#include <stdio.h>
int is_alternating_pattern(int *start, int k) {
int expected=!(*start);
while(k-1)
{
//printf("DEBUG: current element:%d and expecting:%d\n",*start,expected);
if(*(start+1)== expected ) //we will come into this condition until we are keep on satisying the condition.
{
start++;
expected=!(*start);
}
else
return 0;
k--;
}
return 1;
}
int main() {
int n, k, arr[100];
scanf("%d %d", &n, &k);
for (int i = 0; i < n; i++) {
scanf("%d", &arr[i]);
}
int res = is_alternating_pattern(arr, k);
printf("%d", res);
return 0;
}
1: Actually they are very interesting and tough to get caught of an idea to predict the pattern.
2: Initially take a variable: expected and initialize in such a way that expected=!(start), we should have *(start+1)==expected, it means that if we have element in current base address(100) as 1 and in address 104 we are expecting 0, if this is satisfied we increment the start pointer and update the next expected value.
Else return 0.
Input
6 6 1 0 1 0 1 0
Expected Output
1