# Abbreviating the Product of a Range
**Difficulty:** HARD
[External](https://leetcode.com/problems/abbreviating-the-product-of-a-range)
Canonical: https://scaleengineer.com/dsa/problems/abbreviating-the-product-of-a-range
**Patterns:** [Math](https://scaleengineer.com/dsa/patterns/math)
**Companies:** [Avalara](https://scaleengineer.com/companies/avalara)
---
## Problem
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:

1. 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`.
2. 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`.
3. 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`

# Approaches
## Approach 1: Brute Force with BigInteger
This approach uses Java's `BigInteger` class to handle the arbitrarily large product. It's a straightforward simulation of the problem description: calculate the full product, convert it to a string, and then perform the required formatting steps.
**Time:** O(M^2 * (log M)^2), where M is `right`. The product can have up to `O(M log M)` digits. Multiplying an N-digit number with a k-digit number takes `O(N*k)`. Summing this over the range gives a high polynomial complexity, which is too slow for the given constraints. · **Space:** O(M log M) to store the `BigInteger` product, where M is `right`.
**Pros:** Simple to understand and implement.; Correctly handles all cases as it works with the exact large number.
**Cons:** Very slow due to the overhead of `BigInteger` arithmetic. The product can have tens of thousands of digits, making multiplications computationally expensive.; High memory usage to store the large `BigInteger` object.; Likely to result in a 'Time Limit Exceeded' (TLE) error on most online judges for larger inputs.
### Explanation
The algorithm first computes the exact product of all integers in the range `[left, right]` using `BigInteger` to prevent overflow. Once the full product is obtained, it's converted to a string to perform the abbreviation steps.

```java
import java.math.BigInteger;

class Solution {
    public String abbreviateProduct(int left, int right) {
        BigInteger product = BigInteger.ONE;
        for (int i = left; i <= right; i++) {
            product = product.multiply(BigInteger.valueOf(i));
        }

        String s = product.toString();
        int trailingZeros = 0;
        while (trailingZeros < s.length() && s.charAt(s.length() - 1 - trailingZeros) == '0') {
            trailingZeros++;
        }

        String sWithoutZeros = s.substring(0, s.length() - trailingZeros);

        if (sWithoutZeros.length() <= 10) {
            return sWithoutZeros + "e" + trailingZeros;
        } else {
            String prefix = sWithoutZeros.substring(0, 5);
            String suffix = sWithoutZeros.substring(sWithoutZeros.length() - 5);
            return prefix + "..." + suffix + "e" + trailingZeros;
        }
    }
}
```
### Algorithm
*   Initialize a `BigInteger` variable `product` to `1`.
*   Iterate from `left` to `right`, multiplying `product` by each integer.
*   Convert the final `product` to its string representation, let's call it `s`.
*   Count the number of trailing zeros, `C`, by iterating from the end of `s`.
*   Create a new string `s_no_zeros` by removing the trailing zeros from `s`.
*   Check the length of `s_no_zeros`. If it's greater than 10, extract the first 5 digits as the prefix (`pre`) and the last 5 digits as the suffix (`suf`). Then, construct the abbreviated number string as `pre + "..." + suf`.
*   If the length is 10 or less, the number string is just `s_no_zeros`.
*   Finally, append `"e"` and the count of trailing zeros `C` to the number string to get the final result.

## Approach 2: Optimized Calculation with Prefix/Suffix Tracking
This approach avoids using `BigInteger` by breaking down the problem into smaller, manageable parts. It calculates the number of trailing zeros, the prefix, and the suffix separately. This is significantly more efficient as it works with standard data types like `double` and `long` and avoids creating massive numbers.
**Time:** O((right - left) * log(right) + C2), where C2 is the total count of factors of 2. This is because the main loop runs `right - left` times with `log(i)` work inside, and the subsequent loops run `total_c2 - C` times. This is efficient enough for the given constraints. · **Space:** O(1), as we only use a few variables of standard data types to store the state.
**Pros:** Highly efficient in both time and space.; Avoids `BigInteger` and its associated performance costs.; Handles the largest test cases within typical time limits.
**Cons:** More complex logic involving floating-point arithmetic for the prefix and modular arithmetic for the suffix.; Care must be taken with floating-point precision and choosing appropriate modulus values to ensure correctness.
### Explanation
The core idea is to compute the required components of the abbreviated product without ever computing the full product itself. This involves three main parts:

1.  **Count Trailing Zeros (C):** The number of trailing zeros is determined by `min(count(factors of 2), count(factors of 5))`. Since factors of 2 are always more abundant, we only need to count the factors of 5. However, to correctly calculate the product without trailing zeros (`P'`), we need counts of both.

2.  **Calculate Prefix and Suffix:** We need to compute `P' = (product) / 10^C`. To do this without large numbers, we maintain two variables during iteration:
    *   A `double` variable `prefix` to keep track of the most significant digits.
    *   A `long` variable `suffix` to keep track of the least significant digits.
    *   A boolean flag `abbreviate` to determine if the product `P'` has more than 10 digits.

