Abbreviating the Product of a Range
HardPrompt
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
3trailing zeros in1000, and there are0trailing zeros in546.
- For example, there are
- Denote the remaining number of digits in the product as
d. Ifd > 10, then express the product as<pre>...<suf>where<pre>denotes the first5digits of the product, and<suf>denotes the last5digits of the product after removing all trailing zeros. Ifd <= 10, we keep it unchanged.- For example, we express
1234567654321as12345...54321, but1234567is represented as1234567.
- For example, we express
- Finally, represent the product as a string
"<pre>...<suf>eC".- For example,
12345678987600000will be represented as"12345...89876e5".
- For example,
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
2 approaches with complexity analysis and trade-offs.
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.
Algorithm
- Initialize a
BigIntegervariableproductto1. - Iterate from
lefttoright, multiplyingproductby each integer. - Convert the final
productto its string representation, let's call its. - Count the number of trailing zeros,
C, by iterating from the end ofs. - Create a new string
s_no_zerosby removing the trailing zeros froms. - 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 aspre + "..." + suf. - If the length is 10 or less, the number string is just
s_no_zeros. - Finally, append
"e"and the count of trailing zerosCto the number string to get the final result.
Walkthrough
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.
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; } }}Complexity
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`.
Trade-offs
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
BigIntegerarithmetic. The product can have tens of thousands of digits, making multiplications computationally expensive.High memory usage to store the large
BigIntegerobject.Likely to result in a 'Time Limit Exceeded' (TLE) error on most online judges for larger inputs.
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.
Algorithm
- First, calculate the total counts of factors 2 (
total_c2) and 5 (total_c5) in the range[left, right]. The number of trailing zerosCismin(total_c2, total_c5). - Initialize
prefix = 1.0,suffix = 1L, andabbreviate = false. - Iterate
ifromlefttoright. In each step:- Multiply
prefixandsuffixbyiafter stripping its factors of 2 and 5. - To keep
prefixfrom 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
suffixfrom overflowing, if it exceeds a large threshold (e.g.,10^11), setabbreviate = trueand takesuffixmodulo this threshold.
- Multiply
- 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)and5^(total_c5 - C). - During these multiplications, we continue to normalize
prefixandsuffixas before. - Finally, if
abbreviateis true, format the result using the first 5 digits fromprefixand the last 5 digits fromsuffix. Otherwise, the full productP'is represented bysuffix, and we use that.
Walkthrough
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:
-
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. -
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
doublevariableprefixto keep track of the most significant digits. - A
longvariablesuffixto keep track of the least significant digits. - A boolean flag
abbreviateto determine if the productP'has more than 10 digits.
- A
-
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 formP'. Throughout this process, it keeps theprefixandsuffixvariables from becoming too large.
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; }}Complexity
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.
Trade-offs
Pros
Highly efficient in both time and space.
Avoids
BigIntegerand 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.
Solutions
Solution
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; }}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); }};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 )Video walkthrough
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