- Trending Categories
Data Structure
Networking
RDBMS
Operating System
Java
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Number of substrings divisible by 8 and not by 3 in C++
A string of 0-9 is given. For this problem, we need to calculate the number of strings that are divisible by 8 and not by 3. This is a 2 step problem, and we need to do the code one step at a time to solve it, for example
Input
str = "80"
Output
2
Input
str = "7675636788"
Output
15
Approach to Find the Solution
Only numbers with their last 3 digits are divisible by 8, and their sum of digits divisible by 3 are divisible by 8.
Now store the prefix sum of the string so that the sum of digits of prefix module 3 is either 0,1, or 2. Then the string is iterated for all the positions of i. Then count the number of substrings at i, which are divisible by 8—now, subtracting the number of substrings at i, which are divisible by 3 from this value.
|S| X 3 size 2D array is defined, |S| is the size of a string, say dp[i][j].
At any index i dp[i][j]. Starting from index i to 0, the number of substrings have the output j. So 0<=j<=0, since module 3.
We need to iterate over the string to check if one-digit, two-digit, and three-digit numbers are divisible by 8.
To determine if a character is 8 at index, check the number at index.
Divide the number by 8 rather than 3 if there are two digits.
Let us assume the number is divisible by 8. There must be two (1-3) substrings if it is divisible by 8. However, it will also have the substrings, which are divisible by 8. To remove them.
Example
#include <bits/stdc++.h>
using namespace std;
#define MAX 1000
int count (char s[], int len) {
int cur = 0,
dig = 0;
int sum[MAX], dp[MAX][3];
memset (sum, 0, sizeof (sum));
memset (dp, 0, sizeof (dp));
dp[0][0] = 1;
for (int i = 1; i <= len; i++) {
dig = int (s[i - 1]) - 48;
cur += dig;
cur %= 3;
sum[i] = cur;
dp[i][0] = dp[i - 1][0];
dp[i][1] = dp[i - 1][1];
dp[i][2] = dp[i - 1][2];
dp[i][sum[i]]++;
}
int ans = 0, dprev = 0, value = 0, dprev2 = 0;
for (int i = 1; i <= len; i++) {
dig = int (s[i - 1]) - 48;
if (dig == 8) ans++;
if (i - 2 >= 0) {
dprev = int (s[i - 2]) - 48;
value = dprev * 10 + dig;
if ((value % 8 == 0) && (value % 3 != 0)) ans++;
}
// Taking 3 digit number.
if (i - 3 >= 0){
dprev2 = int (s[i - 3]) - 48;
dprev = int (s[i - 2]) - 48;
value = dprev2 * 100 + dprev * 10 + dig;
if (value % 8 != 0) continue;
ans += (i - 2);
ans -= (dp[i - 3][sum[i]]);
}
}
return ans;
}
int main () {
char str[] = "7675636788";
int len = strlen (str);
cout << count (str, len) << endl;
return 0;
}Output
4
Conclusion
In this problem, we learned how to find the number of substrings divisible by 8 and not by 3 along with the c++ code. This code can also be written in java, python, and other languages. To solve this problem, we have reversed a string to find the number of substrings divisible by 8 but not divisible by 3. It is a very straightforward problem once we divide it into 2 parts.
- Related Questions & Answers
- Find permutation of n which is divisible by 3 but not divisible by 6 in C++
- Check if a large number is divisible by 8 or not in C++
- Check if a large number is divisible by 3 or not in java
- Check if a large number is divisible by 3 or not in C++
- Number of Substrings divisible by 6 in a String of Integers in C++
- Check if a large number is divisible by 2, 3 and 5 or not in C++
- Count rotations divisible by 8 in C++
- Number is divisible by 29 or not in C++
- Number of Groups of Sizes Two Or Three Divisible By 3 in C++
- Number of digits to be removed to make a number divisible by 3 in C++
- Check if any permutation of a large number is divisible by 8 in Python
- Check if any permutation of a number is divisible by 3 and is Palindromic in Python
- Possible cuts of a number such that maximum parts are divisible by 3 in C++
- Check if a number is divisible by 23 or not in C++
- Check if a number is divisible by 41 or not in C++