Minimize the String length by removing given consecutive pairs️‍🔥

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Given a string S of size N consisting of characters from ‘0’to ‘9’, the task is to minimize the length of the string where In each minimizing operation, we can remove any two consecutive characters if they are of the form {“12”, “21”, “34”, “43”, “56”, “65”, “78”, “87”, “09”, “90”}.

Examples:

Input: S = “672183”
Output: 2
Explanation:
perform the minimizing operation at index 2 and 3 “672183″, after the removal resultant string will be “6783”
perform the minimizing operation at index 1 and 2 “6783″, after the removal resultant string will be”63″
Since no further minimizing operation can be performed, the minimum possible length of the string is 2.

Input: S = “19532”
Output: 5
Explanation: No operation can be performed

Approach:

Traverse through the string and keep removing the consecutive characters if they are in the above-mentioned form Break if there is no pair of characters found for minimizing return of the final length of the given string

Create a dummy string temp to store the resultant string.
Start traversing the given string and check whether the last character forms a removable pair with the current character then remove both of them.
- To check the pairs, create another boolean function and check whether the two characters form a possible pair for minimizing operation or not.
  - If yes, return true.
  - Otherwise, return false.
If the size of the dummy string temp is greater than 0 and the last character and current character currChar are making pair then remove them.
- Otherwise, add the currChar in the dummy temp string.
Return the length of the temp string.

Below is the implementation of the above approach:

C++

// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether the two characters
// is a possible pair for minimising operation
bool checkPairs(char a, char b)
{
    if ((a == '1' && b == '2') || (a == '2' && b == '1'))
        return true;
    else if ((a == '3' && b == '4')
             || (a == '4' && b == '3'))
        return true;
    else if ((a == '5' && b == '6')
             || (a == '6' && b == '5'))
        return true;
    else if ((a == '7' && b == '8')
             || (a == '8' && b == '7'))
        return true;
    else if ((a == '0' && b == '9')
             || (a == '9' && b == '0'))
        return true;
 
    return false;
}
 
// Function to check the minimum length
// of the string after applying the operation
int minimiseString(string& s)
{
    // If the last character is forming
    // a removable pair with the current
    // character then remove both of them
    string temp = "";
    for (char currChar : s) {
        if (temp.size() > 0
            && checkPairs(temp.back(), currChar)) {
            temp.pop_back();
        }
        else {
            temp.push_back(currChar);
        }
    }
 
    return temp.size();
}
 
// Driver Code
int main()
{
    string S = "672183";
 
    // Function call
    cout << minimiseString(S);
 
    return 0;
}

Java

// Java code to implement the approach
 
import java.util.*;
 
class GFG {
 
    // Function to check whether the two character is a
    // possible pair for minimising operation or not
    private static boolean checkPairs(char a, char b)
    {
        if ((a == '1' && b == '2')
            || (a == '2' && b == '1'))
            return true;
        else if ((a == '3' && b == '4')
                 || (a == '4' && b == '3'))
            return true;
        else if ((a == '5' && b == '6')
                 || (a == '6' && b == '5'))
            return true;
        else if ((a == '7' && b == '8')
                 || (a == '8' && b == '7'))
            return true;
        else if ((a == '0' && b == '9')
                 || (a == '9' && b == '0'))
            return true;
 
        return false;
    }
 
    // Function to check the minimum length
    // of the string after applying the operation
    public static int minimiseString(String s)
    {
        Stack<Character> st = new Stack<>();
        for (int i = 0; i < s.length(); i++) {
 
            // If the last character is
            // forming a removable pair
            // with the current character
            // then remove both of them
            if (!st.empty()
                && checkPairs(st.peek(), s.charAt(i))) {
                st.pop();
            }
            else {
                st.add(s.charAt(i));
            }
        }
        return st.size();
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        String S = "672183";
 
