Roman To Numbers

          
Problem Statement
          example, 2 is written as II in Roman numerals, just two
          one's added together. 12 is written as XII, which is
          simply x + II The number 27 is written as XXVII, which 
          is XX + V + II
          Roman numerals are usually written largest to smallest
          from left to right. However, the numeral for four is
          not IIII. Instead, the number four is written as IV.
          Because the one is before the five we subtract it making
          four. The same principle applies to the number nine,
          which is written as IX. There are six instances where
          subtraction is used:
          I can be placed before V (5) and X (10) to make 4 and 9.
          X can be placed before L (50) and C (100) to make 40 and 90.
          C can be placed before D (500) and M (1000) to make 400 and 900.
          Given an integer, convert it to a roman numeral.
          
          
Example
          Exercise-1
          Input: LXI
          Output : 61
          
          Exercise-2
          Input: MXI
          Output: 1011
          
          

        

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Lets get started

Java code :--

 
         

import java.util.HashMap; 
import java.util.Map; 
import java.util.Scanner; 
 
public class MINtech { 
 
    // Function to convert a Roman numeral to an integer 
    public static int romanToInt(String s) { 
        // Create a map to store the values of Roman numerals 
        Map romanMap = new HashMap<>(); 
        romanMap.put('I', 1); 
        romanMap.put('V', 5); 
        romanMap.put('X', 10); 
        romanMap.put('L', 50); 
        romanMap.put('C', 100); 
        romanMap.put('D', 500); 
        romanMap.put('M', 1000); 
 
        // Initialize variables to store the result and the previous value 
        int result = 0; 
        int prevValue = 0; 
 
        // Iterate through the Roman numeral from right to left 
        for (int i = s.length() - 1; i >= 0; i--) { 
            // Get the numeric value of the current Roman numeral character 
            int value = (int) romanMap.get(s.charAt(i)); 
 
            // Check if the current value is less than the previous value 
            if (value < prevValue) { 
                // If yes, subtract the current value from the result 
                result -= value; 
            } else { 
                // If no, add the current value to the result 
                result += value; 
            } 
 
            // Update the previous value for the next iteration 
            prevValue = value; 
        } 
 
        // Return the final result 
        return result; 
    } 
 
    public static void main(String[] args) { 
        // Create a scanner to take user input 
        Scanner scanner = new Scanner(System.in); 
 
        // Get user input for a Roman numeral 
        System.out.print("Enter a Roman numeral: "); 
        String userInputRoman = scanner.nextLine(); 
 
        // Convert Roman numeral to integer and print the result 
        int integerResult = romanToInt(userInputRoman); 
        System.out.println("Integer: " + integerResult); 
 
        // Close the scanner to avoid resource leaks 
        scanner.close(); 
    } 
} 

         

Output

      Enter a Roman numeral: LXI
      Integer: 61

      Enter a Roman numeral: MXI
      Integer: 1011
  
        

Python code :---

  
       

def roman_to_int(s): 
    # Create a dictionary to map Roman numeral characters to their numeric values 
    roman_map = {'I': 1, 'V': 5, 'X': 10, 'L': 50, 'C': 100, 'D': 500, 'M': 1000} 
 
    # Initialize variables to store the result and the previous value 
    result = 0 
    prev_value = 0 
 
    # Iterate through the Roman numeral from right to left 
    for char in reversed(s): 
        # Get the numeric value of the current Roman numeral character 
        value = roman_map[char] 
 
        # Check if the current value is less than the previous value 
        if value < prev_value: 
            # If yes, subtract the current value from the result 
            result -= value 
        else: 
            # If no, add the current value to the result 
            result += value 
 
        # Update the previous value for the next iteration 
        prev_value = value 
 
    # Return the final result 
    return result 
 
if __name__ == "__main__": 
    # Get user input for a Roman numeral 
    user_input_roman = input("Enter a Roman numeral: ") 
 
    # Convert Roman numeral to integer and print the result 
    integer_result = roman_to_int(user_input_roman) 
    print("Integer:", integer_result) 

       
 

Output

      Enter a Roman numeral: LXI
      Integer: 61

      Enter a Roman numeral: MXI
      Integer: 1011
  
  
   
         

Longest substring without repeating characters

          You are tasked with developing a function that finds 
          the length of the longest substring without repeating 
          characters in a given string s.
          Consider different scenarios involving various 
          characters in the input string.
          Scenario: String with No Repeating Characters 
          Input: "abcdef"
          Expected Output: 6
          Explanation: In this scenario, the input string contains
          no repeating characters, so the entire string itself is the longest substring without repeating characters.
          Scenario: String with Repeating Characters
          Input: "abccba"
          Expected Output: 3
          Explanation: In this case, the longest substring without repeating characters is "abc" with a length of 3. The characters 'c' and 'b' are repeated, so the substring ends at the first occurrence of the repeated character
         
