Check if two expressions with brackets are same

Given two expressions in the form of strings. The task is to compare them and check if they are similar. Expressions consist of lowercase alphabets, ‘+’, ‘-‘ and ‘( )’.

Examples: 

Input  : exp1 = "-(a+b+c)"
         exp2 = "-a-b-c"
Output : Yes

Input  : exp1 = "-(c+b+a)"
         exp2 = "-c-b-a"
Output : Yes

Input  : exp1 = "a-b-(c-d)"
         exp2 = "a-b-c-d"
Output : No

It may be assumed that there are at most 26 operands from ‘a’ to ‘z’ and every operand appears only once.

A simple idea behind is to keep a record of the Global and Local Sign(+/-) through the expression. The Global Sign here means the multiplicative sign at each operand. The resultant sign for an operand is local sign multiplied by the global sign at that operand.

For example, the expression a+b-(c-d) is evaluated as (+)+a(+)+b(-)+c(-)-d => a + b – c + d. The global sign (represented inside bracket) is multiplied to the local sign for each operand.

In the given solution, stack is used to keep record of the global signs. A count vector records the counts of the operands(lowercase Latin letters here). Two expressions are evaluated in opposite manners and finally, it is checked if the all entries in the count vector are zeros. 

Implementation:

C++

// CPP program to check if two expressions
// evaluate to same.
#include <bits/stdc++.h>
using namespace std;

const int MAX_CHAR = 26;

// Return local sign of the operand. For example,
// in the expr a-b-(c), local signs of the operands
// are +a, -b, +c
bool adjSign(string s, int i)
{
 if (i == 0)
  return true;
 if (s[i - 1] == '-')
  return false;
 return true;
};

// Evaluate expressions into the count vector of
// the 26 alphabets.If add is true, then add count
// to the count vector of the alphabets, else remove
// count from the count vector.
void eval(string s, vector<int>& v, bool add)
{
 // stack stores the global sign
 // for operands.
 stack<bool> stk;
 stk.push(true);

 // + means true
 // global sign is positive initially

 int i = 0;
 while (s[i] != '\0') {
  if (s[i] == '+' || s[i] == '-') {
   i++;
   continue;
  }
  if (s[i] == '(') {

   // global sign for the bracket is
   // pushed to the stack
   if (adjSign(s, i))
    stk.push(stk.top());
   else
    stk.push(!stk.top());
  }

  // global sign is popped out which
  // was pushed in for the last bracket
  else if (s[i] == ')')
   stk.pop();

  else {

   // global sign is positive (we use different
   // values in two calls of functions so that
   // we finally check if all vector elements
   // are 0.
   if (stk.top())
    v[s[i] - 'a'] += (adjSign(s, i) ? add ? 1 : -1 :
            add ? -1 : 1);

   // global sign is negative here
   else
    v[s[i] - 'a'] += (adjSign(s, i) ? add ? -1 : 1 :
            add ? 1 : -1);
  }
  i++;
 }
};

// Returns true if expr1 and expr2 represent
// same expressions
bool areSame(string expr1, string expr2)
{
 // Create a vector for all operands and
 // initialize the vector as 0.
 vector<int> v(MAX_CHAR, 0);

 // Put signs of all operands in expr1
 eval(expr1, v, true);

 // Subtract signs of operands in expr2
 eval(expr2, v, false);

 // If expressions are same, vector must
 // be 0.
 for (int i = 0; i < MAX_CHAR; i++)
  if (v[i] != 0)
   return false;

 return true;
}

// Driver code
int main()
{
 string expr1 = "-(a+b+c)", expr2 = "-a-b-c";
 if (areSame(expr1, expr2))
  cout << "Yes\n";
 else
  cout << "No\n";
 return 0;
}

Java

// Java program to check if two expressions
// evaluate to same.
import java.io.*;
import java.util.*;

class GFG
{

    static final int MAX_CHAR = 26;

    // Return local sign of the operand. For example,
    // in the expr a-b-(c), local signs of the operands
    // are +a, -b, +c
    static boolean adjSign(String s, int i)
    {
        if (i == 0)
            return true;
        if (s.charAt(i - 1) == '-')
            return false;
        return true;
    };

    // Evaluate expressions into the count vector of
    // the 26 alphabets.If add is true, then add count
    // to the count vector of the alphabets, else remove
    // count from the count vector.
    static void eval(String s, int[] v, boolean add)
    {

        // stack stores the global sign
        // for operands.
        Stack<Boolean> stk = new Stack<>();
        stk.push(true);

