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2 Keys Keyboard LeetCode Problem 0650 Solution using C++ language
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Efficiently Finding the Minimum Operations to Replicate 'A' on the Screen 'n' Times

This C++ solution tackles the "2 Keys Keyboard" problem, which revolves around determining the fewest operations needed to replicate the character 'A' on the screen exactly 'n' times. The code accomplishes this task through a straightforward and effective approach:

Initialization:
int ans = 0: Initialize 'ans' to zero, serving as a counter to track the minimum number of operations.
int d = 2: Set 'd' to 2, which acts as the potential prime factor under consideration.

Outer While Loop:
while (n greater than 1): The primary loop operates until 'n' is reduced to or below 1, indicating the successful factorization of 'n' into prime factors.

Inner While Loop:
while (n % d == 0): The nested loop examines whether 'd' is a prime factor of 'n.' A successful division confirms 'd' as a prime factor.
ans += d: If 'd' proves to be a prime factor, 'ans' increases by 'd,' signifying 'd' number of "copy" operations.
n /= d: 'n' is divided by 'd' to minimize it to its remaining factors.

Increment 'd':
d++: Following the inner loop, 'd' is incremented by 1, enabling the evaluation of the next number as a potential prime factor.

Continuation of the Outer Loop:
The outer loop persists until 'n' is no longer greater than 1, signifying the complete factorization of 'n' into prime factors.

Result:
Upon exiting the loop, the function returns the value of 'ans,' representing the minimal number of operations (including copying and pasting) required to replicate 'A' on the screen 'n' times.
This approach is highly efficient, relying on the concept of prime factorization by testing potential prime factors. It accurately tallies the number of operations for each prime factor and accumulates them in 'ans,' delivering the total minimum number of operations necessary to achieve the target state.