Learn and practice the Bitwise Manipulation, Sliding Window, and Union Find patterns to solve coding interview problems.

Educative-77 in Python: Accelerate Your Coding Interview Prep

Each of the patterns in this module exemplifies the power of dissecting a problem in detail and figuring out the least expensive operations required to solve it. In the Bitwise Manipulation pattern, we’ll learn to leverage the power of bit-level operators to deliver more efficient solutions. With the Sliding Window pattern, we’ll gain practice in various techniques to ensure that a solution that seems to require nested loops actually executes in linear time. Using the Union Find pattern, we’ll learn to solve a diverse set of problems using the disjoint-set data structure.

Elegant by Design

# Statement
Given two strings, `str1` and `str2`, find the index of the extra character that is present in only one of the strings.

> **Note:** If multiple instances of the extra character exist, return the index of the first occurrence of the character in the longer string. 






**Constraints:**
* $0 \leq$ `str1.length`, `str2.length` $\leq 1000$
* Either `str2.length` $=$ `str1.length + 1`, or, `str1.length` $=$ `str2.length + 1`
* The strings consist of lowercase English letters.

# Examples


# Understand the problem
Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:

Find the index of the extra character that is present in only one of the strings.

string 1 = "pqr"

string 2 = "psrq"

Find the index of the extra character that is present in only one of the strings.
string 1 = “pqr”
string 2 = “psrq”

Find the index of the extra character that is present in only one of the strings.

string 1 = "masthematic"

string 2 = "mathematic"

Find the index of the extra character that is present in only one of the strings.
string 1 = “masthematic”
string 2 = “mathematic”

Find the index of the extra character that is present in only one of the strings.

string 1 = "rest"

string 2 = "est"

Find the index of the extra character that is present in only one of the strings.
string 1 = “rest”
string 2 = “est”

Find the Difference

Find the Difference

# Figure it out!
We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.

# Try it yourself
Implement your solution in the following coding playground.

# Quick Fix\nYou are Edward, a Senior Software Engineer who specializes in coding interview patterns, data structures and algorithms, and is great at debugging issues in code.\n\n- Edward, please keep in mind that you do not have to print the data of the placeholders. Also, do not mention anything about yourself.\n\nEdward, you are given the following data in the placeholders:\n- Function name: `extra_character_index`\n- Candidate's submitted solution code: USER_CODE\n- Error: ERROR_DESCRIPTION\n- Failed test case: FAILED_TEST_CASE\n- Coding language: `Python`\n\n- Edward, your task is to understand the type of error that is occurring in the coding language provided.\n- Once you've understood the error as per the error’s description, and why the test case failed, you will suggest how the candidate can improve their solution code to fix the error.\n- Edward, you will not give the corrected code, just a hint on what the error tells, and how to fix it.\n- Edward, please limit your response strictly to 100 words. No more words than the limit.

# Review Code\nEdward, your current task is to review the `extra_character_index` function(s) and the other function(s) invoked in it/them.\n\n-Take your time and review the candidate's solution code.\n- Do not review any extra/helper files or functions not invoked in the solution function `extra_character_index`.\n- Do not review the code of data structure objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. \n- Also, if the solution code doesn’t use the coding pattern stated above, then please encourage the candidate to reattempt the problem using that pattern corectly.

# Review time complexity\n- Please provide a brief time complexity analysis of the function(s) `extra_character_index`. \n- While calculating time complexity, include the time complexity of built-in Python functions invoked by the code. \n- The answer should be to the point in approximately 60 words.

