What is the time complexity of Abbreviating the Product of a Range?

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

What data structures are used to solve Abbreviating the Product of a Range?

This problem uses Math. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Abbreviating the Product of a Range?

Insight 1: Trailing zeros in a product are determined by the number of pairs of 2 and 5 in the prime factorization of the numbers being multiplied. Counting the number of factors of 2 and 5 is crucial. Insight 2: Directly calculating the product can lead to overflow issues. Using logarithms to estimate the leading digits and modular arithmetic to obtain the trailing digits is essential to avoid integer overflow.

Which companies ask LeetCode 2117. Abbreviating the Product of a Range in interviews?

This problem is frequently asked at Avalara.

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Abbreviating the Product of a Range - LeetCode 2117 Solution

Abbreviating the Product of a Range - Complete Solution Guide

Abbreviating the Product of a Range is LeetCode problem 2117, a Hard 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 two positive integers left and right with left <= right . Calculate the product of all integers in the inclusive range [left, right] . Since the product may be very large, you will abbreviate it following these steps: Count all trailing zeros in the product and remove them. Let us denote this count as C . For example, there are 3 trailing zeros in 1000 , and there are 0 trailing zeros in 546 . Denote the remaining number of digits in the product as d . If d > 10 , then express the

Detailed Explanation

The problem requires calculating the product of all integers within a given inclusive range [left, right]. Since the product can be very large, it needs to be abbreviated. The abbreviation involves counting and removing trailing zeros (denoted as 'C'), and then representing the remaining number based on its digit count. If the digit count ('d') is greater than 10, the number is represented as <pre>...<suf>, where <pre> is the first 5 digits and <suf> is the last 5 digits after removing trailing zeros. If 'd' is less than or equal to 10, the number is kept as is. Finally, the result is formatted as <pre>...<suf>eC or <number>eC.

Solution Approach

The solution approach involves the following steps: first, count the number of 2 and 5 factors present in each number within the given range. The minimum of these counts gives the number of trailing zeros, C. Then, calculate the product without the trailing zeros in two separate ways: Using logarithm and power to get the first 5 digits (pre), and using modular arithmetic to get the last 5 digits (suf). Finally, construct the result string according to the problem specification based on if the number of digits in the product (after removing trailing zeros) is > 10.

Step-by-Step Algorithm

Step 1: Initialize counters `count_2` and `count_5` to 0, and `log_prod` to 0.0. These will store the counts of factors of 2 and 5, and the sum of logarithms of the numbers, respectively.
Step 2: Iterate through the range [left, right]. For each number `i`, increment `log_prod` by `log10(i)`. Also, factorize `i` to count the powers of 2 and 5, incrementing `count_2` and `count_5` accordingly.
Step 3: Calculate `C` as the minimum of `count_2` and `count_5`. This represents the number of trailing zeros.
Step 4: Calculate `log_P_prime` as `log_prod - C`. This gives an approximation of the logarithm of the product after removing trailing zeros.
Step 5: Check if `log_P_prime` is greater than or equal to 10. If true, proceed to abbreviate the product.
Step 6: Calculate the first 5 digits (`pre`) using the fractional part of `log_P_prime`. The fractional part is obtained by subtracting the floor of `log_P_prime` from `log_P_prime`. Then, `pre` is computed as `int(10^(fractional_part) * 10000)`.
Step 7: Calculate the last 5 digits (`suf`) using modular arithmetic. Initialize `suf` to 1 and the modulus `MOD_suf` to 10^10. Iterate through the range [left, right], multiplying `suf` by each number after removing factors of 2 and 5, all modulo `MOD_suf`.
Step 8: Account for the remaining factors of 2 and 5 (rem_2 and rem_5) by multiplying `suf` by `2^(rem_2)` and `5^(rem_5)` modulo `MOD_suf`.
Step 9: Format `suf` to a string with leading zeros if necessary (using `zfill` in Python, `String.format` in Java, or manual zero padding in C++ and C) to ensure it has 5 digits. Then construct result string like: `f"{pre}...{suf_str}e{C}"`
Step 10: If `log_P_prime` is less than 10, calculate the product `p_prime` directly, removing factors of 2 and 5 from each number and storing the remaining product. Then account for the remaining factors of 2 and 5.
Step 11: Construct the result string as `f"{p_prime}e{C}"`.