         
          Exercise-1
          Input:
          pqrsstu
          Output: 4
          
          
          Exercise-2
          Input:
          abbedfgi
          Output: 6
          

        

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Lets get started

Java code :--

 
         

import java.util.HashMap; 
import java.util.Scanner; 
 
public class MINtech { 
 
    public static int lengthOfLongestSubstring(String s) { 
        // Initialize a HashMap to store the last index of each character 
        HashMap lastIdxMap = new HashMap<>(); 
        // Initialize variables to keep track of the start of the current substring and the length of the longest substring 
        int start = 0; 
        int maxLength = 0; 
 
        // Iterate through the characters of the string 
        for (int end = 0; end < s.length(); end++) { 
            // If the current character is already in the substring, update the start index 
            if (lastIdxMap.containsKey(s.charAt(end)) && lastIdxMap.get(s.charAt(end)) >= start) { 
                start = lastIdxMap.get(s.charAt(end)) + 1; 
            } 
 
            // Update the last index of the current character 
            lastIdxMap.put(s.charAt(end), end); 
 
            // Update the length of the longest substring if needed 
            maxLength = Math.max(maxLength, end - start + 1); 
        } 
 
        return maxLength; 
    } 
 
    public static void main(String[] args) { 
        // Get user input for the string 
        Scanner scanner = new Scanner(System.in); 
        System.out.print("Enter any string :- "); 
        String inputString = scanner.nextLine(); 
 
        // Calculate and print the length of the longest substring without repeating characters 
        int result = lengthOfLongestSubstring(inputString); 
        System.out.println("Length of the longest substring without repeating characters :--  " + result); 
    } 
} 

         

Output

      Enter any string :- pqrsstu
      Length of the longest substring without repeating characters :--  4


      Enter any string :- abbedfgi
      Length of the longest substring without repeating characters :--  6
  
        

Python code :---

  
       

def length_of_longest_substring(s): 
    # Initialize a dictionary to store the last index of each character 
    last_index = {} 
    # Initialize variables to keep track of the start of the current substring and the length of the longest substring 
    start = 0 
    max_length = 0 
 
    # Iterate through the characters of the string 
    for end in range(len(s)): 
        # If the current character is already in the substring, update the start index 
        if s[end] in last_index and last_index[s[end]] >= start: 
            start = last_index[s[end]] + 1 
 
        # Update the last index of the current character 
        last_index[s[end]] = end 
 
        # Update the length of the longest substring if needed 
        max_length = max(max_length, end - start + 1) 
 
    return max_length 
 
# Get user input for the string 
input_string = input("Enter any string :-- ") 
 
# Calculate and print the length of the longest substring without repeating characters 
result = length_of_longest_substring(input_string) 
print("Length of the longest substring without repeating characters :--  ", result) 

       
 

Output

      Enter any string :- pqrsstu
      Length of the longest substring without repeating characters :--  4


      Enter any string :- abbedfgi
      Length of the longest substring without repeating characters :--  6
  
   
         

sort colors

          How can an array named "nums" containing objects labeled as
          "Crimson," "Ivory," or "Azure" or objects representing 
          colors ( 0, 1 , or 2), be sorted in-place, such that objects 
          of the same color or label are adjacent, and the colors or 
          labels appear in the specific order of crimson, ivory, and azure?
          For example, consider the input array [2, 0, 1] or  
          ['Azure', 'Crimson', 'Ivory']. The desired output would be 
          [0, 1, 2] or ['Crimson', 'Ivory', 'Azure'], objects are grouped 
          together, followed by all the Is or 'Ivory' objects, and finally, 
          all the 2s or respectively, where all the Os or 'Crimson'
          'Azure' objects.

          Exercise-1
          Input:
             12120120
          Output:
             00111222

        

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Lets get started

Java code :--

 
          

import java.util.Scanner; 
public class MINtech { 
 public static void sortColors(int[] nums) { 
     int left = 0; 
     int right = nums.length - 1; 
     int current = 0; 
 
     while (current <= right) { 
          if (nums[current] == 0) { 
              swap(nums, current, left); 
             left++; 
             current++; 
          } else if (nums[current] == 2) { 
             swap(nums, current, right); 
             right--; 
          } else { 
             current++; 
         } 
      } 
    } 
 
    private static void swap(int[] nums, int i, int j) { 
        int temp = nums[i]; 
        nums[i] = nums[j]; 
        nums[j] = temp; 
    } 
 
    public static void main(String[] args) { 
        Scanner scanner = new Scanner(System.in); 
 
