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Sum of Digit Differences of All Pairs

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Arrays

You are given an array nums consisting of positive integers where all integers have the same number of digits.

The digit difference between two integers is the count of different digits that are in the same position in the two integers.

Return the sum of the digit differences between all pairs of integers in nums.

Example 1:

Input: nums = [13,23,12]

Output: 4

Explanation:
We have the following:
- The digit difference between 13 and 23 is 1.
- The digit difference between 13 and 12 is 1.
- The digit difference between 23 and 12 is 2.
So the total sum of digit differences between all pairs of integers is 1 + 1 + 2 = 4.

Example 2:

Input: nums = [10,10,10,10]

Output: 0

Explanation:
All the integers in the array are the same. So the total sum of digit differences between all pairs of integers will be 0.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] < 10⁹
All integers in nums have the same number of digits.

Try coding it in LeetCode

Sum of Digit Differences of All Pairs

Medium

Asked by:

41 views

Topics:

Arrays

You are given an array nums consisting of positive integers where all integers have the same number of digits.

The digit difference between two integers is the count of different digits that are in the same position in the two integers.

Return the sum of the digit differences between all pairs of integers in nums.

Example 1:

Input: nums = [13,23,12]

Output: 4

Example 2:

Input: nums = [10,10,10,10]

Output: 0

Explanation:
All the integers in the array are the same. So the total sum of digit differences between all pairs of integers will be 0.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] < 10⁹
All integers in nums have the same number of digits.

Try coding it in LeetCode

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:

What is the range of values for the digits within each number in the input array?
Can the input array contain negative numbers?
Are the numbers in the array guaranteed to be non-negative integers?
What is the maximum size of the input array?
Are duplicate numbers allowed within the input array, and if so, how should they be handled in the calculation of the differences?

Brute Force Solution

Approach

We need to find the sum of the absolute differences between the largest and smallest digits of all possible pairs of numbers from a given list. The brute force approach simply looks at every possible pair of numbers and calculates the difference, then adds it to a running total.

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

Take the first number from the list.
Pair it with the second number from the list.
Find the largest digit and the smallest digit in the first number.
Find the largest digit and the smallest digit in the second number.
Calculate the absolute difference between the first number's largest digit and smallest digit.
Calculate the absolute difference between the second number's largest digit and smallest digit.
Add these two differences together to get the pair's digit difference.
Keep a running total of these digit differences.
Now pair the first number with the third number, the fourth number, and so on, repeating steps 3-7 for each pair and adding to the running total.
Once you've paired the first number with every other number in the list, move to the second number.
Pair the second number with the third number, the fourth number, and so on (since we've already paired it with the first number). Repeat steps 3-7 for each pair, adding to the running total.
Continue this process, pairing each number in the list with all the numbers that come after it, until you've considered all possible pairs.
The final running total is the answer: the sum of digit differences for all pairs.

Code Implementation

def sum_of_digit_differences_of_all_pairs(numbers):
    total_digit_difference = 0
    number_of_numbers = len(numbers)

    for first_number_index in range(number_of_numbers):
        # Iterate through each number to form the first half of the pair.
        for second_number_index in range(first_number_index + 1, number_of_numbers):
            first_number = numbers[first_number_index]
            second_number = numbers[second_number_index]

            first_number_string = str(first_number)
            second_number_string = str(second_number)

            first_number_largest_digit = int(first_number_string[0])
            first_number_smallest_digit = int(first_number_string[0])

            # Find the largest and smallest digits in the first number
            for digit_character in first_number_string:
                digit = int(digit_character)
                first_number_largest_digit = max(first_number_largest_digit, digit)
                first_number_smallest_digit = min(first_number_smallest_digit, digit)

            second_number_largest_digit = int(second_number_string[0])
            second_number_smallest_digit = int(second_number_string[0])