Key Insights

Insight 1: Trailing zeros in a product are determined by the number of pairs of 2 and 5 in the prime factorization of the numbers being multiplied. Counting the number of factors of 2 and 5 is crucial.
Insight 2: Directly calculating the product can lead to overflow issues. Using logarithms to estimate the leading digits and modular arithmetic to obtain the trailing digits is essential to avoid integer overflow.
Insight 3: When the total digits exceeds 10, independently calculating the first 5 digits and last 5 digits after zero removal allows for abbreviation.
Insight 4: Modular exponentiation efficiently calculates large powers modulo a given number, which is important for calculating the trailing digits considering the remaining factors of 2 and 5 after zero removal.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Math.

Companies

Asked at: Avalara.

Abbreviating the Product of a Range | LeetCode 2117

Hard

Math

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

Solution Code

Available in:

import java.lang.Math;

class Solution {
    public String abbreviateProduct(int left, int right) {
        int count_2 = 0;
        int count_5 = 0;
        double log_prod = 0.0;

        for (int i = left; i <= right; i++) {
            log_prod += Math.log10(i);
            int num = i;
            while (num % 2 == 0) {
                count_2++;
                num /= 2;
            }
            while (num % 5 == 0) {
                count_5++;
                num /= 5;
            }
        }

        int C = Math.min(count_2, count_5);
        
        double log_P_prime = log_prod - C;
        
        if (log_P_prime >= 10) {
            double fractional_part = log_P_prime - Math.floor(log_P_prime);
            double pre_val = Math.pow(10, fractional_part);
            int pre = (int)(pre_val * 10000);
            
            long suf = 1;
            long MOD_suf = (long)Math.pow(10, 10);
            
            int rem_2 = count_2 - C;
            int rem_5 = count_5 - C;
            
            for (int i = left; i <= right; i++) {
                long num = i;
                while (num % 2 == 0) {
                    num /= 2;
                }
                while (num % 5 == 0) {
                    num /= 5;
                }
                suf = (suf * num) % MOD_suf;
            }
            
            suf = (suf * power(2, rem_2, MOD_suf)) % MOD_suf;
            suf = (suf * power(5, rem_5, MOD_suf)) % MOD_suf;
            
            String suf_str = String.format("%05d", (int)(suf % 100000));
            
            return String.format("%d...%se%d", pre, suf_str, C);
        } else {
            long p_prime = 1;
            for (int i = left; i <= right; i++) {
                long num = i;
                while (num % 2 == 0) {
                    num /= 2;
                }
                while (num % 5 == 0) {
                    num /= 5;
                }
                p_prime *= num;
            }

            int rem_2 = count_2 - C;
            int rem_5 = count_5 - C;
            
            p_prime *= power(2, rem_2, Long.MAX_VALUE);
            p_prime *= power(5, rem_5, Long.MAX_VALUE);
            
            return String.format("%de%d", p_prime, C);
        }
    }
    
    private long power(long base, long exp, long mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % mod;
            base = (base * base) % mod;
            exp /= 2;
        }
        return res;
    }
}

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Problem Info

Description

Problem 2117: Abbreviating the Product of a Range - Detailed explanation and solution approaches.

