# 1[0]1 Pattern Count | Samsung

*Question*

Write a program to find the **1[0]1 pattern count.** Given a string, your task is to find the number of patterns of form 1[0]1 where [0] represents any number of zeroes(minimum requirement is one 0) there should not be any other character exceptÂ 0 in [0] sequence.

*Logic*

- The logic is simple. When we encounter a character other than 1, we continue traversing the string.
- If 1 is encountered then check whether zeros are coming after that. We use
**Zcnt**variable to count the number of zero. - When we encounter 1 after repeating zeros, check whether the Zcnt is positive. If so then increment the count of 1[0]1 pattern count.
- Repeat the same procedure till we reach the end of the string.

*Program*

#include<stdio.h> #include<string.h> int main() { char s[50]; int i,j,t,n,Zcnt=0,count=0; printf("enter the string"); scanf("%s",&s); n=strlen(s); /*ascii of 0 ->48 1-> 49 */ for(i=0;i<n;i++) { t=s[i]; //Storing ascii of character if(t==48 || t>57) continue; //continue if '0' or not a number else { if(t==49) //character is '1' { j=i+1; //move one position forward t=s[j]; while(t==48) //character is '0' { Zcnt++; //increment ZeroCount j++; t=s[j]; } /*character is 1 and Zcnt is greater than zero*/ if(t==49&&Zcnt>0) { count++; //count of 1[0]1 pattern Zcnt=0; //reset ZeroCount } } } i=j; //reset i } printf("%d",count); //total 1[0]1 patterns }

*Analysis*

Time complexity –** O(n)** since we are traversing the whole string.

Space complexity- **O(1)** since we are not using any storage relative to the input.