        System.out.println("Enter the length of the array:"); 
        int length = scanner.nextInt(); 
 
        int[] nums = new int[length]; 
 
        System.out.println("Enter the elements of the array (0, 1, or 2):"); 
        for (int i = 0; i < length; i++) { 
            nums[i] = scanner.nextInt(); 
        } 
 
        sortColors(nums); 
 
        // Print the sorted array 
        System.out.println("Sorted Array:"); 
        for (int num : nums) { 
            System.out.print(num + " "); 
        } 
 
        scanner.close(); 
    } 
} 
 
          
 

Output

      Enter the length of the array:
      8
      Enter the elements of the array (0, 1, or 2):
      1
      2
      1
      2
      0
      1
      2
      0
      Sorted Array:
      0 0 1 1 1 2 2 2 


      Enter the length of the array:
      9
      Enter the elements of the array (0, 1, or 2):
      1  
      2
      0   
      0
      2
      1
      0
      1
      0
      Sorted Array:
      0 0 0 0 1 1 1 2 2 
  
        

Python code :---

  
       

    def MINtech_sort_colors(nums): 
    left, right, current = 0, len(nums) - 1, 0 
 
    while current <= right: 
        if nums[current] == 0: 
            nums[current], nums[left] = nums[left], nums[current] 
            left += 1 
            current += 1 
        elif nums[current] == 2: 
            nums[current], nums[right] = nums[right], nums[current] 
            right -= 1 
        else: 
            current += 1 
 
    if __name__ == "__main__": 
        length = int(input("Enter the length of the array: ")) 
        nums = [int(input(f"Enter element {i + 1} (0, 1, or 2): ")) for i in range(length)] 
 
        MINtech_sort_colors(nums) 
 
        # Print the sorted array 
        print("Sorted Array:", end=" ") 
        for num in nums: 
            print(num, end=" ") 

       
 

Output

      Enter the length of the array: 8
      Enter element 1 (0, 1, or 2): 1
      Enter element 2 (0, 1, or 2): 2
      Enter element 3 (0, 1, or 2): 1
      Enter element 4 (0, 1, or 2): 2
      Enter element 5 (0, 1, or 2): 0
      Enter element 6 (0, 1, or 2): 1
      Enter element 7 (0, 1, or 2): 2
      Enter element 8 (0, 1, or 2): 0
      Sorted Array: 0 0 1 1 1 2 2 2 

      Enter the length of the array: 9
      Enter element 1 (0, 1, or 2): 2
      Enter element 2 (0, 1, or 2): 2
      Enter element 3 (0, 1, or 2): 0
      Enter element 4 (0, 1, or 2): 0
      Enter element 5 (0, 1, or 2): 1
      Enter element 6 (0, 1, or 2): 1
      Enter element 7 (0, 1, or 2): 0
      Enter element 8 (0, 1, or 2): 2
      Enter element 9 (0, 1, or 2): 0
      Sorted Array: 0 0 0 0 1 1 2 2 2 
   
         

Queue Operations using Stacks

Queue Operations

          Write a program to implement a custom queue using two stacks. 
          The queue should support the following three types of queries:
          Enqueue: Thrs query type is denoted by "1 x", where x is an 
          element to be enqueued. It means that you need to insert 
          element x at the end of the queue.
          Dequeue: This query type is denoted by "2". It indicates that
          you should remove the element at the front of the queue.
          Print Front: This query type is denoted by "3". 
          It instructs you to print the element at the front of the 
          queue without removing it.

          Exercise-1
          input:
          1 42,2,1 14,3
          
          output:
          14
          
          Exercise-2
          input:
          1 23,2,1 14,3,2,1 78,3
          Output:
          14
          78

        

Know more

Lets get started

 
      

import java.util.*; 
 
public class MINtech { 
    public static void main(String[] args) { 
        
        // Create a scanner object to read input from the user 
        Scanner sc = new Scanner(System.in); 
        // Read a line of input from the user 
        String str = sc.nextLine();  
         
        // Create a queue to store integers 
        Queue <Integer > queue = new LinkedList< >();  
 
        // Split the input line into an array of queries using commas as a delimiter 
        String[] queries = str.split(","); 
 
        for (int i = 0; i < queries.length; i++) { 
            // Get a single query from the array 
            String query = queries[i]; 
            // Split the query into parts using space as a delimiter 
            String[] parts = query.split(" "); 
           // Parse the first part of the query as an integer to determine the type of operation 
            int type = Integer.parseInt(parts[0]); 
 
            if (type == 1) { 
                // Parse the second part of the query as an inte 
                int x = Integer.parseInt(parts[1]); 
                // Add the integer to the queueger 
                queue.offer(x);  
            } else if (type == 2) { 
                if (!queue.isEmpty()) { 
                    // Remove the element at the front of the queue if the queue is not empty 
                    queue.poll();  
                } 
            } else if (type == 3) { 
                if (!queue.isEmpty()) { 
                    // Print the element at the front of the queue if the queue is not empty 
                    System.out.println(queue.peek()); 
                } 
            } 
        } 
        // Close the scanner to release system resources 
        sc.close(); 
    } 
} 