            # Find the largest and smallest digits in the second number
            for digit_character in second_number_string:
                digit = int(digit_character)
                second_number_largest_digit = max(second_number_largest_digit, digit)
                second_number_smallest_digit = min(second_number_smallest_digit, digit)

            first_number_digit_difference = abs(first_number_largest_digit - first_number_smallest_digit)
            second_number_digit_difference = abs(second_number_largest_digit - second_number_smallest_digit)

            total_digit_difference += first_number_digit_difference + second_number_digit_difference

    return total_digit_difference

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible pairs of numbers in the input list. For a list of n numbers, there are n * (n - 1) / 2 possible pairs. Finding the largest and smallest digits within each number takes constant time. Since the number of pairs is proportional to n squared and all other operations take constant time, the dominant factor is the pair iteration. Thus, the overall time complexity is O(n²).

Space Complexity

O(1) – The provided algorithm iterates through pairs of numbers from the input list but does not create any auxiliary data structures to store intermediate results, such as lists, dictionaries, or sets. It calculates the maximum and minimum digits within each number, but these calculations happen using only a few variables within the loop's scope. The only variables stored are for the loop indices and temporary digit calculations, which take up constant space, regardless of the input list's size N. Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

The key is to focus on each digit position separately. We count how many times each digit appears in each position across all the numbers and then figure out how each digit impacts the total difference.

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

First, go through all the numbers and, for each digit place (like the ones place, tens place, etc.), count how many times each digit (0 through 9) appears.
For each digit place, figure out all the possible pairs of digits that could appear in that place across all the numbers.
For each pair of digits, calculate the difference between them and multiply that difference by how many times each digit appears in that specific place. This will result in the overall contribution of that pair's difference.
Sum up all the contributions from each digit place to get the total sum of digit differences across all pairs of numbers.

Code Implementation

def sum_of_digit_differences(numbers):
    total_sum = 0
    max_digit_length = max(len(str(number)) for number in numbers)

    # Pad numbers with leading zeros for consistent length
    padded_numbers = [str(number).zfill(max_digit_length) for number in numbers]

    for digit_index in range(max_digit_length):
        digit_counts = {digit: 0 for digit in range(10)}

        # Count occurrences of each digit at the current digit place
        for padded_number in padded_numbers:
            digit = int(padded_number[digit_index])
            digit_counts[digit] += 1

        # Iterate through all possible pairs of digits
        for digit1 in range(10):
            for digit2 in range(10):

                #Calculate contribution to the total sum
                difference = abs(digit1 - digit2)
                count1 = digit_counts[digit1]
                count2 = digit_counts[digit2]
                total_sum += difference * count1 * count2

    # Divide by 2 because each pair was counted twice
    return total_sum // 2

Big(O) Analysis

Time Complexity

O(n*log10(max_element)) – The algorithm iterates through each number to count the occurrences of each digit in each position. The number of digits in a number is bounded by log10(max_element), where max_element is the largest number in the input array. The outer loop iterates through the n numbers, and the inner loop iterates through the digits of each number (up to the maximum number of digits), contributing a factor of log10(max_element). Calculating the differences and summing them up takes constant time. Thus, the overall time complexity is O(n * log10(max_element)).

Space Complexity

O(D) – The algorithm uses a count array to store the frequency of each digit (0-9) for each digit place. Since there are a fixed number of digits (10) and the number of digit places, D, depends on the numbers in the input, the auxiliary space used is proportional to the number of digit places, D, which is the maximum number of digits in a number in the input. This results in an array of size 10 * D. The space complexity is therefore O(D), where D is the maximum number of digits in any number in the input array.

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 since there are no pairs to compute the sum of digit differences for.
Array with a single element	Return 0 since a single element cannot form a pair.
Array with maximum allowed size (constrained by memory)	Ensure the algorithm has a time complexity suitable for large arrays (e.g., O(n log n) or better), or utilize a counting sort if applicable given the input constraints.
Array with all identical numbers	The sum of digit differences will be 0 as the digit difference of each pair is 0.
Numbers with very large number of digits causing integer overflow when calculating digit differences	Use appropriate data types (e.g., long long in C++, long in Java, or Python's arbitrary-precision integers) to avoid overflow during digit difference calculation.
Array with numbers having leading zeros	Leading zeros should be removed before calculating digit differences or the algorithm should account for them.
Array containing negative numbers (consider their absolute values for digit difference)	Take the absolute value of each number before computing digit differences.
Numbers exceeding maximum integer limits during intermediate calculations (e.g., summing absolute digit differences across all pairs)	Use a wider integer type (e.g., long long) to store the accumulated sum to prevent overflow during the aggregation of digit differences.

