Function that returns true if
// the expression is invalid
bool isInvalid(vector<int> u, vector<int> v,
               vector<string> op, int n)
{
    // Vector to store parent values
    // of each integer
    vector<int> parent;
 
    // Vector to store rank
    // of each integer
    vector<int> rank;
 
    parent.push_back(-1);
    rank.push_back(-1);
 
    // Initialize parent values for
    // each of the integers
    for (int i = 1; i <= n; i++)
        parent.push_back(i);
 
    // Initialize rank values for
    // each of the integers
    for (int i = 1; i <= n; i++)
        rank.push_back(0);
 
    // Check for = operator and find union
    // for them
    for (int i = 0; i < n; i++)
        if (op[i] == "=")
            findUnion(u[i], v[i], parent, rank);
 
    // Check for != operator
    for (int i = 0; i < n; i++) {
        if (op[i] == "!=") {
            // If the expression is invalid
            if (findParent(u[i], parent)
                == findParent(v[i], parent))
                return true;
        }
    }
 
    // Expression is valid
    return false;
}
 
// Driver code
int main()
{
    vector<int> u;
    vector<int> v;
    vector<string> op;
 
    // Store the first integer
    u.push_back(1);
    u.push_back(2);
    u.push_back(3);
 
    // Store the second integer
    v.push_back(2);
    v.push_back(3);
    v.push_back(1);
 
    // Store the operators
    op.push_back("=");
    op.push_back("=");
    op.push_back("!=");
 
    // Number of expressions
    int n = u.size();
    if (isInvalid(u, v, op, n))
        cout << "Invalid";
    else
        cout << "Valid";
 
    return 0;
}

Java

// Java implementation of 
// the above approach 
import java.util.*;
class GFG{
   
// Function to return the parent 
// of an integer
public static int findParent(int i, 
                             Vector parent)
{
  if ((int)parent.get(i) == i)
    return i;
 
  return findParent((int)parent.get(i), 
                         parent);
}
       
// Find union for both the integers x and y
// using rank method
public static void findUnion(int x, int y, 
                             Vector parent,
                             Vector rank)
{
  int xroot = findParent(x, parent);
  int yroot = findParent(y, parent);
 
  // Union using rank method
  if (xroot != yroot)
  {
    if ((int)rank.get(xroot) > 
        (int)rank.get(yroot)) 
    {
      parent.set(y,x);
      rank.set(xroot, 
               (int)rank.get(xroot) + 1); 
    }
    else if ((int)rank.get(x) < 
             (int)rank.get(y))
    {
      parent.set(x,y);
      rank.set(yroot, 
               (int)rank.get(yroot) + 1); 
    }
    else
    {
      parent.set(y,x);
      rank.set(xroot, 
               (int)rank.get(xroot) + 1);
    }
  }
}
       
// Function that returns true if
// the expression is invalid
public static boolean isInvalid(Vector u, Vector v,
                                Vector op, int n)
{
  // To store parent values
  // of each integer
  Vector parent = new Vector();
 
  // To store rank
  // of each integer
  Vector rank = new Vector();
 
  parent.add(-1);
  rank.add(-1);
 
  // Initialize parent values for
  // each of the integers
  for(int i = 1; i <= n; i++)
    parent.add(i);
 
  // Initialize rank values for
  // each of the integers
  for(int i = 1; i <= n; i++)
    rank.add(0);
 
  // Check for = operator and find union
  // for them
  for(int i = 0; i < n; i++)
    if ((String)op.get(i) == "=")
      findUnion((int)u.get(i), 
                (int)v.get(i), 
                 parent, rank);
 
  // Check for != operator
  for(int i = 0; i < n; i++)
  {
    if ((String)op.get(i) == "!=")
    {
      // If the expression is invalid
      if (findParent((int)u.get(i), parent) == 
          findParent((int)v.get(i), parent))
        return true;
    }
  }
 
