How do you solve Sum of Digit Differences of All Pairs on LeetCode?

Sum of Digit Differences of All Pairs (LeetCode #3153) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * d) time complexity and O(n) space complexity. Key insight: Counting occurrences of each digit at each position allows for efficient calculation of differences.

What is the brute force solution for Sum of Digit Differences of All Pairs?

The brute force approach for Sum of Digit Differences of All Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves comparing every pair of integers in the array and counting the digit differences for each pair. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n * d).

What algorithm or data structure is used to solve Sum of Digit Differences of All Pairs?

Sum of Digit Differences of All Pairs uses the following concepts: Array, Hash Table, Math, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sum of Digit Differences of All Pairs?

Always clarify the constraints and properties of the input (e.g., same number of digits). Think about how to reduce the number of comparisons — look for patterns. Practice explaining your thought process clearly; it helps in interviews.

What are common mistakes when solving Sum of Digit Differences of All Pairs?

Not considering the contribution of each digit position separately. Forgetting to handle cases where all digits are the same, which results in zero differences.

#3153

Sum of Digit Differences of All Pairs

Medium

ArrayHash TableMathCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * d)
Space	O(1)	O(n)

💡

Intuition

Time O(n * d)Space O(n)

Instead of comparing every pair, we can count the occurrences of each digit at each position across all numbers. This allows us to calculate the differences more efficiently by leveraging these counts.

⚙️

Algorithm

3 steps

1Step 1: Initialize a count array for each digit position to count occurrences of digits 0-9.
2Step 2: For each number, update the count for each digit in its respective position.
3Step 3: For each digit position, calculate the contribution of that position to the total digit differences using the counts.

solution.py24 lines

1# Full working Python code
2
3def digit_difference(nums):
4    n = len(nums)
5    digit_length = len(str(nums[0]))
6    count = [[0] * 10 for _ in range(digit_length)]
7
8    # Count occurrences of each digit at each position
9    for num in nums:
10        for i, digit in enumerate(str(num)):
11            count[i][int(digit)] += 1
12
13    total_diff = 0
14
15    # Calculate total differences
16    for i in range(digit_length):
17        for d in range(10):
18            if count[i][d] > 0:
19                total_diff += count[i][d] * (n - count[i][d])
20
21    return total_diff
22
23# Example usage
24print(digit_difference([13, 23, 12]))  # Output: 4

ℹ

Complexity note: The time complexity is O(n * d), where n is the number of integers and d is the number of digits (which is constant for this problem). The space complexity is O(n) for the count array.

1Counting occurrences of each digit at each position allows for efficient calculation of differences.
2The digit differences can be computed independently for each digit position.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.