Given two positive integers N and K. Find the minimum number of digits that can be removed from the number N such that after removals the number is divisible by 10K or print -1 if it is impossible.
Input : N = 10904025, K = 2 Output : 3 Explanation : We can remove the digits 4, 2 and 5 such that the number becomes 10900 which is divisible by 102. Input : N = 1000, K = 5 Output : 3 Explanation : We can remove the digits 1 and any two zeroes such that the number becomes 0 which is divisible by 105 Input : N = 23985, K = 2 Output : -1
Approach : The idea is to start traversing the number from the last digit while keeping a counter. If the current digit is not zero, increment the counter variable, otherwise decrement variable K. When K becomes zero, return counter as answer. After traversing the whole number, check if the current value of K is zero or not. If it is zero, return counter as answer, otherwise return answer as number of digits in N – 1, since we need to reduce the whole number to a single zero which is divisible by any number. Also, if the given number does not contain any zero, return -1 as answer.
Below is the implementation of above approach.
3 3 -1
Time Complexity :Number of digits in the given number.
- Number of digits in 2 raised to power n
- Check if a number can be expressed as x^y (x raised to power y)
- Find last five digits of a given five digit number raised to power five
- Larger of a^b or b^a (a raised to power b or b raised to power a)
- Find the sum of power of bit count raised to the power B
- GCD of a number raised to some power and another number
- Find value of y mod (2 raised to power x)
- K-th digit in 'a' raised to power 'b'
- Print last k digits of a^b (a raised to power b)
- Find unit digit of x raised to power y
- Find multiple of x closest to or a ^ b (a raised to power b)
- Minimum removals to make array sum odd
- Minimum removals from array to make GCD greater
- Minimum removals required to make ranges non-overlapping
- Minimum number of power terms with sum equal to n
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