#
# Unit 5 - Vectors
#
# Practice II
# 

# Solutions should be based only 
# on the material studied thus far. 

# (1)
# Given a non-empty vector vec, 
# produce a vector of all indices 
# where strings appear. 
vec <- c("A", NA, "B", NA, NA, "C")
indices <- c() 
for (ind in 1:length(vec)) {
  if (!is.na(vec[ind])) {
    indices <- c(indices, ind )
  }
}
print(indices)


print(class(NA))
print(class(vec[2]))

vec1 <- c(3, NA, 4, NA)
print(vec1)
print(class(vec1[2]))
# (2)
# Given two non-empty numeric 
# vectors vec1 and vec2, 
# print, for each value in vec1, 
# the value and a vector including 
# all indices in vec2 in which
# the value appears. 
vec1 <- c(3, 5, 4, 2)
vec2 <- c(8, 3, 2, 7)


# (3)  
# Given two equal-length  
# and parallel vectors ids and ages, 
# produce a vector of all ids of persons 
# whose age is under 40, and print 
# the new vector. 
ids <- c(1111, 2222, 3333, 4444)
ages <- c(2 6, 40, 41, 81)
myIDS <- c() 
for (ind in 1:length(ids)) {
  if (ages[ind] < 40) {
    myIDS <- c(myIDS, ids[ind])
  }
}
print(myIDS)


  
# (4)  
# Given a non-empty vector vec, 
# print TRUE if vec is sorted in 
# a non-descending order, 
# or FALSE otherwise.
vec <- c(7, 8, 9, 10)
found <- TRUE
for (ind in 1:(length(vec)-1)) {
  if (vec[ind+1] < vec[ind]) {
    found <- FALSE
    break
  }
}
print(found)


1, 2, 3, 4, 5, 6
1, 1, 2, 3, 4, 5 

6, 5, 4, 3, 2, 1
6, 6, 5, 4, 3, 2 


# (5)
# Given a sorted numeric vector vec, 
# and a given number, 
# insert the number into vec such that
# it remains sorted. 
# 1 
vec <- c(1, 2, 3, 4)
num <- 5 
for (ind in 1:(length(vec) - 1)) {
  if (num <= vec[1]) {
    newVec <- c(num, vec)
    break 
  } else if (num >= vec[length(vec)]) {
    newVec <- c(vec, num)
    break 
  } else if (num >= vec[ind] & num <= vec[ind+1]) {
    section1 <- vec[1:ind]
    section2 <- vec[(ind+1):length(vec)]
    newVec <- c(section1, num, section2)
    break 
  }
}
print(newVec)


# 2
vec <- c(1, 2, 3, 4)
num <- 5 
ind <- 1
newVec <- c() 
while (length(newVec) == 0 & ind <= length(vec)) {
  if (num <= vec[1]) {
    newVec <- c(num, vec)
  } else if (num >= vec[length(vec)]) {
    newVec <- c(vec, num)
  } else if (num >= vec[ind] & num <= vec[ind+1]) {
    section1 <- vec[1:ind]
    section2 <- vec[(ind+1):length(vec)]
    newVec <- c(section1, num, section2)
  }
  ind <- ind + 1 
}
print(newVec)



3     7 8 10 
c(3    5     7 8 10)


# (6)    
# Produce 10 random numbers 
# in range 1:10 and print the 
# most frequent number. 
freqs <- rep(0, times = 10)
for (iteration in 1:10) {
  randNum <- sample(1:10, size = 1)
  print(randNum)
  freqs[randNum] <- freqs[randNum] + 1
}
maxFreq <- -1
for (ind in 1:length(freqs)) {
  if (freqs[ind] > maxFreq) {
    maxFreq <- freqs[ind]
    maxNum <- ind
  }
}
print(maxNum)


# (7)  
# Given two vectors vec1 and vec2, 
# write a program that checks 
# whether vec2 is a section of vec1. 
# If it is the program will print the index 
# where this section begins in vec2. 
# Otherwise the program will print NA. 
# You may assume that vec2 can not appear
# more than once in vec1. 
vec1 <- c(8, 5, 2, 3, 1, 6)
vec2 <- c(2, 3, 1)
found <- TRUE 
if (length(vec1) < length(vec2)) {
  found <- FALSE 
} else {
  found <- FALSE
  for (ind in 1:(length(vec1) - (length(vec2 - 1)))) {
    section <- vec1[ind:(ind+2)]
    allEqual <- TRUE 
    for (ind2 in 1:length(section)) {
      if (section[ind2] != vec2[ind2]) {
        allEqual <- FALSE 
      }
    }
    if (allEqual == TRUE) {
      found <- TRUE
    }
  }
}
print(found)


x <- c(9, 5, 2)
y <- c(9, 5, 2)
allEqual <- TRUE 
for (ind in 1:length(x)) {
  if (x[ind] != y[ind]) {
    allEqual <- FALSE 
  }
}
print(allEqual)