Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Count of occurrences of a “1(0+)1†pattern in a string in C++
We are given a string str containing 0s,1s and other alphabets . It also contains patterns of the form “1(0+)1” where 0+ means any number (>0) of consecutive 0s. The goal is to find such patterns ( “1(0+)1” ) inside string str.
Let us understand with examples
Input − str = “abb010bb10111011”
Output − Count of occurrences of a “1(0+)1” pattern in a string are − 2
Explanation − The patterns inside str are highlighted: “abb010bb10111011”, “abb010bb10111011”
Input − str = “01001011001001100”
Output − Count of occurrences of a “1(0+)1” pattern in a string are − 4
Explanation − The patterns inside str are highlighted: “01001011001001100”, “01001011001001100”, “01001011001001100”, “01001011001001100”
Approach used in the below program is as follows
It can be seen that all patterns start and end with 1. We will mark the first 1 using a flag variable check=1 and skip all 0s.
For any other character (neither 0 nor 1 ) set check as 0.
If we find another 1 and flag check is 1 then see if the previous value is 0. If yes then increment count as previous 0 is in between two ones. For any non 0 or 1 set check again as 0.
Take input string as str.
Function Pattern_occurrences(string str, int length) takes the string and its length and returns the count of occurrences of a “1(0+)1” pattern in a string
Take the initial count as 0.
Take flag variable check as 0 initially.
Traverse str using for loop from index i=0 to i<length.
If current character str[i] is 1 and check is 0 then set check as 1 and continue.
If current character str[i] is 1 and check is 1. Then it is second 1. Check if previous value str[i-1] is 0. If yes then pattern found. Increment count.
If the current character is neither 0 nor 1 then it will never be part of the pattern. Set check as 0. Now next encountered 1 will be considered as the start of the next pattern (if exists).
At the end count will have a number of such patterns inside str.
Return count as result.
Example
#include<iostream>
using namespace std;
int Pattern_occurrences(string str, int length){
int count = 0;
bool check = 0;
for (int i = 0; i < length ; i++){
if (str[i] == '1' && check == 1){
if (str[i - 1] == '0'){
count++;
}
}
if (str[i] == '1' && check == 0){
check = 1;
continue;
}
if (str[i] != '0' && str[i] != '1'){
check = 0;
}
}
return count;
}
int main(){
string str = "01010111011";
int length = str.length();
cout<<"Count of occurrences of a “1(0+)1” pattern in a string are: "<< Pattern_occurrences(str, length);
return 0;
}Output
If we run the above code it will generate the following output −
Count of occurrences of a “1(0+)1” pattern in a string are: 3
103 Views
To Continue Learning Please Login