LeetCode #2117 — HARD

Abbreviating the Product of a Range

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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 product as <pre>...<suf> where <pre> denotes the first 5 digits of the product, and <suf> denotes the last 5 digits of the product after removing all trailing zeros. If d <= 10, we keep it unchanged.
- For example, we express 1234567654321 as 12345...54321, but 1234567 is represented as 1234567.
Finally, represent the product as a string "<pre>...<suf>eC".
- For example, 12345678987600000 will be represented as "12345...89876e5".

Return a string denoting the abbreviated product of all integers in the inclusive range [left, right].

Example 1:

Input: left = 1, right = 4
Output: "24e0"
Explanation: The product is 1 × 2 × 3 × 4 = 24.
There are no trailing zeros, so 24 remains the same. The abbreviation will end with "e0".
Since the number of digits is 2, which is less than 10, we do not have to abbreviate it further.
Thus, the final representation is "24e0".

Example 2:

Input: left = 2, right = 11
Output: "399168e2"
Explanation: The product is 39916800.
There are 2 trailing zeros, which we remove to get 399168. The abbreviation will end with "e2".
The number of digits after removing the trailing zeros is 6, so we do not abbreviate it further.
Hence, the abbreviated product is "399168e2".

Example 3:

Input: left = 371, right = 375
Output: "7219856259e3"
Explanation: The product is 7219856259000.

Constraints:

1 <= left <= right <= 10⁴

Step 01

Brute Force Baseline

Problem summary: 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 product as <pre>...<suf> where <pre> denotes the first 5 digits of the product, and <suf> denotes the last 5 digits of the product after removing all trailing zeros. If d <= 10, we keep it unchanged. For example, we express 1234567654321 as 12345...54321, but 1234567 is represented as 1234567. Finally, represent the product as a string "<pre>...<suf>eC". For example, 12345678987600000 will be

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

1
4

Example 2

2
11

Example 3

371
375

Core Insight

What unlocks the optimal approach

Calculating the number of trailing zeros, the last five digits, and the first five digits can all be done separately.
Use a prime factorization property to find the number of trailing zeros. Use modulo to find the last 5 digits. Use a logarithm property to find the first 5 digits.
The number of trailing zeros C is nothing but the number of times the product is completely divisible by 10. Since 2 and 5 are the only prime factors of 10, C will be equal to the minimum number of times 2 or 5 appear in the prime factorization of the product.
Iterate through the integers from left to right. For every integer, keep dividing it by 2 as long as it is divisible by 2 and C occurrences of 2 haven't been removed in total. Repeat this process for 5. Finally, multiply the integer under modulo of 10^5 with the product obtained till now to obtain the last five digits.
The product P can be represented as P=10^(x+y) where x is the integral part and y is the fractional part of x+y. Using the property "if S = A * B, then log(S) = log(A) + log(B)", we can write x+y = log_10(P) = sum(log_10(i)) for each integer i in [left, right]. Once we obtain the sum, the first five digits can be represented as floor(10^(y+4)).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2117: Abbreviating the Product of a Range
class Solution {

    public String abbreviateProduct(int left, int right) {
        int cnt2 = 0, cnt5 = 0;
        for (int i = left; i <= right; ++i) {
            int x = i;
            for (; x % 2 == 0; x /= 2) {
                ++cnt2;
            }
            for (; x % 5 == 0; x /= 5) {
                ++cnt5;
            }
        }
        int c = Math.min(cnt2, cnt5);
        cnt2 = cnt5 = c;
        long suf = 1;
        double pre = 1;
        boolean gt = false;
        for (int i = left; i <= right; ++i) {
            for (suf *= i; cnt2 > 0 && suf % 2 == 0; suf /= 2) {
                --cnt2;
            }
            for (; cnt5 > 0 && suf % 5 == 0; suf /= 5) {
                --cnt5;
            }
            if (suf >= (long) 1e10) {
                gt = true;
                suf %= (long) 1e10;
            }
            for (pre *= i; pre > 1e5; pre /= 10) {
            }
        }
        if (gt) {
            return (int) pre + "..." + String.format("%05d", suf % (int) 1e5) + "e" + c;
        }
        return suf + "e" + c;
    }
}

// Accepted solution for LeetCode #2117: Abbreviating the Product of a Range
func abbreviateProduct(left int, right int) string {
	cnt2, cnt5 := 0, 0
	for i := left; i <= right; i++ {
		x := i
		for x%2 == 0 {
			cnt2++
			x /= 2
		}
		for x%5 == 0 {
			cnt5++
			x /= 5
		}
	}
	c := int(math.Min(float64(cnt2), float64(cnt5)))
	cnt2 = c
	cnt5 = c
	suf := int64(1)
	pre := float64(1)
	gt := false
	for i := left; i <= right; i++ {
		for suf *= int64(i); cnt2 > 0 && suf%2 == 0; {
			cnt2--
			suf /= int64(2)
		}
		for cnt5 > 0 && suf%5 == 0 {
			cnt5--
			suf /= int64(5)
		}
		if float64(suf) >= 1e10 {
			gt = true
			suf %= int64(1e10)
		}
		for pre *= float64(i); pre > 1e5; {
			pre /= 10
		}
	}
	if gt {
		return fmt.Sprintf("%05d...%05de%d", int(pre), int(suf)%int(1e5), c)
	}
	return fmt.Sprintf("%de%d", suf, c)
}

