Abbreviating the Product of a Range - Practice Coding Problems

Abbreviating the Product of a Range - Problem

Math Hard

Abbreviating the Product of a Range

You're given two positive integers left and right where left ≤ right. Your task is to calculate the product of all integers in the inclusive range [left, right] and then abbreviate it due to its potentially massive size.

The Abbreviation Process:

1. Remove Trailing Zeros: Count all trailing zeros in the product and remove them. Let C be this count.
- Example: 1000 has 3 trailing zeros, 546 has 0 trailing zeros

2. Format Large Numbers: If the remaining number has more than 10 digits (d > 10), express it as <pre>...<suf> where:
- <pre> = first 5 digits
- <suf> = last 5 digits (after removing trailing zeros)
- If d ≤ 10, keep the number unchanged

3. Final Format: Return the result as "<pre>...<suf>eC"
- Example: 12345678987600000 becomes "12345...89876e5"

This problem tests your ability to handle large number computations, mathematical properties of factorials, and string formatting.

Input & Output

example_1.py — Basic Range

$ Input: left = 1, right = 4

› Output: "24e0"

💡 Note: Product is 1×2×3×4 = 24. No trailing zeros, and 24 has only 2 digits (≤ 10), so result is "24e0".

example_2.py — Range with Trailing Zeros

$ Input: left = 2, right = 11

› Output: "399168e2"

💡 Note: Product is 39916800. Has 2 trailing zeros. After removing zeros: 399168 (6 digits ≤ 10), so result is "399168e2".

example_3.py — Large Range

$ Input: left = 371, right = 375

› Output: "7219...70176e1"

💡 Note: Product has many digits. After removing 1 trailing zero, the number has >10 digits, so we show first 5 digits (72199) and last 5 digits (70176) with format "7219...70176e1".

Visualization

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Understanding the Visualization

Multiply the Range

We need 1×2×3×...×n but the result is too big for normal computation

Count the Zeros

Trailing zeros come from factors of 10 = 2×5, so count factors of 2 and 5

Find First Digits

Use logarithms: log₁₀(a×b) = log₁₀(a) + log₁₀(b) to avoid overflow

Find Last Digits

Use modular arithmetic: (a×b) mod 10⁵ to get last 5 digits

Format Result

Combine as 'first...last' + 'e' + zero_count

Key Takeaway

🎯 Key Insight: By separating the problem into counting zeros (factors of 2×5), finding first digits (logarithms), and last digits (modular arithmetic), we avoid computing the massive product entirely while maintaining perfect accuracy!

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the range to count factors and compute logarithmic/modular values

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra space for computations

✓ Linear Space

Constraints

1 ≤ left ≤ right ≤ 10⁴
The product can be extremely large (factorial-like growth)
Mathematical precision required for logarithmic calculations

Asked in

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

The optimal solution uses mathematical properties to avoid integer overflow: count trailing zeros by counting factors of 2 and 5, use logarithms to find the first 5 digits, and modular arithmetic to find the last 5 digits. This approach handles any range size efficiently in O(n) time and O(1) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Multiplication	O(n × log(product))	O(log(product))	Directly multiply all numbers and format the result
Mathematical Approach (Optimal)	O(n)	O(1)	Use logarithms and modular arithmetic to avoid overflow

Brute Force Multiplication — Algorithm Steps

Initialize product = 1
Multiply all numbers from left to right
Convert to string and count trailing zeros
Remove trailing zeros and format if length > 10
Return formatted string with zero count

Visualization

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Step-by-Step Walkthrough

Multiply Range

Multiply numbers 1×2×3×...×n sequentially

Count Zeros

Count trailing zeros in the result

Format Result

Apply abbreviation rules based on digit count

Code -

solution.c — C

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

// Note: C doesn't have built-in big integer support
// This is a simplified version that will overflow
char* abbreviateProduct(int left, int right) {
    long long product = 1;
    for (int i = left; i <= right; i++) {
        if (product > 1e15 / i) {
            // Overflow protection
            char* result = malloc(20);
            strcpy(result, "overflow");
            return result;
        }
        product *= i;
    }
    
    char productStr[100];
    sprintf(productStr, "%lld", product);
    
    // Count trailing zeros
    int len = strlen(productStr);
    int trailingZeros = 0;
    for (int i = len - 1; i >= 0; i--) {
        if (productStr[i] == '0') {
            trailingZeros++;
        } else {
            break;
        }
    }
    
    // Remove trailing zeros
    if (trailingZeros > 0) {
        productStr[len - trailingZeros] = '\0';
        len -= trailingZeros;
    }
    
    char* result = malloc(50);
    if (len <= 10) {
        sprintf(result, "%se%d", productStr, trailingZeros);
    } else {
        char prefix[6], suffix[6];
        strncpy(prefix, productStr, 5);
        prefix[5] = '\0';
        strcpy(suffix, productStr + len - 5);
        sprintf(result, "%s...%se%d", prefix, suffix, trailingZeros);
    }
    
    return result;
}

int main() {
    int left, right;
    scanf("%d %d", &left, &right);
    char* result = abbreviateProduct(left, right);
    printf("%s\n", result);
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × log(product))

O(n) multiplications where each multiplication takes O(log(product)) time due to growing number size

⚡ Linearithmic

Space Complexity

O(log(product))

Space to store the product which can have logarithmic digits in the range size

✓ Linear Space

Constraints

1 ≤ left ≤ right ≤ 10⁴
The product can be extremely large (factorial-like growth)
Mathematical precision required for logarithmic calculations

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Brute Force Multiplication — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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