Duplicate Invoice Detector - Problem

You work in the accounting department and need to detect potential duplicate invoices to prevent double payments. You're given a list of invoices, where each invoice contains three fields:

amount - the invoice amount (decimal)
vendor - the vendor name (string)
date - the invoice date (string in YYYY-MM-DD format)

Two invoices are considered potential duplicates if they match on any two of the three fields. For example:

Same amount + same vendor (different dates)
Same amount + same date (different vendors)
Same vendor + same date (different amounts)

Return a list of all invoices that have at least one potential duplicate in the dataset. Each invoice should appear only once in the result, even if it matches multiple other invoices.

Note: Each invoice is represented as [amount, vendor, date] where amount is a float.

Input & Output

Example 1 — Basic Duplicate Detection

$ Input: invoices = [[1000.0, "Apple", "2024-01-15"], [1000.0, "Google", "2024-01-15"], [500.0, "Apple", "2024-01-16"]]

› Output: [[1000.0, "Apple", "2024-01-15"], [1000.0, "Google", "2024-01-15"]]

💡 Note: First two invoices share same amount ($1000) and same date (2024-01-15), so both are flagged as potential duplicates. Third invoice shares no 2+ fields with others.

Example 2 — Multiple Duplicate Groups

$ Input: invoices = [[800.0, "Microsoft", "2024-01-10"], [800.0, "Microsoft", "2024-01-11"], [600.0, "Oracle", "2024-01-10"], [700.0, "Oracle", "2024-01-10"]]

› Output: [[800.0, "Microsoft", "2024-01-10"], [800.0, "Microsoft", "2024-01-11"], [600.0, "Oracle", "2024-01-10"], [700.0, "Oracle", "2024-01-10"]]

💡 Note: First two invoices match on amount+vendor. Last two invoices match on vendor+date. All four invoices have duplicates.

Example 3 — No Duplicates Found

$ Input: invoices = [[1000.0, "Apple", "2024-01-15"], [2000.0, "Google", "2024-01-16"], [1500.0, "Microsoft", "2024-01-17"]]

› Output: []

💡 Note: No two invoices share 2+ matching fields, so no duplicates detected.

Constraints

1 ≤ invoices.length ≤ 1000
-10⁶ ≤ amount ≤ 10⁶
1 ≤ vendor.length ≤ 50
date is in YYYY-MM-DD format

Visualization

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Asked in

G Google 35 a Amazon 28 M Microsoft 22 A Apple 18

The key insight is creating field pair fingerprints to detect duplicates efficiently. Instead of comparing every invoice pair (O(n²)), we use hash maps to store fingerprints like "amount_vendor" and detect collisions in O(1) time. Best approach: Hash Map with Fingerprints. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force Comparison

⏱️ Time: O(n²) Space: O(n)

For each invoice, compare it against every other invoice to see if they share exactly two fields. If a match is found, mark both invoices as duplicates.

Hash Map with Field Pair Fingerprints

⏱️ Time: O(n) Space: O(n)

Generate three possible fingerprints for each invoice (amount+vendor, amount+date, vendor+date) and use hash maps to quickly find invoices sharing the same fingerprints.

Brute Force Comparison — Algorithm Steps

Step 1: Take each invoice and compare with all other invoices
Step 2: For each pair, count matching fields (amount, vendor, date)
Step 3: If exactly 2 fields match, both invoices are potential duplicates
Step 4: Collect all flagged invoices, removing duplicates from result

Visualization

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Step-by-Step Walkthrough

Pick First Invoice

Start with invoice[0] and compare with all others

Count Matches

Check amount, vendor, date fields - count matching ones

Flag Duplicates

If 2+ fields match, both invoices are potential duplicates

Code -

solution.c — C

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

// Simplified implementation for demonstration
char* solution(char* input) {
    // Parse invoices and compare pairs
    // Return JSON string of duplicates
    return "[]";
}

int main() {
    char input[10000];
    fgets(input, sizeof(input), stdin);
    input[strcspn(input, "\n")] = 0;
    
    char* result = solution(input);
    printf("%s\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops compare every invoice pair, n×n comparisons total

⚠ Quadratic Growth

Space Complexity

O(n)

Set to store result invoices, up to n elements

⚡ Linearithmic Space

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Medium Frequency

~15 min Avg. Time

892 Likes

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Duplicate Invoice Detector - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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