absoluteValuesSumMinimization.R

# 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.
# a = list(2, 4, 7)

absoluteValuesSumMinimization <- function(a) {
  a = as.vector(unlist(a))
  
  diffsum <- lapply(a, function(x) {
    diffSum = sum(abs(a - x))
    return(diffSum)
  })
  
  diffsum <- as.vector(unlist(diffsum))
  min(a[which(diffsum == min(diffsum))])
}