Counting Coin Change Sum problem using Recursion and Momoization in C language

Counting Coin Change Sum problem using Momoization in C language

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <assert.h>
int *cache(int n);
int canSum(int target, int numbers[], int len, int *memo);
int main(int argc, char const *argv[])
{
    assert(canSum(11, (int[]){1, 4}, 2, cache(11)));
    printf("Test 1: Passed\n");
    assert(canSum(17, (int[]){3, 4, 5, 6}, 4, cache(17)));
    printf("Test 2: Passed\n");
    assert(canSum(70, (int[]){7, 14}, 2, cache(100)));
    printf("Test 3: Passed\n");

    return 0;
}

int canSum(int target, int numbers[], int len, int *memo)
{
    if (memo[target] == target)
    {
        return memo[target];
    }
    int remainder = -1;
    if (target == 0)
    {
        return true;
    }
    if (target < 0)
    {
        return false;
    }

    for (int i = 0; i < len; i++)
    {
        /**
         * *This is for debugging purpose.
         */

        // printf("Try [%d]:[%d - %d] = %d \n", i, target, numbers[i], target - numbers[i]);
        remainder = target - numbers[i];
        if (remainder == 0)
        {
            return true;
        }
        if (canSum(remainder, numbers, len, memo) == true)
        {
            memo[target] = true;
            return memo[target];
        }
    }
    memo[target] = false;
    return memo[target];
}

int *cache(int n)
{
    int *cache = (int *)malloc(sizeof(int) * n);
    for (int i = 0; i < n; i++)
    {
        cache[i] = -1;
    }
    return cache;
}