Given an integer N, the task is to check if the Digit Cube limit of an integer arrives at a fixed point or in a limit cycle.
A Digit Cube Limit is a number which repeatedly arrives at a point if its value is computed as the sum of cubes of its digits, i.e. they have the following properties:
- Arrive at a fixed point if it is an Armstrong number.
- Arrive at a limit cycle if it is repeating in a cycle.
Input: N = 3
Output: Reached to fixed point 153
F(3) = 3 * 3 * 3 = 27
F(27) = 2*2*2 + 7*7*7 = 351
F(351) = 3*3*3 + 5*5*5 + 1*1*1 = 153
F(153) = 1*1*1 + 5*5*5 + 3*3*3 = 153
Since a fixed point(= 153) was obtained, which is an Armstrong number of order 3. Below is the illustration:
Input: N = 4
Output: Arrived at cycle
F(4) = 4 * 4 * 4 = 64
F(64) = 6 * 6 * 6 + 4 * 4 * 4 = 280
F(280) = 2 * 2 * 2 + 8 * 8 * 8 + 0 * 0 * 0 = 520
F(520) = 5 * 5 * 5 + 2 * 2 * 2 + 0*0*0 = 133
F(133) = 1*1*1 + 3*3*3 + 3*3*3 = 55
F(55) = 5*5*5 + 5*5*5 = 250
F(250) = 5*5*5 + 2*2*2 + 0*0*0 = 133
A cycle between
133 -> 55 -> 250 -> 133 is obtained.
Below is the illustration of the same:
Approach: Follow the steps below to solve the problem:
- Create a hashMap to store the sum of the cube of digits of the number while iteration.
- Iterate for the next values for the sum of cube of digits of a number and Check if the next sum of the cube of digits is already present in the hashMap or not.
- If the number is already in the hash-map then, check if the number is an Armstrong number or not. If found to be true, then the number reaches a fixed point.
- Otherwise, If the number is not Armstrong number then continue.
Below is the implementation of the above approach:
3 reaches to a fixed point: 153
Time Complexity: O(N)
Auxiliary Space: O(N)
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