Detect Alternating Pattern

Code

#include <stdio.h>

int is_alternating_pattern(int *mem, int k) {
    // Write your pointer logic here
    int* p = mem;

    for(int i = 0; i < k - 1; ++i){
        if(*p == *(p+1)){
            return 0;
        }
        ++p;
    }
    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;
}

Solving Approach

/*
Questions:
- can the segment size k be less than n ?
- are the values in array 0 or 1?
- k >= 2
- cases where k > n
- empty arrays or k = 0

Plan:
- Pointer arithmetic: no arr indexing
- Alternating pattern: for size k, starting at mem[0], check if adjacent pairs
are alternating
- Implementation
    - pointer to traverse the array
    - each position from 0 to k - 1, compare *p and *(p+1)
    - if any adjacent pairs are equal, return 0
    - if all pairs alternate then return 1
*/