2117. Abbreviating the Product of a Range LeetCode Solution

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Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Abbreviating the Product of a Range solution in C++
4. Abbreviating the Product of a Range solution in Java
5. Abbreviating the Product of a Range solution in Python
6. Additional Resources

Problem Statement of Abbreviating the Product of a Range

You are given two positive integers left and right with left 10, then express the product as

... where 
 denotes the first 5 digits of the product, and  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 "
...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

    
Complexity Analysis
    

        
Time Complexity: O(\texttt{right} - \texttt{left})
        
Space Complexity: O(1)
    

    
2117. Abbreviating the Product of a Range LeetCode Solution in C++
    
class Solution {
 public:
  string abbreviateProduct(int left, int right) {
    constexpr long maxSuf = 100'000'000'000;
    double prod = 1.0;
    long suf = 1;
    int countDigits = 0;
    int countZeros = 0;

    for (int num = left; num <= right; ++num) {
      prod *= num;
      while (prod >= 1.0) {
        prod /= 10;
        ++countDigits;
      }
      suf *= num;
      while (suf % 10 == 0) {
        suf /= 10;
        ++countZeros;
      }
      if (suf > maxSuf)
        suf %= maxSuf;
    }

    if (countDigits - countZeros <= 10) {
      const long tens = pow(10, countDigits - countZeros);
      return to_string(static_cast<long>(prod * tens + 0.5)) + 'e' +
             to_string(countZeros);
    }

    const string pre = to_string(static_cast<long>(prod * pow(10, 5)));
    string sufStr = to_string(suf);
    sufStr = sufStr.substr(sufStr.length() - 5);
    return pre + "..." + sufStr + 'e' + to_string(countZeros);
  }
};
/* code provided by PROGIEZ */
2117. Abbreviating the Product of a Range LeetCode Solution in Java
class Solution {
  public String abbreviateProduct(int left, int right) {
    final long maxSuf = 100_000_000_000L;
    double prod = 1.0;
    long suf = 1;
    int countDigits = 0;
    int countZeros = 0;

    for (int num = left; num <= right; ++num) {
      prod *= num;
      while (prod >= 1.0) {
        prod /= 10;
        ++countDigits;
      }
      suf *= num;
      while (suf % 10 == 0) {
        suf /= 10;
        ++countZeros;
      }
      if (suf > maxSuf)
        suf %= maxSuf;
    }

    if (countDigits - countZeros <= 10) {
      final long tens = (long) Math.pow(10, countDigits - countZeros);
      return String.valueOf((long) (prod * tens + 0.5)) + "e" + String.valueOf(countZeros);
    }

    final String pre = String.valueOf((long) (prod * Math.pow(10, 5)));
    String sufStr = String.valueOf(suf);
    sufStr = sufStr.substring(sufStr.length() - 5);
    return pre + "..." + sufStr + "e" + String.valueOf(countZeros);
  }
}
// code provided by PROGIEZ
2117. Abbreviating the Product of a Range LeetCode Solution in Python
class Solution:
  def abbreviateProduct(self, left: int, right: int) -> str:
    prod = 1.0
    suf = 1
    countDigits = 0
    countZeros = 0

    for num in range(left, right + 1):
      prod *= num
      while prod >= 1.0:
        prod /= 10
        countDigits += 1
      suf *= num
      while suf % 10 == 0:
        suf //= 10
        countZeros += 1
      if suf > 10**8:
        suf %= 10**8

    if countDigits - countZeros <= 10:
      tens = 10**(countDigits - countZeros)
      return str(int(prod * tens + 0.5)) + 'e' + str(countZeros)

    pre = str(int(prod * 10 ** 5))
    suf = str(suf)[-5:]
    return pre + '...' + suf + 'e' + str(countZeros)
 # code by PROGIEZ
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