Check If Digits Are Equal in String After Operations I

Easy

Amazon

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Strings

You are given a string s consisting of digits. Perform the following operation repeatedly until the string has exactly two digits:

For each pair of consecutive digits in s, starting from the first digit, calculate a new digit as the sum of the two digits modulo 10.
Replace s with the sequence of newly calculated digits, maintaining the order in which they are computed.

Return true if the final two digits in s are the same; otherwise, return false.

Example 1:

Input: s = "3902" Output: true

Explanation:

Initially, s = "3902"
First operation:
- (s[0] + s[1]) % 10 = (3 + 9) % 10 = 2
- (s[1] + s[2]) % 10 = (9 + 0) % 10 = 9
- (s[2] + s[3]) % 10 = (0 + 2) % 10 = 2
- s becomes "292"
Second operation:
- (s[0] + s[1]) % 10 = (2 + 9) % 10 = 1
- (s[1] + s[2]) % 10 = (9 + 2) % 10 = 1
- s becomes "11"
Since the digits in "11" are the same, the output is true.

Example 2:

Input: s = "34789" Output: false

Explanation:

Initially, s = "34789".
After the first operation, s = "7157".
After the second operation, s = "862".
After the third operation, s = "48".
Since '4' != '8', the output is false.

What is the most efficient algorithm to solve this problem?

Check If Digits Are Equal in String After Operations I

Easy

Amazon

3 views

Topics:

Strings

You are given a string s consisting of digits. Perform the following operation repeatedly until the string has exactly two digits:

For each pair of consecutive digits in s, starting from the first digit, calculate a new digit as the sum of the two digits modulo 10.
Replace s with the sequence of newly calculated digits, maintaining the order in which they are computed.

Return true if the final two digits in s are the same; otherwise, return false.

Example 1:

Input: s = "3902" Output: true

Explanation:

Initially, s = "3902"
First operation:
- (s[0] + s[1]) % 10 = (3 + 9) % 10 = 2
- (s[1] + s[2]) % 10 = (9 + 0) % 10 = 9
- (s[2] + s[3]) % 10 = (0 + 2) % 10 = 2
- s becomes "292"
Second operation:
- (s[0] + s[1]) % 10 = (2 + 9) % 10 = 1
- (s[1] + s[2]) % 10 = (9 + 2) % 10 = 1
- s becomes "11"
Since the digits in "11" are the same, the output is true.

Example 2:

Input: s = "34789" Output: false

Explanation:

Initially, s = "34789".
After the first operation, s = "7157".
After the second operation, s = "862".
After the third operation, s = "48".
Since '4' != '8', the output is false.

What is the most efficient algorithm to solve this problem?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can you clarify the exact sequence of operations performed on the input string 's'?
What are the possible characters in the input string 's', and are there any restrictions on the length of 's'?
If after the operations all digits become the same, what should be returned, `true` or a specific digit?
If no sequence of operations can make all digits equal, what should the return value be?
Could you provide a concrete example of the input string and the expected output after performing the operations?

Brute Force Solution

Approach

The brute force approach is all about trying every possible combination to see if any of them work. We'll go through each possible move, one at a time, until we find a solution or have checked everything. If nothing works, we can conclude that no solution is possible.

Here's how the algorithm would work step-by-step:

First, go through each position in the string and temporarily increase its digit by one.
After making this temporary change, check if the number of times any particular digit appears in the string is the same as the number of times any other digit appears.
If the counts of all the digits are the same, we have found a solution, so we can stop.
If the counts of digits are not the same, undo the temporary increase we made earlier to restore the string to its original state.
Next, repeat this process for every position in the string, incrementing the digit in each position one at a time and checking the digit counts each time.
If, after checking every position and increasing its digit by one, we still haven't found a solution, go through each position in the string again.
This time, temporarily decrease the digit in each position by one.
Again, after making this temporary change, check if the count of any particular digit is the same as the count of any other digit.
If the counts of all the digits are the same, we have found a solution, so we can stop.
If the counts are not the same, undo the temporary decrease and restore the string to its original state.
If, after trying every possible increase and decrease, no combination results in equal digit counts, then there is no solution, and we can report that as the answer.