  // Expression is valid
  return false;
}
 
// Driver code
public static void main(String[] args) 
{
  Vector u = new Vector();
  Vector v = new Vector();
  Vector op = new Vector();
 
  // Store the first integer
  u.add(1);
  u.add(2);
  u.add(3);
 
  // Store the second integer
  v.add(2);
  v.add(3);
  v.add(1);
 
  // Store the operators
  op.add("=");
  op.add("=");
  op.add("!=");
 
  // Number of expressions
  int n = u.size();
 
  if (isInvalid(u, v, op, n))
    System.out.print("Invalid");
  else
    System.out.print("Valid");
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 implementation of the approach 
 
# Function to return the parent of an integer 
def findParent(i, parent): 
 
    if parent[i] == i:
        return i 
    return findParent(parent[i], parent) 
 
# Find union for both the integers 
# x and y using rank method 
def findUnion(x, y, parent, rank): 
 
    xroot = findParent(x, parent) 
    yroot = findParent(y, parent) 
 
    # Union using rank method 
    if xroot != yroot: 
        if rank[xroot] > rank[yroot]: 
            parent[y] = x 
            rank[xroot] += 1
         
        elif rank[x] < rank[y]: 
            parent[x] = y 
            rank[yroot] += 1
         
        else:
            parent[y] = x 
            rank[xroot] += 1
 
# Function that returns true if 
# the expression is invalid 
def isInvalid(u, v, op, n): 
 
    # Vector to store parent values 
    # of each integer 
    parent = []
 
    # Vector to store rank 
    # of each integer 
    rank = [] 
 
    parent.append(-1) 
    rank.append(-1) 
 
    # Initialize parent values for 
    # each of the integers 
    for i in range(1, n + 1): 
        parent.append(i) 
 
    # Initialize rank values for 
    # each of the integers 
    for i in range(1, n + 1): 
        rank.append(0) 
 
    # Check for = operator and 
    # find union for them 
    for i in range(0, n): 
        if op[i] == "=": 
            findUnion(u[i], v[i], parent, rank) 
 
    # Check for != operator 
    for i in range(0, n): 
        if op[i] == "!=": 
             
            # If the expression is invalid 
            if (findParent(u[i], parent) ==
                findParent(v[i], parent)): 
                return True
         
    # Expression is valid 
    return False
 
# Driver code 
if __name__ == "__main__": 
 
    u = [1, 2, 3] 
    v = [2, 3, 1]
    op = ["=", "=", "!="] 
 
    # Number of expressions 
    n = len(u) 
    if isInvalid(u, v, op, n): 
        print("Invalid") 
    else:
        print("Valid") 
 
# This code is contributed by Rituraj Jain

C#

// C# implementation of the approach 
using System; 
using System.Collections;
using System.Collections.Generic;  
 
class GFG{ 
   
// Function to return the parent of an integer
static int findParent(int i, ArrayList parent)
{
    if ((int)parent[i] == i)
        return i;
   
    return findParent((int)parent[i], parent);
}
  
// Find union for both the integers x and y
// using rank method
static void findUnion(int x, int y, 
                      ArrayList parent,
                      ArrayList rank)
{
    int xroot = findParent(x, parent);
    int yroot = findParent(y, parent);
  
    // Union using rank method
    if (xroot != yroot)
    {
        if ((int)rank[xroot] > (int)rank[yroot]) 
        {
            parent[y] = x;
            rank[xroot] = (int)rank[xroot] + 1;
        }
        else if ((int)rank[x] < (int)rank[y])
        {
            parent[x] = y;
            rank[yroot] = (int)rank[yroot] + 1;
        }
        else
        {
            parent[y] = x;
            rank[xroot] = (int)rank[xroot] + 1;
        }
    }
}
  
// Function that returns true if
// the expression is invalid
static bool isInvalid(ArrayList u, ArrayList v,
                      ArrayList op, int n)
{
   
    // To store parent values
    // of each integer
    ArrayList parent = new ArrayList();
  