# Accepted solution for LeetCode #2117: Abbreviating the Product of a Range
class Solution:
    def abbreviateProduct(self, left: int, right: int) -> str:
        cnt2 = cnt5 = 0
        for x in range(left, right + 1):
            while x % 2 == 0:
                cnt2 += 1
                x //= 2
            while x % 5 == 0:
                cnt5 += 1
                x //= 5
        c = cnt2 = cnt5 = min(cnt2, cnt5)
        pre = suf = 1
        gt = False
        for x in range(left, right + 1):
            suf *= x
            while cnt2 and suf % 2 == 0:
                suf //= 2
                cnt2 -= 1
            while cnt5 and suf % 5 == 0:
                suf //= 5
                cnt5 -= 1
            if suf >= 1e10:
                gt = True
                suf %= int(1e10)
            pre *= x
            while pre > 1e5:
                pre /= 10
        if gt:
            return str(int(pre)) + "..." + str(suf % int(1e5)).zfill(5) + "e" + str(c)
        return str(suf) + "e" + str(c)

// Accepted solution for LeetCode #2117: Abbreviating the Product of a Range
/**
 * [2117] Abbreviating the Product of a Range
 *
 * 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:
 * <ol>
 * 	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 product as <pre>...<suf> where <pre> denotes the first 5 digits of the product, and <suf> denotes the last 5 digits of the product after removing all trailing zeros. If d <= 10, we keep it unchanged.
 *
 * 		For example, we express 1234567654321 as 12345...54321, but 1234567 is represented as 1234567.
 *
 *
 * 	Finally, represent the product as a string "<pre>...<suf>eC".
 *
 * 		For example, 12345678987600000 will be represented as "12345...89876e5".
 *
 *
 * </ol>
 * Return a string denoting the abbreviated product of all integers in the inclusive range [left, right].
 *  
 * Example 1:
 *
 * Input: left = 1, right = 4
 * Output: "24e0"
 * Explanation: The product is 1 &times; 2 &times; 3 &times; 4 = 24.
 * There are no trailing zeros, so 24 remains the same. The abbreviation will end with "e0".
 * Since the number of digits is 2, which is less than 10, we do not have to abbreviate it further.
 * Thus, the final representation is "24e0".
 *
 * Example 2:
 *
 * Input: left = 2, right = 11
 * Output: "399168e2"
 * Explanation: The product is 39916800.
 * There are 2 trailing zeros, which we remove to get 399168. The abbreviation will end with "e2".
 * The number of digits after removing the trailing zeros is 6, so we do not abbreviate it further.
 * Hence, the abbreviated product is "399168e2".
 *
 * Example 3:
 *
 * Input: left = 371, right = 375
 * Output: "7219856259e3"
 * Explanation: The product is 7219856259000.
 *
 *  
 * Constraints:
 *
 * 	1 <= left <= right <= 10^4
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/abbreviating-the-product-of-a-range/
// discuss: https://leetcode.com/problems/abbreviating-the-product-of-a-range/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn abbreviate_product(left: i32, right: i32) -> String {
        String::new()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2117_example_1() {
        let left = 1;
        let right = 4;

        let result = "24e0".to_string();

        assert_eq!(Solution::abbreviate_product(left, right), result);
    }

    #[test]
    #[ignore]
    fn test_2117_example_2() {
        let left = 2;
        let right = 11;

        let result = "399168e2".to_string();

        assert_eq!(Solution::abbreviate_product(left, right), result);
    }

    #[test]
    #[ignore]
    fn test_2117_example_3() {
        let left = 371;
        let right = 375;

        let result = "7219856259e3".to_string();

        assert_eq!(Solution::abbreviate_product(left, right), result);
    }
}

// Accepted solution for LeetCode #2117: Abbreviating the Product of a Range
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2117: Abbreviating the Product of a Range
// class Solution {
// 
//     public String abbreviateProduct(int left, int right) {
//         int cnt2 = 0, cnt5 = 0;
//         for (int i = left; i <= right; ++i) {
//             int x = i;
//             for (; x % 2 == 0; x /= 2) {
//                 ++cnt2;
//             }
//             for (; x % 5 == 0; x /= 5) {
//                 ++cnt5;
//             }
//         }
//         int c = Math.min(cnt2, cnt5);
//         cnt2 = cnt5 = c;
//         long suf = 1;
//         double pre = 1;
//         boolean gt = false;
//         for (int i = left; i <= right; ++i) {
//             for (suf *= i; cnt2 > 0 && suf % 2 == 0; suf /= 2) {
//                 --cnt2;
//             }
//             for (; cnt5 > 0 && suf % 5 == 0; suf /= 5) {
//                 --cnt5;
//             }
//             if (suf >= (long) 1e10) {
//                 gt = true;
//                 suf %= (long) 1e10;
//             }
//             for (pre *= i; pre > 1e5; pre /= 10) {
//             }
//         }
//         if (gt) {
//             return (int) pre + "..." + String.format("%05d", suf % (int) 1e5) + "e" + c;
//         }
//         return suf + "e" + c;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.