Given a number n, the task is to find the largest perfect cube that can be formed by deleting minimum digits(possibly 0) from the number.
X is called a perfect cube if X = Y3 for some Y.
Input : 4125 Output : 125 Explanation 125 = 53. We can form 125 by deleting digit 4 from 4125 Input : 876 Output :8 Explanation 8 = 23. We can form 8 by deleting digits 7 and 6 from 876
We can generate cubes of all numbers till from 1 to N1/3 (We don’t consider 0 as 0 is not considered as a perfect cube). We iterate the cubes from largest to the smallest.
Now if we look at the number n given to us, then we know that this number contains only log(n) + 1 digits, thus we can efficiently approach the problem if we treat this number n as a string hereafter.
While iterating on the perfect cubes, we check if the perfect cube is a subsequence of the number n when its represented as a string.If this is the case then the deletions required for changing the number n to the current perfect cube is:
No of deleted digits = No of digits in number n - Number of digits in current perfect cube
Since we want the largest cube number we traverse the array of preprocessed cubes in reverse order.
Largest Cube that can be formed from 4125 is 125 Largest Cube that can be formed from 876 is 8
Time Complexity of the above algorithm is O(N1/3log(N) log(N) is due to the fact that the number of digits in N are Log(N) + 1.
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