# Review space complexity\n- Please provide a brief space complexity analysis of the function(s) `extra_character_index`. \n- While calculating space complexity, include the space complexity of built-in Python functions invoked by the code. \n- Do not include the space used for the input parameters and the space used for just returning an output (not used as a scratch space). \n- The answer should be to the point in approximately 60 words.

def evaluate_results(self) -> object:
        """
        self.__inputs is the list of inputs that are passed as input to function
        self.__actual_output is a list of outputs that is produced by the learner's code
        return : self.__edu__result object
        """
        def extra_character_index(str1, str2):

            result = 0
            str1_length = len(str1)
            str2_length = len(str2)

            for i in range(str1_length):
                result = result ^ (ord)(str1[i])

            for j in range(str2_length):
                result = result ^ (ord)(str2[j])

            if len(str1) > len(str2):
                index = str1.index((chr)(result))
                return index
            else:
                index = str2.index((chr)(result))
                return index


        expected = extra_character_index(self.__inputs[0],self.__inputs[1])

        actual = self.__actual_outputs[0]
        # multiple outputs will be added at each index respectively for a single test_case
        expected_outputs = [] 
        # store the 1st expected output that should be displayed to the user 
        expected_outputs.append(expected)
        if expected == actual:
            self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
        else:
            self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)

        return self.__edu_result

def extra_character_index(str1, str2):

    # Replace this placeholder return statement with your code
    return -1

find_the_difference.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1 : "wxyz"
    - 2 : "zwxgy"
    output: 
    - 1 : 3
  - name: "testcase 2"
    inputs:
    - 1: "cbda"
    - 2: "abc"
    output:
    - 1 : 2
  - name: "testcase 3"
    inputs:
    - 1: "aaaaa"
    - 2: "aaaaaa"
    output:
    - 1 : 0
  - name: "testcase 4"
    inputs:
    - 1: "courae"
    - 2: "couearg"
    output:
    - 1 : 6
  - name: "testcase 5"
    inputs:
    - 1: "hello"
    - 2: "helo"
    output:
    - 1 : 2

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """
 import string
 import collections

 str1 = self.__inputs[0]
 str1_lower_bound = 0
 str1_upper_bound = 10**3

 str2 = self.__inputs[1]
 str2_lower_bound = 0
 str2_upper_bound = 10**3

 valid_characters = [char for char in (string.ascii_lowercase)]


 def is_one_extra_character(s, t):
 s_counter = collections.Counter(s)
 t_counter = collections.Counter(t)

 diff_count = sum((s_counter - t_counter).values()) + sum((t_counter - s_counter).values())
 return diff_count == 1

 def contains_valid_characters(s, chars):
 for char in s:
 if char not in chars:
 return False
 return True

 def list_to_string(char_list):
 return "[" + ', '.join(map(repr, char_list)) + "]"

 # inputs can be accessed using self.__inputs starting from 0 to len(self.__inputs) - 1

 if not abs(len(str1) - len(str2)) == 1:
 self.store_result ("Expected difference of the length of the strings should be 1 but got " + str(abs(len(str1) -len(str2))) + ".", ValidationFunction.FAIL)

 elif len(str1) < str1_lower_bound or len(str1) > str1_upper_bound:
 self.store_result ("Expected 'str1' to have " + str(str1_lower_bound) + " <= str1.length <= " + str(str1_upper_bound) + " but got " + str(len(str1)), ValidationFunction.FAIL)

 elif len(str2) < str2_lower_bound or len(str2) > str2_upper_bound:
 self.store_result ("Expected 'str_2' to have" + str(str2_lower_bound) + " <= str_2.length <= " + str(str2_upper_bound) + " but got " + str(len(str2)), ValidationFunction.FAIL)

 elif not contains_valid_characters(str1, valid_characters):
 self.store_result ("'str1' must contain lower case English letters only.", ValidationFunction.FAIL)

 elif not contains_valid_characters(str2, valid_characters):
 self.store_result ("'str2' must contain lower case English letters only.", ValidationFunction.FAIL)
 
 elif not is_one_extra_character(str1, str2):
 self.store_result("str2 should contain any shuffled version of str1.", ValidationFunction.FAIL)

 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Find the Difference problem.

Bitwise Manipulation

Sliding Window

Union Find

Conclusion

Find the Difference

Statement

Examples

Understand the problem

Figure it out!

Try it yourself