perf: improve Project Euler problem 030 solution 1 (TheAlgorithms#6267)

Improve solution (locally 3+ times - from 3+ seconds to ~1 second)

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>

Author: MaximSmolskiy

project_euler/problem_030/sol1.py

@@ -1,4 +1,4 @@
-""" Problem Statement (Digit Fifth Power ): https://projecteuler.net/problem=30
+""" Problem Statement (Digit Fifth Powers): https://projecteuler.net/problem=30
 
 Surprisingly there are only three numbers that can be written as the sum of fourth
 powers of their digits:
@@ -13,26 +13,32 @@
 Find the sum of all the numbers that can be written as the sum of fifth powers of their
 digits.
 
-(9^5)=59,049‬
-59049*7=4,13,343 (which is only 6 digit number )
-So, number greater than 9,99,999 are rejected
-and also 59049*3=1,77,147 (which exceeds the criteria of number being 3 digit)
-So, n>999
-and hence a bound between (1000,1000000)
+9^5 = 59049
+59049 * 7 = 413343 (which is only 6 digit number)
+So, numbers greater than 999999 are rejected
+and also 59049 * 3 = 177147 (which exceeds the criteria of number being 3 digit)
+So, number > 999
+and hence a number between 1000 and 1000000
 """
 
 
-def digitsum(s: str) -> int:
+DIGITS_FIFTH_POWER = {str(digit): digit**5 for digit in range(10)}
+
+
+def digits_fifth_powers_sum(number: int) -> int:
     """
-    >>> all(digitsum(str(i)) == (1 if i == 1 else 0) for i in range(100))
-    True
+    >>> digits_fifth_powers_sum(1234)
+    1300
     """
-    i = sum(pow(int(c), 5) for c in s)
-    return i if i == int(s) else 0
+    return sum(DIGITS_FIFTH_POWER[digit] for digit in str(number))
 
 
 def solution() -> int:
-    return sum(digitsum(str(i)) for i in range(1000, 1000000))
+    return sum(
+        number
+        for number in range(1000, 1000000)
+        if number == digits_fifth_powers_sum(number)
+    )
 
 
 if __name__ == "__main__":