Thats algorithm make a union between 2 int8 arrays without to repeat values

array_union.c

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

bool contains(char l, char * v, char c);
char * do_union(char n1, char n2, char * v1, char * v2);
int main()
{
	char n1, n2;
	printf("type length of array 1 and array 2:\n ex: 2 3 ");
	scanf("%2" SCNu8 " %2" SCNu8, (uint8_t*)&n1, (uint8_t*)&n2);

	char v1[n1], v2[n2];
	for (int i = 0; i < n1; i++) {
		printf("type array 1 %i position: ", i+1);
		scanf("%2" SCNu8, (uint8_t*)&v1[i]);
	}

	for (int i = 0; i < n2; i++) {
		printf("type array 2 %i position: ", i+1);
		scanf("%2" SCNu8, (uint8_t*)&v2[i]);
	}

	printf("Merged array: ");
	char *v3 = do_union(n1, n2, v1, v2);
	for (int i = 0; i < n1+n2; i++) {
		if (v3[i] >= 0) {
			printf("%i ", v3[i]);
		}
	}
	puts("");

	free(v3);
	return 0;
}

bool contains(char l, char * v, char c) {
	for (int i = 0; i < l; i++) {
		if (c == v[i]) return true;
	}

	return false;
}

char * do_union(char n1, char n2, char * v1, char * v2) {
	int size = n1 + n2, i = 0;
	char * v3 = (char *)malloc(size);
	for (; i < n1; i++) {
		v3[i] = v1[i];
	}

	for (int j = 0; j < n2; j++) {
		if (!contains(size, v3, v2[j])) {
			v3[i++] = v2[j];
		} else {
			v3[i++] = -1;
		}
	}

	return v3;
}