<Codesignal> absoluteValuesSumMinimization

Medium

Given a sorted array of integers a, your task is to determine which element of a is closest to all other values of a. In other words, find the element x in a, which minimizes the following sum:

abs(a[0] - x) + abs(a[1] - x) + ... + abs(a[a.length - 1] - x)

(where abs denotes the absolute value)

If there are several possible answers, output the smallest one.

Example

• For a = [2, 4, 7], the output should be absoluteValuesSumMinimization(a) = 4.

  • for x = 2, the value will be abs(2 - 2) + abs(4 - 2) + abs(7 - 2) = 7.
  • for x = 4, the value will be abs(2 - 4) + abs(4 - 4) + abs(7 - 4) = 5.
  • for x = 7, the value will be abs(2 - 7) + abs(4 - 7) + abs(7 - 7) = 8.

  The lowest possible value is when x = 4, so the answer is 4.

• For a = [2, 3], the output should be absoluteValuesSumMinimization(a) = 2.

  • for x = 2, the value will be abs(2 - 2) + abs(3 - 2) = 1.
  • for x = 3, the value will be abs(2 - 3) + abs(3 - 3) = 1.

  Because there is a tie, the smallest x between x = 2 and x = 3 is the answer.

Input/Output

• [execution time limit] 0.5 seconds (c)

• [input] array.integer a

  A non-empty array of integers, sorted in ascending order.

  Guaranteed constraints:
  1 ≤ a.length ≤ 1000,
  -10^6 ≤ a[i] ≤ 10^6.

• [output] integer

  • An integer representing the element from a that minimizes the sum of its absolute differences with all other elements.

[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

// 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;
// }
//
//
int absoluteValuesSumMinimization(arr_integer a)
{
	int min=1000000000, minValue;

	for(int i=0;i<a.size;i++)
	{
		int sum=0;

		for(int j=0;j<a.size;j++)
			sum+=abs(a.arr[i]-a.arr[j]);

		if(sum<min)
		{
			min=sum;
			minValue=a.arr[i];
		}
	}

	return minValue;
}