Topics

Math

Companies

Avalara

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2117

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

11 steps

1
Step 1: Initialize counters `count_2` and `count_5` to 0, and `log_prod` to 0.0. These will store the counts of factors of 2 and 5, and the sum of logarithms of the numbers, respectively.
2
Step 2: Iterate through the range [left, right]. For each number `i`, increment `log_prod` by `log10(i)`. Also, factorize `i` to count the powers of 2 and 5, incrementing `count_2` and `count_5` accordingly.
3
Step 3: Calculate `C` as the minimum of `count_2` and `count_5`. This represents the number of trailing zeros.
4
Step 4: Calculate `log_P_prime` as `log_prod - C`. This gives an approximation of the logarithm of the product after removing trailing zeros.
5
Step 5: Check if `log_P_prime` is greater than or equal to 10. If true, proceed to abbreviate the product.
6
Step 6: Calculate the first 5 digits (`pre`) using the fractional part of `log_P_prime`. The fractional part is obtained by subtracting the floor of `log_P_prime` from `log_P_prime`. Then, `pre` is computed as `int(10^(fractional_part) * 10000)`.
7
Step 7: Calculate the last 5 digits (`suf`) using modular arithmetic. Initialize `suf` to 1 and the modulus `MOD_suf` to 10^10. Iterate through the range [left, right], multiplying `suf` by each number after removing factors of 2 and 5, all modulo `MOD_suf`.
8
Step 8: Account for the remaining factors of 2 and 5 (rem_2 and rem_5) by multiplying `suf` by `2^(rem_2)` and `5^(rem_5)` modulo `MOD_suf`.
9
Step 9: Format `suf` to a string with leading zeros if necessary (using `zfill` in Python, `String.format` in Java, or manual zero padding in C++ and C) to ensure it has 5 digits. Then construct result string like: `f"{pre}...{suf_str}e{C}"`
10
Step 10: If `log_P_prime` is less than 10, calculate the product `p_prime` directly, removing factors of 2 and 5 from each number and storing the remaining product. Then account for the remaining factors of 2 and 5.
11
Step 11: Construct the result string as `f"{p_prime}e{C}"`.

💡

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(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the loop iterating through the range [left, right]. The operations inside the loop, such as factoring, are constant-time operations. Therefore, the overall time complexity is O(n). The space complexity is constant because only a fixed number of variables are used to store intermediate results, regardless of the input size.

📖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

Using BigInteger (Java) or similar libraries

Instead of logarithms and modular arithmetic, we could use BigInteger (in Java) or similar large integer libraries in other languages to calculate the full product without overflow. Then, we would manually count trailing zeros, remove them, and extract the prefix and suffix as required.

Time: O(n + k) where n is the range and k is the number of digits after multiplication.Space: O(k) to store the large integer.

Trade-offs: This approach can be simpler to implement initially but is generally slower because BigInteger operations are more expensive than native integer operations. This is more memory-intensive if the number of digits are very large.

✨

Key Insights

1Insight 1: Trailing zeros in a product are determined by the number of pairs of 2 and 5 in the prime factorization of the numbers being multiplied. Counting the number of factors of 2 and 5 is crucial.
2Insight 2: Directly calculating the product can lead to overflow issues. Using logarithms to estimate the leading digits and modular arithmetic to obtain the trailing digits is essential to avoid integer overflow.
3Insight 3: When the total digits exceeds 10, independently calculating the first 5 digits and last 5 digits after zero removal allows for abbreviation.
4Insight 4: Modular exponentiation efficiently calculates large powers modulo a given number, which is important for calculating the trailing digits considering the remaining factors of 2 and 5 after zero removal.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: left and right are close, such that the product is small and the number of digits after zero removal is less than 10. The solution correctly handles this by computing the actual product and appending "eC".
Edge case 2: Large range where the product is extremely large. This is correctly handled by using logarithms to obtain the leading digits, and modular arithmetic to obtain trailing digits, avoiding potential overflow during calculation of the overall product.
Edge case 3: When there are more factors of 2 than 5, or vice-versa. The code correctly calculates C as the minimum, and keeps track of the remaining factors which are later incorporated into the trailing digits calculation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow when directly calculating the product of numbers within the range. The product can become extremely large even for moderately sized ranges, leading to incorrect results if not handled properly.

✗Common mistake 2: Incorrect calculation of the number of trailing zeros. Failing to recognize that trailing zeros are determined by the minimum number of 2s and 5s as factors is a common error.

✗Common mistake 3: Incorrect string formatting. Forgetting to pad the suffix with leading zeros if it has less than 5 digits can lead to an incorrect abbreviated product.

✅

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: Scientific simulations often deal with extremely large or small numbers. Abbreviating products in this manner is useful for presenting results in a human-readable format without losing precision.

🏦Real-world scenario 2: Cryptography involves calculations with very large numbers, and abbreviated representations can be useful in debugging or monitoring the intermediate results of cryptographic operations.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: October 16, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

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

@TechLeadJun 29

❓

Frequently Asked Questions

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Abbreviating the Product of a Range"?

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

🤖 Need help with this problem?