Given a positive integer N and K where and . The task is to check whether any permutation of digits of N equals any power of K. If possible return “True” otherwise return “False“.
Input: N = 96889010407, K = 7 Output: True Explanation: 96889010407 = 713 Input : N = 123456789, K = 4 Output : False
Approach: The Naive approach is to generate all permutation of digits of N and then check one by one if any of them is divisible of any power of K.
Efficient Approach: We know that total numbers of all power of K will not be more than logK(1018), for eg: if K = 2 then there will be atmost 64 numbers of power of K. We generate all power of K and store it in array.
Now we iterate all numbers from array and check where it contains all digits of N or not.
Below is the implementation of above approach:
Time Complexity: O(logK(1018)2)
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- K difference permutation
- Permutation Coefficient
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