What is the time complexity of Sum of Digit Differences of All Pairs?

The optimal solution has a time complexity of O(d*n).

What data structures are used to solve Sum of Digit Differences of All Pairs?

This problem uses Array, Hash Table, Math, Counting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Sum of Digit Differences of All Pairs?

Insight 1: Calculating the digit difference between each pair directly would result in O(n^2 * d) time complexity where n is the number of elements and d is the number of digits in each number, which is not optimal. Insight 2: Instead of comparing each pair directly, we can iterate through each digit position and count the occurrences of each digit (0-9) at that position. Then, for each position, the sum of differences can be calculated by counting how many numbers have been processed * the number of numbers that have the current number at that position.

Which companies ask LeetCode 3153. Sum of Digit Differences of All Pairs in interviews?

This problem is frequently asked at Turing.

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Sum of Digit Differences of All Pairs - LeetCode 3153 Solution

Sum of Digit Differences of All Pairs - Complete Solution Guide

Sum of Digit Differences of All Pairs is LeetCode problem 3153, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 1 3 and 2 3 is 1. - The digit difference between 1 3 and 1 2 is 1. - T

Detailed Explanation

The problem asks us to calculate the sum of digit differences between all pairs of numbers in a given array `nums`. The digit difference between two numbers is the number of digits that are different in the same position. All numbers in the array have the same number of digits. For example, if `nums = [13, 23, 12]`, the digit differences are: (13, 23) -> 1 (first digit different), (13, 12) -> 1 (second digit different), (23, 12) -> 2 (both digits different). The sum is 1 + 1 + 2 = 4.

Solution Approach

The solution iterates through each digit position of the numbers in the input array. For each position, it counts the occurrences of each digit (0-9) using a frequency array `counts`. Then, it iterates through the `counts` array to calculate the total digit differences for that specific position. The core idea is that for a given digit value, it multiplies the count of that digit value with the numbers that have already been seen (and are different, since they have other digit values). This sum is accumulated into the total result.

Step-by-Step Algorithm

Step 1: Determine the number of digits in each number. This is done by finding the length of the string representation of the first number.
Step 2: Initialize the `total_diff` variable to 0. This variable will store the sum of the digit differences between all pairs.
Step 3: Iterate through each digit position from right to left (least significant digit to most significant digit).
Step 4: For each digit position, create a `counts` array of size 10 (for digits 0-9) and initialize all elements to 0.
Step 5: Iterate through each number in the input array `nums`.
Step 6: Extract the digit at the current position from the current number. This is done by using integer division and modulo operation.
Step 7: Increment the count for that digit in the `counts` array.
Step 8: After iterating through all numbers in `nums`, calculate the digit difference for the current position.
Step 9: Iterate through the `counts` array. Maintain a `n_processed` variable that is initialized to 0. In each count, multiply the current count with `n_processed`, add this number to `diff_at_pos`, and increment `n_processed` by the current count.
Step 10: Add the `diff_at_pos` to the `total_diff`.
Step 11: Multiply `power_of_10` by 10.
Step 12: After iterating through all digit positions, return the `total_diff`.

Key Insights

Insight 1: Calculating the digit difference between each pair directly would result in O(n^2 * d) time complexity where n is the number of elements and d is the number of digits in each number, which is not optimal.
Insight 2: Instead of comparing each pair directly, we can iterate through each digit position and count the occurrences of each digit (0-9) at that position. Then, for each position, the sum of differences can be calculated by counting how many numbers have been processed * the number of numbers that have the current number at that position.
Insight 3: Using a counter array (frequency map) for each digit position allows us to efficiently determine the differences without pairwise comparisons, reducing the complexity.

Complexity Analysis

Time Complexity: O(d*n)

Space Complexity: O(1)

Topics

This problem involves: Array, Hash Table, Math, Counting.

Companies

Asked at: Turing.