        // + means true
        // global sign is positive initially

        int i = 0;
        while (i < s.length())
        {
            if (s.charAt(i) == '+' || s.charAt(i) == '-')
            {
                i++;
                continue;
            }
            if (s.charAt(i) == '(')
            {

                // global sign for the bracket is
                // pushed to the stack
                if (adjSign(s, i))
                    stk.push(stk.peek());
                else
                    stk.push(!stk.peek());
            }

            // global sign is popped out which
            // was pushed in for the last bracket
            else if (s.charAt(i) == ')')
                stk.pop();

            else
            {

                // global sign is positive (we use different
                // values in two calls of functions so that
                // we finally check if all vector elements
                // are 0.
                if (stk.peek())
                    v[s.charAt(i) - 'a'] += (adjSign(s, i) ?
                               add ? 1 : -1 : add ? -1 : 1);

                // global sign is negative here
                else
                    v[s.charAt(i) - 'a'] += (adjSign(s, i) ?
                                add ? -1 : 1 : add ? 1 : -1);
            }
            i++;
        }
    };

    // Returns true if expr1 and expr2 represent
    // same expressions
    static boolean areSame(String expr1, String expr2)
    {

        // Create a vector for all operands and
        // initialize the vector as 0.
        int[] v = new int[MAX_CHAR];

        // Put signs of all operands in expr1
        eval(expr1, v, true);

        // Subtract signs of operands in expr2
        eval(expr2, v, false);

        // If expressions are same, vector must
        // be 0.
        for (int i = 0; i < MAX_CHAR; i++)
            if (v[i] != 0)
                return false;

        return true;
    }

    // Driver Code
    public static void main(String[] args)
    {

        String expr1 = "-(a+b+c)", expr2 = "-a-b-c";
        if (areSame(expr1, expr2))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}

// This code is contributed by
// sanjeev2552

Python3

# Python3 program to check if two expressions
# evaluate to same.
MAX_CHAR = 26;

# Return local sign of the operand. For example,
# in the expr a-b-(c), local signs of the operands
# are +a, -b, +c
def adjSign(s, i):
  if (i == 0):
    return True;
  if (s[i - 1] == '-'):
    return False;
  return True;

# Evaluate expressions into the count vector of
# the 26 alphabets.If add is True, then add count
# to the count vector of the alphabets, else remove
# count from the count vector.
def eval(s, v, add):

  # stack stores the global sign
  # for operands.
  stk = []
  stk.append(True);

  # + means True
  # global sign is positive initially
  i = 0;

  while (i < len(s)):

    if (s[i] == '+' or s[i] == '-'):
      i += 1
      continue;

    if (s[i] == '('):

      # global sign for the bracket is
      # pushed to the stack
      if (adjSign(s, i)):
        stk.append(stk[-1]);
      else:
        stk.append(not stk[-1]);

    # global sign is popped out which
    # was pushed in for the last bracket
    elif (s[i] == ')'):
      stk.pop();
    else:

      # global sign is positive (we use different
      # values in two calls of functions so that
      # we finally check if all vector elements
      # are 0.
      if (stk[-1]):
        v[ord(s[i]) - ord('a')] += (1 if add else -1) if adjSign(s, i) else (-1 if add else 1)

      # global sign is negative here
      else:
        v[ord(s[i]) - ord('a')] += (-1 if add else 1) if adjSign(s, i) else (1 if add else -1)

    i += 1

# Returns True if expr1 and expr2 represent
# same expressions
def areSame(expr1, expr2):

  # Create a vector for all operands and
  # initialize the vector as 0.
  v = [0 for i in range(MAX_CHAR)];

  # Put signs of all operands in expr1
  eval(expr1, v, True);

  # Subtract signs of operands in expr2
  eval(expr2, v, False);

  # If expressions are same, vector must
  # be 0.
  for i in range(MAX_CHAR):
    if (v[i] != 0):
      return False;
  return True;

# Driver Code
if __name__=='__main__':
  expr1 = "-(a+b+c)"
  expr2 = "-a-b-c";
  if (areSame(expr1, expr2)):
    print("Yes");
  else:
    print("No");

    # This code is contributed by rutvik_56.

C#

// C# program to check if two expressions
// evaluate to same.
using System;
using System.Collections.Generic;
public class GFG
{

  static readonly int MAX_CHAR = 26;

  // Return local sign of the operand. For example,
  // in the expr a-b-(c), local signs of the operands
  // are +a, -b, +c
  static bool adjSign(String s, int i)
  {
    if (i == 0)
      return true;
    if (s[i-1] == '-')
      return false;
    return true;
  }

  // Evaluate expressions into the count vector of
  // the 26 alphabets.If add is true, then add count
  // to the count vector of the alphabets, else remove
  // count from the count vector.
  static void eval(String s, int[] v, bool add)
  {