Empty or null input array

How to Handle:

Return 0 since there are no pairs to compute the sum of digit differences for.

Array with a single element

How to Handle:

Return 0 since a single element cannot form a pair.

Array with maximum allowed size (constrained by memory)

How to Handle:

Ensure the algorithm has a time complexity suitable for large arrays (e.g., O(n log n) or better), or utilize a counting sort if applicable given the input constraints.

Array with all identical numbers

How to Handle:

The sum of digit differences will be 0 as the digit difference of each pair is 0.

Numbers with very large number of digits causing integer overflow when calculating digit differences

How to Handle:

Use appropriate data types (e.g., long long in C++, long in Java, or Python's arbitrary-precision integers) to avoid overflow during digit difference calculation.

Array with numbers having leading zeros

How to Handle:

Leading zeros should be removed before calculating digit differences or the algorithm should account for them.

Array containing negative numbers (consider their absolute values for digit difference)

How to Handle:

Take the absolute value of each number before computing digit differences.

Numbers exceeding maximum integer limits during intermediate calculations (e.g., summing absolute digit differences across all pairs)

How to Handle:

Use a wider integer type (e.g., long long) to store the accumulated sum to prevent overflow during the aggregation of digit differences.

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:

What is the range of values for the digits within each number in the input array?
Can the input array contain negative numbers?
Are the numbers in the array guaranteed to be non-negative integers?
What is the maximum size of the input array?
Are duplicate numbers allowed within the input array, and if so, how should they be handled in the calculation of the differences?

Brute Force Solution

Approach

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

Take the first number from the list.
Pair it with the second number from the list.
Find the largest digit and the smallest digit in the first number.
Find the largest digit and the smallest digit in the second number.
Calculate the absolute difference between the first number's largest digit and smallest digit.
Calculate the absolute difference between the second number's largest digit and smallest digit.
Add these two differences together to get the pair's digit difference.
Keep a running total of these digit differences.
Now pair the first number with the third number, the fourth number, and so on, repeating steps 3-7 for each pair and adding to the running total.
Once you've paired the first number with every other number in the list, move to the second number.
Pair the second number with the third number, the fourth number, and so on (since we've already paired it with the first number). Repeat steps 3-7 for each pair, adding to the running total.
Continue this process, pairing each number in the list with all the numbers that come after it, until you've considered all possible pairs.
The final running total is the answer: the sum of digit differences for all pairs.

Code Implementation

def sum_of_digit_differences_of_all_pairs(numbers):
    total_digit_difference = 0
    number_of_numbers = len(numbers)

    for first_number_index in range(number_of_numbers):
        # Iterate through each number to form the first half of the pair.
        for second_number_index in range(first_number_index + 1, number_of_numbers):
            first_number = numbers[first_number_index]
            second_number = numbers[second_number_index]

            first_number_string = str(first_number)
            second_number_string = str(second_number)

            first_number_largest_digit = int(first_number_string[0])
            first_number_smallest_digit = int(first_number_string[0])

            # Find the largest and smallest digits in the first number
            for digit_character in first_number_string:
                digit = int(digit_character)
                first_number_largest_digit = max(first_number_largest_digit, digit)
                first_number_smallest_digit = min(first_number_smallest_digit, digit)

            second_number_largest_digit = int(second_number_string[0])
            second_number_smallest_digit = int(second_number_string[0])