3.  **Algorithm:** The method calculates `P'` by first multiplying all numbers in the range after stripping them of all their factors of 2 and 5. Then, it multiplies back the necessary number of factors of 2 and 5 to form `P'`. Throughout this process, it keeps the `prefix` and `suffix` variables from becoming too large.

```java
class Solution {
    public String abbreviateProduct(int left, int right) {
        long c2 = 0, c5 = 0;
        for (int i = left; i <= right; i++) {
            int temp = i;
            while (temp % 2 == 0) {
                c2++;
                temp /= 2;
            }
            while (temp % 5 == 0) {
                c5++;
                temp /= 5;
            }
        }

        long C = Math.min(c2, c5);
        long rem_c2 = c2 - C;
        long rem_c5 = c5 - C;

        double prefix = 1.0;
        long suffix = 1L;
        boolean abbreviate = false;
        long suffix_mod = 100000000000L; // 10^11, a safe modulus

        for (int i = left; i <= right; i++) {
            int temp = i;
            while (temp % 2 == 0) temp /= 2;
            while (temp % 5 == 0) temp /= 5;
            
            prefix *= temp;
            suffix *= temp;

            while (prefix >= 100000.0) {
                prefix /= 10.0;
            }
            if (suffix > suffix_mod) {
                abbreviate = true;
                suffix %= suffix_mod;
            }
        }

        for (int i = 0; i < rem_c2; i++) {
            prefix *= 2;
            suffix *= 2;
            while (prefix >= 100000.0) {
                prefix /= 10.0;
            }
            if (abbreviate || suffix > suffix_mod) {
                abbreviate = true;
                suffix %= suffix_mod;
            }
        }

        for (int i = 0; i < rem_c5; i++) {
            prefix *= 5;
            suffix *= 5;
            while (prefix >= 100000.0) {
                prefix /= 10.0;
            }
            if (abbreviate || suffix > suffix_mod) {
                abbreviate = true;
                suffix %= suffix_mod;
            }
        }

        String result;
        if (abbreviate) {
            String preStr = String.valueOf((int) prefix);
            String sufStr = String.format("%05d", suffix % 100000);
            result = preStr + "..." + sufStr;
        } else {
            result = String.valueOf(suffix);
        }

        return result + "e" + C;
    }
}
```
### Algorithm
*   First, calculate the total counts of factors 2 (`total_c2`) and 5 (`total_c5`) in the range `[left, right]`. The number of trailing zeros `C` is `min(total_c2, total_c5)`.
*   Initialize `prefix = 1.0`, `suffix = 1L`, and `abbreviate = false`.
*   Iterate `i` from `left` to `right`. In each step:
    *   Multiply `prefix` and `suffix` by `i` after stripping its factors of 2 and 5.
    *   To keep `prefix` from growing too large and losing precision, normalize it by repeatedly dividing by 10 until it's less than a certain threshold (e.g., `100000.0`).
    *   To keep `suffix` from overflowing, if it exceeds a large threshold (e.g., `10^11`), set `abbreviate = true` and take `suffix` modulo this threshold.
*   After the loop, we have calculated the product of numbers stripped of their 2 and 5 factors. Now we must multiply back the excess factors of 2 and 5, i.e., `2^(total_c2 - C)` and `5^(total_c5 - C)`.
*   During these multiplications, we continue to normalize `prefix` and `suffix` as before.
*   Finally, if `abbreviate` is true, format the result using the first 5 digits from `prefix` and the last 5 digits from `suffix`. Otherwise, the full product `P'` is represented by `suffix`, and we use that.

# Solutions
### Java

```java
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)1 e10) {
        gt = true;
        suf %= (long)1 e10;
      }
      for (pre *= i; pre > 1 e5; pre /= 10) {
      }
    }
    if (gt) {
      return (int)pre + "..." + String.format("%05d", suf % (int)1 e5) + "e" +
             c;
    }
    return suf + "e" + c;
  }
}

```

### CPP

```cpp
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 = min(cnt2, cnt5);
    cnt2 = cnt5 = c;
    long long suf = 1;
    long double pre = 1;
    bool gt = false;
    for (int i = left; i <= right; ++i) {
      for (suf *= i; cnt2 && suf % 2 == 0; suf /= 2) {
        --cnt2;
      }
      for (; cnt5 && suf % 5 == 0; suf /= 5) {
        --cnt5;
      }
      if (suf >= 1e10) {
        gt = true;
        suf %= (long long)1e10;
      }
      for (pre *= i; pre > 1e5; pre /= 10) {
      }
    }
    if (gt) {
      char buf[10];
      snprintf(buf, sizeof(buf), "%0*lld", 5, suf % (int)1e5);
      return to_string((int)pre) + "..." + string(buf) + "e" + to_string(c);
    }
    return to_string(suf) + "e" + to_string(c);
  }
};

```

### Python

```python
import numpy class Solution : def abbreviateProduct ( self , left : int , right : int ) -> str : cnt2 = cnt5 = 0 z = numpy . float128 ( 0 ) for x in range ( left , right + 1 ): z += numpy . log10 ( x ) while x % 2 == 0 : x //= 2 cnt2 += 1 while x % 5 == 0 : x //= 5 cnt5 += 1 c = cnt2 = cnt5 = min ( cnt2 , cnt5 ) suf = y = 1 gt = False for x in range ( left , right + 1 ): while cnt2 and x % 2 == 0 : x //= 2 cnt2 -= 1 while cnt5 and x % 5 == 0 : x //= 5 cnt5 -= 1 suf = suf * x % 100000 if not gt : y *= x gt = y >= 1e10 if not gt : return str ( y ) + "e" + str ( c ) pre = int ( pow ( 10 , z - int ( z ) + 4 )) return str ( pre ) + "..." + str ( suf ). zfill ( 5 ) + "e" + str ( c )
```