Find largest factor of N such that NF is less than K in Java10 Sept 2024 | 4 min read Given two numbers, N and K, our task is to determine the lowest value X such that N < X*K. Example 1: Input: int num = 8 int K = 7 Output: The largest factor of N is 2. Explanation: 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. Example 2: Input: int num = 999999733, int K = 999999732 Output: The largest factor of N is 999999733. Explanation: 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: Na�ve 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. Algorithm: Step 1: Initialize the variables k and num. Step 2: Go over [1, K] repeatedly. Step 3: Declare the temporary variable that will be used to hold the needed response. Step 4: Verify that (n % i) = 0 for each integer i. Step 4.1: Continue updating the max variable, which keeps track of the greatest N divisor traversed up to i. Step 5: Num/maxi, which must be returned, is the required answer. Implementation:FileName: LargestFactorofN.java Output: The largest factor of N is 2 Complexity Analysis: 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. Algorithm: Step 1: Create a function to determine the largest number factor that stays inside K's value. Step 2: Create a variable called res to hold the outcome; it should start at 0. Step 3: Repeats step 1 through step j, up to the square root of num. Step 4: It determines whether j is a factor of num for each j by determining whether num modulo j equals 0. Step 5: It compares j with K if j is a factor. It updates res to the maximum of its current value and j if j is less than or equal to K. Step 6: It also determines whether num divided by j is equal to or less than K. If this is the case, it indicates that num/j is a factor that is either equal to or less than K, in which case res is updated as needed. Step 7: Finally, the function returns the largest factor (res) that is less than or equal to K. Implementation:FileName: EfficientLargestFactorN.java Output: Complexity Analysis: 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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