Maximum and Minimum element of a linked list which is divisible by a given number k Archives

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:


<strong>Input :</strong> N = 10904025, K = 2
<strong>Output :</strong> 3
<strong>Explanation :</strong> We can remove the digits 4, 2 and 5 such that the number 
becomes 10900 which is divisible by 10<sup>2</sup>.

<strong>Input :</strong> N = 1000, K = 5
<strong>Output :</strong> 3
<strong>Explanation :</strong> We can remove the digits 1 and any two zeroes such that the
number becomes 0 which is divisible by 10<sup>5</sup><strong>Input :</strong> N = 23985, K = 2
<strong>Output :</strong> -1

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 105Input : 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.

#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.

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