<Codesignal> Are Similar?

Medium

Two arrays are called similar if one can be obtained from another by swapping at most one pair of elements in one of the arrays.

Given two arrays a and b, check whether they are similar.

Example

• For a = [1, 2, 3] and b = [1, 2, 3], the output should be
  areSimilar(a, b) = true.

  The arrays are equal, no need to swap any elements.

• For a = [1, 2, 3] and b = [2, 1, 3], the output should be
  areSimilar(a, b) = true.

  We can obtain b from a by swapping 2 and 1 in b.

• For a = [1, 2, 2] and b = [2, 1, 1], the output should be
  areSimilar(a, b) = false.

  Any swap of any two elements either in a or in b won't make a and b equal.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] array.integer a

  Array of integers.

  Guaranteed constraints:
  3 ≤ a.length ≤ 10^5,
  1 ≤ a[i] ≤ 1000.

• [input] array.integer b

  Array of integers of the same length as a.

  Guaranteed constraints:
  b.length = a.length,
  1 ≤ b[i] ≤ 1000.

• [output] boolean

  • true if a and b are similar, false otherwise.

[C] Syntax Tips

// Prints help message to the console
// Returns a string
char * helloWorld(char * name) {
    char * answer = malloc(strlen(name) + 8);
    printf("This prints to the console when you Run Tests");
    strcpy(answer, "Hello, ");
    strcat(answer, name);
    return answer;
}

---

Solution 1

// Arrays are already defined with this interface:
// typedef struct arr_##name {
//   int size;
//   type *arr;
// } arr_##name;
//
// arr_##name alloc_arr_##name(int len) {
//   arr_##name a = {len, len > 0 ? malloc(sizeof(type) * len) : NULL};
//   return a;
// }
//
//
bool areSimilar(arr_integer a,arr_integer b)
{
	int index;
	bool isSame=true;

	for(int i=0;i<a.size;i++)
		if(a.arr[i]!=b.arr[i])
		{
			isSame=false;
			index=i;
			break;
		}

	if(isSame)
		return true;

	for(int i=index+1;i<a.size;i++)
		if(a.arr[i]!=b.arr[i])
		{
			int temp=a.arr[i];
			a.arr[i]=a.arr[index];
			a.arr[index]=temp;
			break;
		}

	for(int i=0;i<a.size;i++)
		if(a.arr[i]!=b.arr[i])
			return false;

	return true;
}

Solution 2

// Arrays are already defined with this interface:
// typedef struct arr_##name {
//   int size;
//   type *arr;
// } arr_##name;
//
// arr_##name alloc_arr_##name(int len) {
//   arr_##name a = {len, len > 0 ? malloc(sizeof(type) * len) : NULL};
//   return a;
// }
//
//
bool areSimilar(arr_integer a,arr_integer b)
{
	bool different=false;

	for(int i=0;i<a.size;i++)
		if(a.arr[i]!=b.arr[i])
		{
			int j=i+1, temp;
			if(different)
				return false;
			different=true;
			while(b.arr[j]!=a.arr[i] || b.arr[i]!=a.arr[j])
			{
				j++;
				if(j>=b.size)
					return false;
			}
			temp=b.arr[i];
			b.arr[i]=b.arr[j];
			b.arr[j]=temp;
		}

	return true;
}