Puzzle DC29

Delete the element at index N, move elements down to fill the gap

[E-7] Write a function that deletes the element at index N in an array, and fills the gap by moving elements above N downward by one. Put a 0 in the last element. If index N does not exist, do nothing to the array. Return 1 if the deletion was performed, 0 if not.

   0    1    2    3    4    5    6    7    8    9
  10   11   12   13   14
delete at index:15
Failure

   0    1    2    3    4    5    6    7    8    9
  10   11   12   13   14
delete at index:14
Success

   0    1    2    3    4    5    6    7    8    9
  10   11   12   13    0
delete at index:6
Success

   0    1    2    3    4    5    7    8    9   10
  11   12   13    0    0
delete at index:

If asked to delete one of the zeros at the end of the array, the function does so, even though this has no effect. Here is a testing framework:

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

int deleteElement( int size, int arr[], int n )
{
. . .
}

void printArray( int size, int arr[] )
{
  const int N = 10;
  int j;
  
  for ( j=0; j < size; j++ )
  {
    if ( j%N == N-1 )
      printf("%4d\n", arr[j] );
    else
      printf("%4d ", arr[j] );    
  }
}

void fillArrayInOrder( int size, int arr[], int first )
{
  int j;

  for ( j=0; j < size; j++ )
  {
    arr[j] = first+j;
  }
}

int main()
{
  const int SIZE =  25;
  int x[SIZE];
  char input[32];
  int del ;

  fillArrayInOrder( x, SIZE, 0 );
  printArray( SIZE, x  );
  
  printf("\ndelete at index:");
  gets( input );
  while ( strcmp( input, "q") != 0 )
  {
    del = atoi( input );
    if ( deleteElement( SIZE, x, del ) )
      printf("Success\n\n");
    else
      printf("Failure\n\n");
      
    printArray( SIZE, x );
    printf("\ndelete at index:");
    gets( input );
  }
  
  	
  return 0;
}