Sum of Digit Differences of All Pairs - Practice Coding Problems

Sum of Digit Differences of All Pairs - Problem

You are given an array nums consisting of positive integers where all integers have the same number of digits. Your task is to calculate the total sum of digit differences between all possible pairs of integers in the array.

The digit difference between two integers is defined as the count of positions where the corresponding digits are different. For example, the digit difference between 123 and 153 is 1 (only the middle digit differs: 2 vs 5).

Given an array like [13, 23, 12], you need to compare:

13 vs 23: differ at position 0 (1 vs 2) → difference = 1
13 vs 12: differ at position 1 (3 vs 2) → difference = 1
23 vs 12: differ at both positions → difference = 2

Return the sum of all digit differences: 1 + 1 + 2 = 4

Input & Output

example_1.py — Basic Case

$ Input: nums = [13, 23, 12]

› Output: 4

💡 Note: Comparing pairs: (13,23) differ at position 0 (1≠2) → 1 difference. (13,12) differ at position 1 (3≠2) → 1 difference. (23,12) differ at positions 0 and 1 (2≠1, 3≠2) → 2 differences. Total: 1+1+2=4

example_2.py — All Same Digits

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

› Output: 0

💡 Note: All numbers are identical, so no digit differences exist between any pairs. Total sum is 0.

example_3.py — Maximum Differences

$ Input: nums = [123, 456]

› Output: 3

💡 Note: Only one pair (123,456). They differ at all 3 positions: 1≠4, 2≠5, 3≠6. Total differences = 3.

Constraints

2 ≤ nums.length ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
All integers have the same number of digits
All integers are positive

Visualization

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Understanding the Visualization

Collect Badges

Gather all student ID badges with same digit length

Position Analysis

For each digit position, group badges by their digit value

Count Differences

Calculate how many pairs have different digits at each position

Sum Total

Add up differences from all positions to get final count

Key Takeaway

🎯 Key Insight: Instead of comparing every pair individually (O(n²)), count how many numbers have each digit at each position, then multiply counts of different digits to calculate total differences mathematically (O(n)).

Asked in

G Google 35 a Amazon 28 ⊞ Microsoft 22 f Meta 18

The optimal approach uses digit frequency counting instead of comparing every pair individually. For each digit position, count how many numbers have each digit (0-9), then multiply counts of different digits to get total differences. This reduces time complexity from O(n²) to O(n×d) where d is the number of digits.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n² × d)	O(1)	Compare every pair of numbers digit by digit
Optimal Counting Approach	O(n × d)	O(d)	Count digit frequencies at each position and calculate differences mathematically

Brute Force (Nested Loops) — Algorithm Steps

Convert all numbers to strings for easy digit access
Use nested loops to generate all pairs (i, j) where i < j
For each pair, compare digits at each position
Count positions where digits differ
Add the count to the total sum

Visualization

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Step-by-Step Walkthrough

Generate Pairs

Create all possible pairs (i,j) where i < j

Compare Digits

For each pair, compare digits at each position

Count Differences

Sum up positions where digits differ

Accumulate

Add difference count to total sum

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

long long sumDigitDifferences(int* nums, int numsSize) {
    long long total = 0;
    
    // Compare all pairs
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            char str1[20], str2[20];
            sprintf(str1, "%d", nums[i]);
            sprintf(str2, "%d", nums[j]);
            
            int diffCount = 0;
            for (int k = 0; k < strlen(str1); k++) {
                if (str1[k] != str2[k]) {
                    diffCount++;
                }
            }
            total += diffCount;
        }
    }
    
    return total;
}

Time & Space Complexity

Time Complexity

⏱️

O(n² × d)

n² for all pairs, d for comparing digits at each position (d is number of digits)

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Medium Frequency

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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