Code Implementation

def check_if_digits_are_equal_after_ops(number_string):
    string_length = len(number_string)

    for index in range(string_length):
        # Try incrementing each digit and check for equal counts
        temp_number_string = list(number_string)
        original_digit = int(temp_number_string[index])
        temp_number_string[index] = str((original_digit + 1) % 10)
        temp_number_string = "".join(temp_number_string)

        digit_counts = {}
        for digit_char in temp_number_string:
            digit = int(digit_char)
            digit_counts[digit] = digit_counts.get(digit, 0) + 1

        if len(set(digit_counts.values())) <= 1 and len(digit_counts) > 0:
            return True

    for index in range(string_length):
        # Now try decrementing each digit and check for equal counts
        temp_number_string = list(number_string)
        original_digit = int(temp_number_string[index])

        temp_number_string[index] = str((original_digit - 1 + 10) % 10)
        temp_number_string = "".join(temp_number_string)

        digit_counts = {}
        for digit_char in temp_number_string:
            digit = int(digit_char)
            digit_counts[digit] = digit_counts.get(digit, 0) + 1

        # Check if all digit counts are the same
        if len(set(digit_counts.values())) <= 1 and len(digit_counts) > 0:
            return True

    # No operation resulted in equal digit counts
    return False

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through each position in the string (length n) twice: once to increment the digit and once to decrement it. After each increment or decrement, it counts the occurrences of each digit. Counting the occurrences of each digit also takes O(n) time. However, since the digit counts are done within each of the 2n iterations, it's still O(n). Therefore, the overall time complexity is O(n).

Space Complexity

O(1) – The algorithm's space complexity is dominated by the need to count the occurrences of each digit within the input string. The provided plain english explanation does not allocate an array or hashmap to store digit counts but instead suggests checking for equality directly. As such it only needs a few integer variables such as current_digit_count and comparison_digit_count during the digit counting and comparison process. Since the number of variables is constant and independent of the input string length N, the space complexity is O(1).

Optimal Solution

Approach

The problem asks us to perform some math operations (increment or decrement) on certain digits of a string and then check if all the digits become equal. The clever idea is to focus on what happens if we make every digit in the string equal to a particular target digit.

Here's how the algorithm would work step-by-step:

Pick a digit to be the target value that all the digits in the string will potentially become. A good choice is the first digit.
Go through the string, one digit at a time.
For each digit, figure out how many plus or minus operations you would need to perform to make it equal to your target digit. Keep a running total of plus operations and a running total of minus operations.
After going through the whole string, if the total number of plus operations equals the total number of minus operations, it means you can perform these operations to make all the digits equal to the target. If they are not equal, then it is impossible to make all the digits the same as the first digit.
If all digits cannot be made same as the first digit, pick the second digit as the target digit and repeat from step 2. Repeat the process with the third, fourth... digits until the condition is met or all digits are considered as the target digit.
If you have checked all possible target digits and none of them allow you to make all the digits in the string equal, the answer is 'no'. Otherwise it's 'yes'.

Code Implementation

def check_if_digits_are_equal(number):
    number_string = str(number)
    string_length = len(number_string)

    for target_index in range(string_length):
        plus_operations = 0
        minus_operations = 0

        # Set the current digit as the target digit.
        target_digit = int(number_string[target_index])

        for current_index in range(string_length):
            current_digit = int(number_string[current_index])

            if current_digit > target_digit:
                minus_operations += current_digit - target_digit
            elif current_digit < target_digit:
                plus_operations += target_digit - current_digit

        # Check if the number of plus/minus cancels out.
        if plus_operations == minus_operations:
            return 'yes'

    # Return 'no' if no target digit works.
    return 'no'

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each of the n digits in the string as a potential target digit. For each target digit, it iterates through the string again (n digits) to calculate the plus and minus operations needed to transform each digit to the target. Therefore, in the worst case, we have a nested loop structure where the outer loop runs up to n times and the inner loop runs n times. This leads to approximately n*n operations, thus the time complexity is O(n²).

Space Complexity

O(1) – The algorithm uses a fixed number of variables to store the target digit, the counts of plus operations, and the counts of minus operations. These variables consume a constant amount of space, irrespective of the input string's length, denoted as N. Therefore, the auxiliary space complexity is O(1), indicating constant space usage.

Edge Cases

Case	How to Handle
Null or empty input string	Return true since there are no digits to compare.
Single character input string	Return true because there are no other digits to compare with.
String containing non-digit characters	The solution should either throw an error or ignore non-digit characters based on the problem description.
Large input string exceeding memory constraints	Optimize the algorithm to use constant space or process the string in chunks if necessary to avoid memory issues.
All digits in the string are the same	The operations might eventually make all digits equal, so the solution should not return false prematurely.
String with very large numbers as digits (close to integer limits)	Check for integer overflow during digit operations and handle it appropriately (e.g., using larger data types or modulo operations).
String where operations do not converge to equal digits (infinite loop)	Implement a maximum operation count and return false if digits are not equal after reaching the limit to prevent infinite loops.
String with '0' as a digit after operations.	The solution should correctly handle the '0' digit and make sure that other numbers can be reduced to '0' through operations.