How to remove elements in an array in order to another array

Question

I'm trying to write a function in C, that removes the second array's elements from the first array if the first array fully contains the second array in the exact same order.

I know it's complicated to ask in words, so here are some examples:

If Array-1 is hello and Array-2 is lo, then the output should be hel.

If Array-1 is hello hello and Array-2 is lo, then the output should be hel hel.

If Array-1 is hello and Array-2 is call (the first array does not contain all of the second array), then the output should be hello. (It shouldn't be changed.)

I wrote some code but when I try 2nd example (the hello hello one), it gives me hello hel, not hel hel.


char removeText (char a[], char b[], int lengthA, int lengthB) {

  int indexB = 0, i=0;
  char d[lengthA];

  for (; i<lengthA; i++) {

    if (a[i]==b[indexB]) {
      indexB++;
      if (indexB==lengthB) {
        for (int k=0, j=0; k<i-lengthB; j++, k++) {
          d[k] = a[j];
        }
      }
    }
    else {
      i -= indexB;
      indexB = 0;
    }
  }

  printf("%s", d);

  if(indexB!=lengthB) {
    return *a;
  }

  return *d;
}

int main(void) {

  char a[] = "hello hello";
  char b[] = "lo";
  int c = 11;
  int d = 2;
  
  removeText(a, b, c, d);

  return 0;
}

The output should be given with return. The printf("%s", d); part is just for trying if the code works or not.

I know what's wrong in my code. The

if (indexB==lengthB) {
    for (int k=0, j=0; k<i-lengthB; j++, k++) {
        d[k] = a[j];
    }
}

part causes the error but how can I fix it?

If the first array is "banana" and the second is "ana"; would the result be "b" or "bna"? — Brendan
– Brendan, Commented Feb 18, 2021 at 14:50

Mathieu · Accepted Answer · 2021-02-18 16:38:15Z

1

Your problem is in the lines:

for (int k=0, j=0; k<i-lengthB; j++, k++) {
    d[k] = a[j];
}

Each time you find a pattern in a, you copy all a data into d, cancelling what have been done before.

To highlight this, you should modify your code with some debug info:

printf("copy to d from a, indexd for %d to %d\n", 0, i-lengthB)
for (int k=0, j=0; k<i-lengthB; j++, k++) {
    d[k] = a[j];
}

You could have noticed too that k and j are always equal.

One more question, how do you code if the pattern to remove is not at the end of input, or is not present in input ? Try it.

To solve your problem, I would advice a different approach: have a counter for ignored character in a and copy the character when pattern is not found, something like:

char * removeText (char a[], char b[], int lengthA, int lengthB) {

  int indexB = 0, i=0;
  int ignored = 0;
  char d[lengthA] = "";

  for (; i+ignored<lengthA; i++) {

    if (a[i+ignored]==b[indexB]) {
      indexB++;
      if (indexB==lengthB) {
        ignored += lengthB;
        printf("ignored = %d\n", ignored);
      }
    }
    else {
      i -= indexB;
      indexB = 0;
      printf("d[%d] = a[%d] (%c)\n", i, i+ignored , a[i+ignored ]);
      d[i] = a[i+ignored];
    }
  }
  printf("%s", d);
  strcpy(a, d);
  return a;
}

edited Feb 18, 2021 at 16:38

answered Feb 18, 2021 at 14:35

Mathieu

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2 Comments

Efe Over a year ago

Thanks for the answer. It worked but there is one little problem. When I try just hello and lo, the output is hal�, but it should be just hal. What causes that strange character? The lengthA is 5 and lengthB is 2, as needed.

Mathieu Over a year ago

I think it's because the \0 is missing from d. One solution to fight this problem is to initialize d to "" (post edited)

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