Detect Alternating Pattern

Code

#include <stdio.h>

int is_alternating_pattern(int *mem, int k) {
    // Write your pointer logic here
    if(k <= 2) return 1;

    int *temp1 = mem;         // first element
    int *temp2 = mem + 1;     // second element
    int loop_cnt = 0;
    if(k%2 == 0){             // if number of elements are even
        loop_cnt = (k/2) - 1;

        // same loop to check at both pointers
        for(int i = 0; i < loop_cnt; i++){
            if( (*temp1 != *(temp1+2)) ||
                (*temp2 != *(temp2+2)) ){
                return 0;
            }
            temp1+=2;
            temp2+=2;
        }
    }else{                    // if number of elements are odd
        loop_cnt = k/2;
        // at pointer one
        for(int i = 0; i < loop_cnt; i++){
            if( (*temp1 != *(temp1+2)) ){
                return 0;
            }
            temp1+=2;
        }
        // at second pointer
        for(int i = 0; i < loop_cnt-1; i++){
            if( (*temp2 != *(temp2+2)) ){
                return 0;
            }
            temp2+=2;
        }
    }

    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