## Find largest factor of N such that NF is less than K in JavaGiven two numbers, N and K, our task is to determine the lowest value X such that N < X*K.
int num = 8 int K = 7
The largest factor of N is 2.
For the given number, numbers less than K divisible by N are 1, 2, and 4. So the minimum value of X is 2 such that 8 < 2*7 = 14. Hence, the largest factor of N is 2.
int num = 999999733, int K = 999999732
The largest factor of N is 999999733.
Since the integer 999999733 is a prime number, it can be divided by both 1 and the number itself; because K is smaller than 999999733. Since 999999733 < 999999733*999999732, it can be determined that 999999733 is the minimum value of X. Hence, the largest factor of N is 999999733. ## Approach: Nave ApproachEquation K * X = N can be used to represent the given problem expression. The primary objective of this equation is to reduce X. Therefore, we must determine the largest K that divides N.
## Implementation:
The largest factor of N is 2
The Time Complexity of the above code is O(K), and the Space complexity is O(1). ## Approach: Efficient ApproachThis approach, which iterates through factors up to the number's square root, effectively determines the largest factor of a given number num that is less than or equal to a set threshold K. It does this by using fundamental mathematical characteristics.
Step 7: Finally, the function returns the largest factor (res) that is less than or equal to K. ## Implementation:
The Time Complexity of the above code is O(sqrt(N)) and the Space Complexity is O(1), where 'N' represents the input number. |

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