            # Find the largest and smallest digits in the second number
            for digit_character in second_number_string:
                digit = int(digit_character)
                second_number_largest_digit = max(second_number_largest_digit, digit)
                second_number_smallest_digit = min(second_number_smallest_digit, digit)

            first_number_digit_difference = abs(first_number_largest_digit - first_number_smallest_digit)
            second_number_digit_difference = abs(second_number_largest_digit - second_number_smallest_digit)

            total_digit_difference += first_number_digit_difference + second_number_digit_difference

    return total_digit_difference

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, go through all the numbers and, for each digit place (like the ones place, tens place, etc.), count how many times each digit (0 through 9) appears.
For each digit place, figure out all the possible pairs of digits that could appear in that place across all the numbers.
For each pair of digits, calculate the difference between them and multiply that difference by how many times each digit appears in that specific place. This will result in the overall contribution of that pair's difference.
Sum up all the contributions from each digit place to get the total sum of digit differences across all pairs of numbers.

Code Implementation

def sum_of_digit_differences(numbers):
    total_sum = 0
    max_digit_length = max(len(str(number)) for number in numbers)

    # Pad numbers with leading zeros for consistent length
    padded_numbers = [str(number).zfill(max_digit_length) for number in numbers]

    for digit_index in range(max_digit_length):
        digit_counts = {digit: 0 for digit in range(10)}

        # Count occurrences of each digit at the current digit place
        for padded_number in padded_numbers:
            digit = int(padded_number[digit_index])
            digit_counts[digit] += 1

        # Iterate through all possible pairs of digits
        for digit1 in range(10):
            for digit2 in range(10):

                #Calculate contribution to the total sum
                difference = abs(digit1 - digit2)
                count1 = digit_counts[digit1]
                count2 = digit_counts[digit2]
                total_sum += difference * count1 * count2

    # Divide by 2 because each pair was counted twice
    return total_sum // 2

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 since there are no pairs to compute the sum of digit differences for.
Array with a single element	Return 0 since a single element cannot form a pair.
Array with maximum allowed size (constrained by memory)	Ensure the algorithm has a time complexity suitable for large arrays (e.g., O(n log n) or better), or utilize a counting sort if applicable given the input constraints.
Array with all identical numbers	The sum of digit differences will be 0 as the digit difference of each pair is 0.
Numbers with very large number of digits causing integer overflow when calculating digit differences	Use appropriate data types (e.g., long long in C++, long in Java, or Python's arbitrary-precision integers) to avoid overflow during digit difference calculation.
Array with numbers having leading zeros	Leading zeros should be removed before calculating digit differences or the algorithm should account for them.
Array containing negative numbers (consider their absolute values for digit difference)	Take the absolute value of each number before computing digit differences.
Numbers exceeding maximum integer limits during intermediate calculations (e.g., summing absolute digit differences across all pairs)	Use a wider integer type (e.g., long long) to store the accumulated sum to prevent overflow during the aggregation of digit differences.

Empty or null input array

How to Handle:

Return 0 since there are no pairs to compute the sum of digit differences for.

Array with a single element

How to Handle:

Return 0 since a single element cannot form a pair.

Array with maximum allowed size (constrained by memory)

How to Handle:

Ensure the algorithm has a time complexity suitable for large arrays (e.g., O(n log n) or better), or utilize a counting sort if applicable given the input constraints.

Array with all identical numbers

How to Handle:

The sum of digit differences will be 0 as the digit difference of each pair is 0.

Numbers with very large number of digits causing integer overflow when calculating digit differences

How to Handle:

Use appropriate data types (e.g., long long in C++, long in Java, or Python's arbitrary-precision integers) to avoid overflow during digit difference calculation.

Array with numbers having leading zeros

How to Handle:

Leading zeros should be removed before calculating digit differences or the algorithm should account for them.

Array containing negative numbers (consider their absolute values for digit difference)

How to Handle:

Take the absolute value of each number before computing digit differences.

Numbers exceeding maximum integer limits during intermediate calculations (e.g., summing absolute digit differences across all pairs)

How to Handle:

Use a wider integer type (e.g., long long) to store the accumulated sum to prevent overflow during the aggregation of digit differences.