Sum of Digit Differences of All Pairs | LeetCode 3153

Medium

Array Hash Table Math Counting

Time: O(d*n)Space: O(1)

Solution Code

Available in:

class Solution {
    public long sumDigitDifferences(int[] nums) {
        int n = nums.length;
        if (n <= 1) {
            return 0;
        }
        
        int num_digits = String.valueOf(nums[0]).length();
        
        long total_diff = 0;
        int power_of_10 = 1;
        
        for (int i = 0; i < num_digits; i++) {
            int[] counts = new int[10];
            for (int num : nums) {
                int digit = (num / power_of_10) % 10;
                counts[digit] += 1;
            }
            
            long diff_at_pos = 0;
            long n_processed = 0;
            for (int count : counts) {
                diff_at_pos += (long)count * n_processed;
                n_processed += count;
            }
            
            total_diff += diff_at_pos;
            power_of_10 *= 10;
        }
        
        return total_diff;
    }
}

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Problem Info

Description

Problem 3153: Sum of Digit Differences of All Pairs - Detailed explanation and solution approaches.

Topics

Array Hash Table Math Counting

Companies

Turing

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:3153

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

12 steps

1
Step 1: Determine the number of digits in each number. This is done by finding the length of the string representation of the first number.
2
Step 2: Initialize the `total_diff` variable to 0. This variable will store the sum of the digit differences between all pairs.
3
Step 3: Iterate through each digit position from right to left (least significant digit to most significant digit).
4
Step 4: For each digit position, create a `counts` array of size 10 (for digits 0-9) and initialize all elements to 0.
5
Step 5: Iterate through each number in the input array `nums`.
6
Step 6: Extract the digit at the current position from the current number. This is done by using integer division and modulo operation.
7
Step 7: Increment the count for that digit in the `counts` array.
8
Step 8: After iterating through all numbers in `nums`, calculate the digit difference for the current position.
9
Step 9: Iterate through the `counts` array. Maintain a `n_processed` variable that is initialized to 0. In each count, multiply the current count with `n_processed`, add this number to `diff_at_pos`, and increment `n_processed` by the current count.
10
Step 10: Add the `diff_at_pos` to the `total_diff`.
11
Step 11: Multiply `power_of_10` by 10.
12
Step 12: After iterating through all digit positions, return the `total_diff`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(d*n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operations are the nested loops: one iterating through the digits (d) and the other iterating through the input array (n). The space is constant because the `counts` array's size is fixed.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute-Force Approach

Compare each pair of numbers in the array and calculate the digit difference for each pair. Sum the differences.

Time: O(n^2 * d), where n is the number of integers and d is the number of digits.Space: O(1)

Trade-offs: This approach is simple to understand but is less efficient than the optimized solution, especially for large input arrays. The optimized approach uses O(n*d). Use brute force when memory is extremely restricted but n is small.

✨

Key Insights

1Insight 1: Calculating the digit difference between each pair directly would result in O(n^2 * d) time complexity where n is the number of elements and d is the number of digits in each number, which is not optimal.
2Insight 2: Instead of comparing each pair directly, we can iterate through each digit position and count the occurrences of each digit (0-9) at that position. Then, for each position, the sum of differences can be calculated by counting how many numbers have been processed * the number of numbers that have the current number at that position.
3Insight 3: Using a counter array (frequency map) for each digit position allows us to efficiently determine the differences without pairwise comparisons, reducing the complexity.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: If the input array `nums` has only one element or is empty, the sum of digit differences is 0. The code handles this by checking `if n <= 1: return 0`.
Edge case 2: If all the numbers in the array are identical, the sum of digit differences should be 0. The solution handles this case correctly because the counts for each digit position will result in no differences, resulting in a `total_diff` of 0.
Edge case 3: All numbers consisting of single digits (e.g. [1,2,3]): works fine. One inner loop computes (0*count[0] + count[1] + ...) and each contribution is precisely the number of differences between an earlier number, and later numbers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case where the input array has only one or zero elements, resulting in incorrect results.

✗Common mistake 2: Incorrectly calculating the digit difference at each position, leading to incorrect results. The `n_processed` is required for correctness.

✗Common mistake 3: Not handling potential integer overflow issues when calculating the sum of differences, especially in languages like Java or C++. While the constraints on number size are tight, using long for the running total and individual counts helps to eliminate errors.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Data cleaning/deduplication: In databases, you might want to find records that are 'mostly' similar to each other. Digit difference is a simplified version of more general string similarity metrics.

🏦Real-world scenario 2: Image processing: Comparing pixel values and identifying patterns in images might involve analyzing the differences between pixel values, which can be conceptually similar to digit differences (although image pixel values are often not integers).

🎮Real-world scenario 3: Genetic sequence comparison: Comparing DNA/RNA sequences often involves identifying the differences between individual elements of the sequences.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 2, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Sum of Digit Differences of All Pairs"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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