code-string-approach.cpp

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 * Copyright 2024 The OpenGenus.org Authors. All Rights Reserved.
 *
 * Code for the book "DAILY 43: Algorithmic Problems for Coding Interviews: Easy level, C++ edition"
 *
 * Licensed under the GNU General Public License, Version 3.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 *     http://www.gnu.org/licenses/gpl-3.0.html
 *
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 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
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 * For details about the book, please visit: https://www.amazon.com/dp/B0CZJNBLQS
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#include <iostream>
#include <string>
#include <cmath>
using namespace std;

bool isBeauty(string s1, string s2) {
    int num1 = stoi(s1);
    int num2 = stoi(s2);
    return num2 == 0 ? false : !(num1 % num2);
}

int divisorSubstrings(int num, int k) {
    string str = to_string(num);
    int i = 0, j = 0, beautyCount = 0, digits = str.size();
    
    while (j < digits) {
        string window = str.substr(i, j - i + 1);
        if (j - i + 1 < k)
            ++j;
        else if (j - i + 1 == k) {
            if (isBeauty(str, window))
                ++beautyCount;
            ++i;
            ++j;
        }
    }
    return beautyCount;
}

int main() {
    // Example test cases
    int num1 = 120;
    int k1 = 2;
    cout << "Number of divisor substrings: " << divisorSubstrings(num1, k1) << endl; // Expected output: 2

    int num2 = 430043;
    int k2 = 2;
    cout << "Number of divisor substrings: " << divisorSubstrings(num2, k2) << endl; // Expected output: 2

    // Additional test cases
    int num3 = 555;
    int k3 = 1;
    cout << "Number of divisor substrings: " << divisorSubstrings(num3, k3) << endl; // Expected output: 3

    int num4 = 111;
    int k4 = 3;
    cout << "Number of divisor substrings: " << divisorSubstrings(num4, k4) << endl; // Expected output: 1

    return 0;
}