    // To store rank
    // of each integer
    ArrayList rank = new ArrayList();
  
    parent.Add(-1);
    rank.Add(-1);
  
    // Initialize parent values for
    // each of the integers
    for(int i = 1; i <= n; i++)
        parent.Add(i);
  
    // Initialize rank values for
    // each of the integers
    for(int i = 1; i <= n; i++)
        rank.Add(0);
  
    // Check for = operator and find union
    // for them
    for(int i = 0; i < n; i++)
        if ((string)op[i] == "=")
            findUnion((int)u[i], 
                      (int)v[i],
                      parent, rank);
  
    // Check for != operator
    for(int i = 0; i < n; i++)
    {
        if ((string)op[i] == "!=")
        {
           
            // If the expression is invalid
            if (findParent((int)u[i], parent) == 
                findParent((int)v[i], parent))
                return true;
        }
    }
  
    // Expression is valid
    return false;
}
 
// Driver Code 
public static void Main(string[] args) 
{ 
    ArrayList u = new ArrayList();
    ArrayList v = new ArrayList();
    ArrayList op = new ArrayList();
  
    // Store the first integer
    u.Add(1);
    u.Add(2);
    u.Add(3);
  
    // Store the second integer
    v.Add(2);
    v.Add(3);
    v.Add(1);
  
    // Store the operators
    op.Add("=");
    op.Add("=");
    op.Add("!=");
  
    // Number of expressions
    int n = u.Count;
   
    if (isInvalid(u, v, op, n))
        Console.Write("Invalid");
    else
        Console.Write("Valid");
} 
} 
 
// This code is contributed by rutvik_56

Javascript

<script>
// Javascript implementation of the approach
 
 
// Function to return the parent of an integer
function findParent(i, parent)
{
    if (parent[i] == i)
        return i;
    return findParent(parent[i], parent);
}
 
// Find union for both the integers x and y
// using rank method
function findUnion(x, y, parent, rank) {
    let xroot = findParent(x, parent);
    let yroot = findParent(y, parent);
 
    // Union using rank method
    if (xroot != yroot) {
        if (rank[xroot] > rank[yroot]) {
            parent[y] = x;
            rank[xroot]++;
        }
        else if (rank[x] < rank[y]) {
            parent[x] = y;
            rank[yroot]++;
        }
        else {
            parent[y] = x;
            rank[xroot]++;
        }
    }
}
 
// Function that returns true if
// the expression is invalid
function isInvalid(u, v, op, n) {
    // Vector to store parent values
    // of each integer
    let parent = [];
 
    // Vector to store rank
    // of each integer
    let rank = [];
 
    parent.push(-1);
    rank.push(-1);
 
    // Initialize parent values for
    // each of the integers
    for (let i = 1; i <= n; i++)
        parent.push(i);
 
    // Initialize rank values for
    // each of the integers
    for (let i = 1; i <= n; i++)
        rank.push(0);
 
    // Check for = operator and find union
    // for them
    for (let i = 0; i < n; i++)
        if (op[i] == "=")
            findUnion(u[i], v[i], parent, rank);
 
    // Check for != operator
    for (let i = 0; i < n; i++) {
        if (op[i] == "!=") {
            // If the expression is invalid
            if (findParent(u[i], parent)
                == findParent(v[i], parent))
                return true;
        }
    }
 
    // Expression is valid
    return false;
}
 
// Driver code
 
let u = [];
let v = [];
let op = [];
 
// Store the first integer
u.push(1);
u.push(2);
u.push(3);
 
// Store the second integer
v.push(2);
v.push(3);
v.push(1);
 
// Store the operators
op.push("=");
op.push("=");
op.push("!=");
 
// Number of expressions
let n = u.length;
if (isInvalid(u, v, op, n))
    document.write("Invalid");
else
    document.write("Valid");
 
// This code is contributed by _saurabh_jaiswal
</script>

Output:

Invalid

Time Complexity: O(nlogn)

Auxiliary Space: O(n)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!