How to make competitive coding solutions more efficient (BIT wise operations)?

Question

How do I make my code more efficient (in time) pertaining to a competitive coding question (source: codechef starters 73 div 4):

(Problem) Chef has an array A of length N. Chef wants to append a non-negative integer X to the array A such that the bitwise OR of the entire array becomes = Y .

Determine the minimum possible value of X. If no possible value of X exists, output -1.

Input Format The first line contains a single integer T — the number of test cases. Then the test cases follow. The first line of each test case contains two integers N and Y — the size of the array A and final bitwise OR of the array A. The second line of each test case contains N space-separated integers A_1, A_2, ..., A_N denoting the array A. Please don't judge me for my choice of language .

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int* binary_number(int n)     // returns pointer to a array of length 20(based on given constrains)     representing binary 
{
    int* ptc;
    ptc = (int*) malloc(20*sizeof(int));
    for(int i = 0; i < 20; i++)
    {
        if((n / (int) pow(2,19-i)) > 0){*(ptc + i) = 1;}
        else {*(ptc + i) = 0;}
        n = n % (int) pow(2,19-i) ;
    }
    return ptc;
}

int or_value(int* ptc, int n)    // Takes in pointers containing 1 or zero and gives the logical OR 
{
    for(int k = 0; k < n; n++)
    {
        if(*ptc == *(ptc + 20*k)){continue;}        // pointers are 20 units apart
        else{return 1;break;}
    }

    return *ptc;

}


int main(void) {

int t; scanf("%d", &t);

for (int i = 0; i < t; i++)
{
    int n, y;
    scanf("%d %d", &n, &y);
    int a[n];
    
    for(int j = 0; j < n ; j++)
    {
        scanf("%d", &a[j]);
    }
    
    int b[20*n];
    for (int j = 0; j < n; j++)
    {
        for (int k = 0; k < 20; k++)
        {
            b[20*j + k] = *(binary_number(a[n])+k);
        }
    }
    
    int c = 0;
    int p = 0;
    for (int j = 0; j < 20; j++)
    {
        
        if ((*(binary_number(y) + j) == 1) && (or_value((&b[0] + j),n) == 0)){c = c + pow(2,19 - j);}
        else if ((*(binary_number(y) + j) == 0) && (or_value((&b[0] + j),n) == 1)){p = 1; break;}
    }

    if (p==1){printf("-1");}
    else {printf("%d\n", c);}
}

return 0;
}