    // stack stores the global sign
    // for operands.
    Stack<Boolean> stk = new Stack<Boolean>();
    stk.Push(true);

    // + means true
    // global sign is positive initially
    int i = 0;
    while (i < s.Length)
    {
      if (s[i] == '+' || s[i] == '-')
      {
        i++;
        continue;
      }
      if (s[i] == '(')
      {

        // global sign for the bracket is
        // pushed to the stack
        if (adjSign(s, i))
          stk.Push(stk.Peek());
        else
          stk.Push(!stk.Peek());
      }

      // global sign is popped out which
      // was pushed in for the last bracket
      else if (s[i] == ')')
        stk.Pop();

      else
      {

        // global sign is positive (we use different
        // values in two calls of functions so that
        // we finally check if all vector elements
        // are 0.
        if (stk.Peek())
          v[s[i] - 'a'] += (adjSign(s, i) ?
                            add ? 1 : -1 : add ? -1 : 1);

        // global sign is negative here
        else
          v[s[i] - 'a'] += (adjSign(s, i) ?
                            add ? -1 : 1 : add ? 1 : -1);
      }
      i++;
    }
  }

  // Returns true if expr1 and expr2 represent
  // same expressions
  static bool areSame(String expr1, String expr2)
  {

    // Create a vector for all operands and
    // initialize the vector as 0.
    int[] v = new int[MAX_CHAR];

    // Put signs of all operands in expr1
    eval(expr1, v, true);

    // Subtract signs of operands in expr2
    eval(expr2, v, false);

    // If expressions are same, vector must
    // be 0.
    for (int i = 0; i < MAX_CHAR; i++)
      if (v[i] != 0)
        return false;

    return true;
  }

  // Driver Code
  public static void Main(String[] args)
  {

    String expr1 = "-(a+b+c)", expr2 = "-a-b-c";
    if (areSame(expr1, expr2))
      Console.WriteLine("Yes");
    else
      Console.WriteLine("No");
  }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
    // Javascript program to check if two expressions
    // evaluate to same.

    let MAX_CHAR = 26;

    // Return local sign of the operand. For example,
    // in the expr a-b-(c), local signs of the operands
    // are +a, -b, +c
    function adjSign(s, i)
    {
        if (i == 0)
            return true;
        if (s[i - 1] == '-')
            return false;
        return true;
    }

    // Evaluate expressions into the count vector of
    // the 26 alphabets.If add is true, then add count
    // to the count vector of the alphabets, else remove
    // count from the count vector.
    function eval(s, v, add)
    {

        // stack stores the global sign
        // for operands.
        let stk = [];
        stk.push(true);

        // + means true
        // global sign is positive initially

        let i = 0;
        while (i < s.length)
        {
            if (s[i] == '+' || s[i] == '-')
            {
                i++;
                continue;
            }
            if (s[i] == '(')
            {

                // global sign for the bracket is
                // pushed to the stack
                if (adjSign(s, i))
                    stk.push(stk[stk.length - 1]);
                else
                    stk.push(!stk[stk.length - 1]);
            }

            // global sign is popped out which
            // was pushed in for the last bracket
            else if (s[i] == ')')
                stk.pop();

            else
            {

                // global sign is positive (we use different
                // values in two calls of functions so that
                // we finally check if all vector elements
                // are 0.
                if (stk[stk.length - 1])
                    v[s[i] - 'a'] += (adjSign(s, i) ?
                               add ? 1 : -1 : add ? -1 : 1);

                // global sign is negative here
                else
                    v[s[i] - 'a'] += (adjSign(s, i) ?
                                add ? -1 : 1 : add ? 1 : -1);
            }
            i++;
        }
    };

    // Returns true if expr1 and expr2 represent
    // same expressions
    function areSame(expr1, expr2)
    {

        // Create a vector for all operands and
        // initialize the vector as 0.
        let v = new Array(MAX_CHAR);
        v.fill(0);

        // Put signs of all operands in expr1
        eval(expr1, v, true);

        // Subtract signs of operands in expr2
        eval(expr2, v, false);

        // If expressions are same, vector must
        // be 0.
        for (let i = 0; i < MAX_CHAR; i++)
            if (v[i] != 0)
                return false;

        return true;
    }

    let expr1 = "-(a+b+c)", expr2 = "-a-b-c";
    if (areSame(expr1, expr2))
      document.write("YES");
    else
      document.write("NO");

    // This code is contributed by suresh07.
</script>

Output

Yes

Time Complexity: O(n)
Auxiliary Space: O(n) 

This article is contributed by Amol Mejari. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.