TheAlgorithms · sarthaka1310 · Oct 5, 2020 · Oct 5, 2020
diff --git a/project_euler/problem_50/__init__.py b/project_euler/problem_50/__init__.py
diff --git a/project_euler/problem_50/solution_50.py b/project_euler/problem_50/solution_50.py
@@ -0,0 +1,67 @@
+#!/usr/bin/env python3
+
+"""
+Consecutive prime sum
+Problem 50
+
+The prime 41, can be written as the sum of six consecutive primes:
+
+41 = 2 + 3 + 5 + 7 + 11 + 13
+This is the longest sum of consecutive primes that adds to a prime below one-hundred.
+
+The longest sum of consecutive primes below one-thousand that adds to a prime,
+contains 21 terms, and is equal to 953.
+
+Which prime, below 1 million, can be written as the sum of the most consecutive primes?
+"""
+
+
+def solution() -> int:
+    """
+    Returns solution to problem.
+
+    Algorithm:
+    > Construct a "Sieve of Eratosthenes" to get all primes till million
+    (This will also serve as O(1) primality check later)
+    > Now make the largest size window and slide over primes,
+    until we encounter the sum of slide being a prime.
+
+    >>> solution()
+    997651
+    """
+
+    # To avoid redundant exponentiation
+    million = 10 ** 6
+
+    sieve = [True] * million
+    primes = []
+    # Creation of Sieve
+    for number in range(2, million):
+        if sieve[number]:
+            primes.append(number)
+            for multiple in range(number * number, million, number):
+                sieve[multiple] = False
+
+    # Cumulative sum of primes for increased efficiency when calculating sum over window
+    cumulative_sum = [2]
+    for i in range(1, len(primes)):
+        cumulative_sum.append(cumulative_sum[i - 1] + primes[i])
+
+    # Find size of largest window with smallest primes adding to more than million
+    largest_size = 0
+    while cumulative_sum[largest_size] < million:
+        largest_size += 1
+
+    for size in range(largest_size, 1, -1):
+        for start in range(0, len(primes) - size + 1):
+            # Sum over window of size 'size' from index 'start'
+            prime_sum = (
+                cumulative_sum[start + size - 1] - cumulative_sum[start] + primes[start]
+            )
+
+            if prime_sum < million and sieve[prime_sum]:
+                return prime_sum
+
+
+if __name__ == "__main__":
+    print(solution())