Find longest sub-array with no repetitions - c

Question

The task is to find the longest contiguous sub-array with all elements distinct.
Example Input {4, 3, 1, 3, 2, 1, 0} Output {3, 2, 1, 0}
Algorithm
Extract first sub Array (here 431)
Extract second sub Array (here 31)
Compare number of elements and keep the array with the biggest number (keep 431)
Return to 2
Problem The output is incorrect
/* Free old array and replace it by the new array
* If we only want to free old array and replace it by a new array
* Function will free old array and replace it by a new array with size equal to maximum size it can have
* Maximum size is the size of the input array
*/
int* newArray(int* oldArray,int* newArray, int sizeArray, int sizeFArray)
{
if (newArray == NULL) {
int* temp = malloc(sizeFArray * sizeof(int));
if (temp == NULL)
exit(1);
return temp;
} else {
memcpy(oldArray, newArray, sizeArray);
return oldArray;
}
printf("Error");
exit(1);
}
//int isAvailable(int* array , int size, int number) checks if number is available in array (return 0 if true, 1 if false)
//printArray(int* array, int size) is a simple function to print an array
void subArray(int* inputArray, int sizeInputArray)
{
int* candidate = malloc(sizeInputArray * sizeof(int));
if (candidate == NULL)
exit(1);
int sizeCandidate = 0;
int* newCandidate = malloc(sizeInputArray * sizeof(int));
if (newCandidate == NULL)
exit(1);
int sizeNewCandidate = 0;
//We will first fill the candidate
while (sizeCandidate < sizeInputArray && isAvailable(candidate, sizeCandidate, *(inputArray + sizeCandidate)) != 0) {
*(candidate + sizeCandidate) = *(inputArray + sizeCandidate);
sizeCandidate++;
}
int index = 1;
//Check all potential new candidates
//If new candidate holds more elements than the current candidate
//Current candidate will be replaced by new candidate
//Else we will redo the process and check using the next candidate if availble
for (int i = 1; i < sizeInputArray; i++) {
if(isAvailable(newCandidate, sizeNewCandidate, *(inputArray + i)) == 0) {
if (sizeNewCandidate > sizeCandidate) {
candidate = newArray(candidate, newCandidate, sizeNewCandidate, sizeInputArray);
newCandidate = newArray(newCandidate, NULL, 0, 0);
sizeCandidate = sizeNewCandidate;
sizeNewCandidate = 0;
i = ++index;
} else {
newCandidate = newArray(newCandidate, NULL, 0, sizeInputArray);
sizeNewCandidate = 0;
i = ++index;
}
} else {
*(newCandidate + sizeNewCandidate) = *(inputArray + i);
sizeNewCandidate++;
}
}
printArray(candidate, sizeCandidate);
}

Answer 1

I hope this code looks more compact and has clear comments:
#include <stdio.h>
int a[] = { 4, 3, 1, 3, 2, 1, 0 };
int check(int a[], int i, int j)
{
for (int k = i; k < j; k++)
for (int l = k + 1; l < j; l++)
if (a[k] == a[l])
return 0;
return 1;
}
int main()
{
int s = 0; // start position of the best candidate
int m = 1; // length of the best candidate
int n = sizeof(a) / sizeof(0); // length of the array
for (int i = 0; i < n; i++) { // for every start position
for (int j = i + m + 1; j <= n; j++) { // for every lengh if it more than the best one
if (check(a, i, j)) { // check if it contains repetitions
if (j - i > m) { // if no repetions
s = i; // update the candidate
m = j - i; // and length
}
}
else
break;
}
}
printf("{%d", a[s]);
for(int i = s + 1; i < s + m; ++i)
printf(", %d", a[i]);
printf("}\n");
return 0;
}
This works and gives the correct output.

Answer 2

I don't understand the complexity of the code.
#include <stdio.h>
int main()
{
// int x[] = { 4, 3, 1, 3, 2, 1, 0 };
int x[] = { 4, 3, 1, 3, 2, 5, 0 };
int offset;
int cur_offset = 0;
int max = 0;
int max_offset = 0;
for (int i = 1; i < sizeof(x) / sizeof(int); i++) {
for (int j = i-1; j >= cur_offset; j--) {
if (x[i] == x[j]) {
if (max <= i - j) {
max = i - j;
max_offset = j + 1;
} else if (max_offset == cur_offset) {
max = i - max_offset;
}
cur_offset = j + 1;
break;
}
}
}
if (max < sizeof(x) / sizeof(int) - cur_offset) {
max_offset = cur_offset;
max = sizeof(x) / sizeof(int) - max_offset;
}
printf("%d", x[max_offset]);
for (int i = max_offset + 1; i < max_offset + max; i++)
printf(", %d", x[i]);
printf("\n");
}