        // Function call
        System.out.println(minimiseString(S));
    }
}

Python3

# Python code to implement the approach
 
# Function to check whether the two characters
# is a possible pair for minimising operation
def checkPairs(a,b):
    if((a == '1' and b == '2') or (a == '2' and b == '1')):
        return True
    elif((a == '3' and b == '4') or (a == '4' and b == '3')):
        return True
    elif((a == '5' and b == '6') or (a == '6' and b == '5')):
        return True
    elif((a == '7' and b == '8') or (a == '8' and b == '7')):
        return True
    elif((a == '0' and b == '9') or (a == '9' and b == '0')):
        return True
     
    return False
 
# Function to check the minimum length
# of the string after applying the operation
def minimiseString(s):
    # If the last character is forming
    # a removable pair with the current
    # character then remove both of them
    temp=""
    for currChar in s:
        if(len(temp)>0 and checkPairs(temp[len(temp)-1],currChar)):
            temp = temp.rstrip(temp[-1])
        else:
            temp=temp+currChar
     
    return len(temp)
     
# Driver Code
S="672183"
 
# Function call
print(minimiseString(S))
 
# This code is contributed by Pushpesh Raj.

C#

using System;
using System.Text;
using System.Collections.Generic;
 
public class GFG 
{
   
  // Function to check whether the two character is a
  // possible pair for minimising operation or not
  public static bool checkPairs(char a, char b)
  {
    if ((a == '1' && b == '2')
        || (a == '2' && b == '1'))
      return true;
    else if ((a == '3' && b == '4')
             || (a == '4' && b == '3'))
      return true;
    else if ((a == '5' && b == '6')
             || (a == '6' && b == '5'))
      return true;
    else if ((a == '7' && b == '8')
             || (a == '8' && b == '7'))
      return true;
    else if ((a == '0' && b == '9')
             || (a == '9' && b == '0'))
      return true;
 
    return false;
  }
 
  // Function to check the minimum length
  // of the string after applying the operation
  public static int minimiseString(string s)
  {
    Stack<char> st = new Stack<char>();
    for (int i = 0; i < s.Length; i++) {
 
      // If the last character is
      // forming a removable pair
      // with the current character
      // then remove both of them
      if (st.Count != 0
          && checkPairs(st.Peek(), s[i])) {
        st.Pop();
      }
      else {
        st.Push(s[i]);
      }
    }
    return st.Count;
  }
 
  // Driver Code
  static public void Main()
  {
    string S = "672183";
 
    // Function call
    Console.WriteLine(minimiseString(S));
  }
}
 
// This code is contributed by Rohit Pradhan

Javascript

// JavaScript code to implement the approach
 
// Function to check whether the two character is a
// possible pair for minimising operation or not
function checkPairs(a, b){
    if ((a == '1' && b == '2')
        || (a == '2' && b == '1'))
        return true;
    else if ((a == '3' && b == '4')
             || (a == '4' && b == '3'))
        return true;
    else if ((a == '5' && b == '6')
             || (a == '6' && b == '5'))
        return true;
    else if ((a == '7' && b == '8')
             || (a == '8' && b == '7'))
        return true;
    else if ((a == '0' && b == '9')
             || (a == '9' && b == '0'))
        return true;
 
    return false;
}
 
// Function to check the minimum length
// of the string after applying the operation
function minimizeString(s){
    var st = [];
    for(let i=0;i<s.length;i++){
        // If the last character is
        // forming a removable pair
        // with the current character
        // then remove both of them
        if(st.length!=0 && checkPairs(st[st.length-1], s[i])){
            st.pop();
        }
        else{
            st.push(s[i]);
        }
    }
    return st.length;
}
 
let S = "672183";
// Function call
console.log(minimizeString(S));
 
// This code is contributed by lokesh.

Output

Time Complexity: O(N) // since we are traversing the entire string using a for loop hence the loops run till the length of the stri
Auxiliary Space: O(N) // since we are using a stack to store the characters hence the space taken is equal to the length of the string