      
 

Output

      1 14,2,1 41,3
      41
      
      1 23,2,1 14,3,2,1 78,3
      14
      78

      1 42,2,1 14,3
      14
  
        

Reversing the given string

We are going to develop a function that will reverse the given
string while preserving the positions of capital letters and 
punctuation marks and spaces
For example --- 
Input : 
  MINtech is the best   
Output : 
 tsebeht si hce tNIM

        

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Lets get started

 
      

        /* now we  are going to create a function that will 
   reverse the given string whithout touching the 
   positions of capital letters . and punctuation marks */ 
 
import java.util.* ; 
public class MINtech{ 
  // Define a static method named "MINtechJAVA" which takes a string as input 
  public static String MINtechJAVA(String str){ 
      // Convert the input string to a character array 
      char[] arr = str.toCharArray() ; 
       
      // Initialize two pointers, "left" and "right", to reverse the string 
      int left = 0 ; 
      int right = arr.length -1 ; 
       
      // Start a loop that continues until "left" is less than "right" 
      while(left < right){
         // Move the "left" pointer to the right while it's not a letter 
          while(left < right && !Character.isLetter(arr[left])){ 
              left++ ; 
          } 
          // Move the "right" pointer to the left while it's not a letter 
          while(left < right && !Character.isLetter(arr[right])){ 
              right -- ; 
          } 
 
         // Swap the characters at the "left" and "right" positions 
         char temp = arr[left] ; 
         arr[left] = arr[right] ; 
         arr[right] = temp ; 
 
         // Move the pointers inward 
         right -- ; 
         left ++ ; 
         } 
         // Convert the character array back to a string and return it 
         return new String(arr) ; 
         } 
     // Main method to execute the program 
	 public static void main(String[] args) { 
	     
		// Create a Scanner object to read user input 
		Scanner sc = new Scanner(System.in); 
		System.out.println("Enter a string that you want to reverse : "); 
		// Read a line of input from the user 
		String str = sc.nextLine(); 
		 
		 // Call the "MINtechJAVA" function and print the reversed string 
		System.out.println(MINtechJAVA(str)); 
	} 
} 
     
 

Output

      Enter a string that you want to reverse : 
      MINtech is amzing
      gnizmas ih cetNIM

      Enter a string that you want to reverse : 
      I Love MINtech
      h cetN IMevoLI

  https://www.toprevenuegate.com/rzm38ght?key=d841e7a69b77ba94bcb8e38c0573aab0
        

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Time format coneversion from 12H to 24H

https://www.highcpmrevenuegate.com/s4ttwmua2u?key=c02c9252219c2ddb8d72b452016a4796

You have been given a time in 12 Hours AM/Pm format . 
And we have convert the given time into  24 Hour (military) format . 
For exmaple : -- 
    12:00:00 AM -----------> 00:00:00
    01:00:00 PM -----------> 13:00:00
    06:56:59 PM -----------> 18:56:59
        

https://www.highcpmrevenuegate.com/s4ttwmua2u?key=c02c9252219c2ddb8d72b452016a4796

Lets get started

 
      

import java.util.* ; 
import java.text.* ; 
 
public class MINtech{ 
    public static String MINtechJava(String time) { 
        try { 
            SimpleDateFormat originalFormat = new SimpleDateFormat("hh:mm:ss a"); 
            SimpleDateFormat targetFormat = new SimpleDateFormat("HH:mm:ss"); 
             
            Date date = originalFormat.parse(time) ; 
            return targetFormat.format(date) ; 
        } catch(Exception e) { 
            return "invalid input format" ; 
        } 
         
    } 
    public static void main (String[] args) { 
        Scanner sc = new Scanner(System.in); 
        System.out.println("Enter your in hh:mm:ss AM/PM"); 
        String time = sc.nextLine() ; 
         
        System.out.println( MINtechJava(time)); 
    } 
} 
      

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Output

      Enter your in hh:mm:ss AM/PM
      6:45:45 PM
      18:45:45
      

      Enter your in hh:mm:ss AM/PM
      9:9:53 AM
      09:09:53

      Enter your in hh:mm:ss AM/PM
      2:98:78 PM
      15:39:18

  
        

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`