Dangerous String - Cplusplus

A string contains ‘0’ and ‘1’.You have to find out whether the string is dangerous or not. The string will be dangerous in the following two conditions.

There are at least 7 continuous occurrences of ‘1’.
There are at least 7 continuous occurrences of ‘0’.

Examples:

Input : 1000000001
Output : YES
Explanation: It has more than 7 continuous occurrences of '0'.

Input :  001001
Output :  NO
Explanation:String length is less the 7.

Approach: Follow the following steps:

Find the length of the string, If the length of the string is less than 7 then print “NO” because there is no chance of 7 continuous ‘0’ or ‘1’.
If the length of the string is exactly 7 then if anyone in ‘0’ or ‘1’ has the frequency 7 then print YES.
If the string length is greater than 7 then iterate the complete string to find 7 continuous ‘1’ and ‘0’.

Below is the implementation of the above approach.

// CPP program to find the string is
// Dangerous or not
// #include <bits/stdc++.h > 
#include <iostream>
using namespace std;

// Function to find string
// Dangerous or not
void Dangerous(string str)
{
    int freq[2] = { 0 }, flag = 0;
    
    // If string length less than 7
    if (str.length() < 7) 
    {
        cout << "NO";
    }

    // If string length is equal to 7
    if (str.length() == 7) {
        for (int i = 0; i < 7; i++) 
        {
            freq[str[i] - '0']++;
        }

        if (freq[0] == 7 or freq[1] == 7)
            cout << "YES";
        else
            cout << "NO";
    }

    // If string length is greater then 7
    else if (str.length() > 7) {
        for (int i = 0; i < str.length() - 6; i++) {
            if (str[i] == '0' and str[i + 1] == '0' 
            and str[i + 2] == '0' and str[i + 3] == '0' 
            and str[i + 4] == '0' and str[i + 5] == '0' 
            and str[i + 6] == '0') 
            {
                cout << "YES";
                flag = 1;
                break;
            }

            if (str[i] == '1' and str[i + 1] == '1' 
            and str[i + 2] == '1' and str[i + 3] == '1' 
            and str[i + 4] == '1' and str[i + 5] == '1' 
            and str[i + 6] == '1') 
            {
                cout << "YES";
                flag = 1;
                break;
            }
        }

        if (flag == 0)
            cout << "NO";
    }
}

// Driver Program.
int main()
{
    string str = "1000000001";
    Dangerous(str);
    return 0;
}

Output:
YES

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