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 **10 ^{K}** or print -1 if it is impossible.

**Examples:**

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.

### C++ program for Minimum removals in a number to be divisible by **10**^{K}

^{K}

#include <bits/stdc++.h> using namespace std; int countDigitsToBeRemoved(int N, int K) { string s = to_string(N); int res = 0; int f_zero = 0; for (int i = s.size() - 1; i >= 0; i--) { if (K == 0) return res; if (s[i] == '0') { f_zero = 1; K--; } else res++; } if (!K) return res; else if (f_zero) return s.size() - 1; return -1; } int main() { int N = 10904025, K = 2; cout << countDigitsToBeRemoved(N, K) << endl; N = 1000, K = 5; cout << countDigitsToBeRemoved(N, K) << endl; N = 23985, K = 2; cout << countDigitsToBeRemoved(N, K) << endl; return 0; }

**Time Complexity :**Number of digits in the given number.