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Distribution of sweets by age of students

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Given two integer arrays and where stores the ages of different students and the stores the number of chocolates in the package (the full array represents the number of packages). Candy can be distributed to students in such a way that:

  1. Each student should receive only one box of candy.
  2. All students of the same age should receive the same amount of candy.
  3. The older student should receive more sweets than all the younger students.

The challenge is to determine if candy can be dispensed in this manner. If possible, print more print ,

Examples :

Input: ages [] = {5, 15, 10}, packs [] = {2, 2, 2, 3, 3, 4}
Output: YES
There are 3 students with age 5, 15 and 10.And there are 6 packets of candies containing 2, 2, 2, 3, 3, 4 candies respectively.
We will give one packet containing 2 candies to the student of age 5, one packet containing 3 candies to student with age 10 and give the packet containing 4 candies to student age 15

Input: ages [] = {5, 5, 6, 7}, packs [] = {5, 4, 6, 6}
Output: NO

Fit :

  • Make 2 frequency arrays, one of which will store the number of students with a specific age and the other — number of bags with a certain amount of sweets.
  • Then browse the frequency array for ages starting at the youngest age, and for each age in ascending order, try to find bags of sweets that are greater than or equal to the number of students for the selected age (starting with smallest number of candies in a bag)
  • If the above case fails then answer otherwise repeat the steps above until all students will not receive candy and will print finally.

Below is the implementation of the above approach:

# Python3 implementation of the approach

 
# Function for checking the correctness of distribution enia

def check_distribution (n, k, age, candy):

 

# Consider maximum age all students + 1

mxage = max (age) + 1

  

  # Destroying max candy + 1

mxcandy = max (candy) + 1

fr1 = [ 0 ] * mxage

fr2 = [ 0 ] * mxcandy

 

  # create frequency array

# age of students

for j in range (n):

fr1 [age [j]] + = 1

 

# Frequency array creation

  # candy bags

for j in range (k):

fr2 [candy [j]] + = 1

 

# a pointer to tell if we have reached

# end of frequency array of sweets

  k = 0

 

# Flag to tell if distribution is possible

Tf = True

for j in range (mxage):

if (fr1 [j] = = 0 ):

continue

  

  # A flag to tell if we can select some

  # candy bags for students aged j

flag = False

while (k & lt; mxcandy):

 

# If there are more packages

# than or equal to the number of students

age no j, then we can select these

# student packages

if (fr1 [j] " = fr2 [k]):

  flag = True

  break

k  + = 1

  

# Start search from k + 1 in next operation

k = k + 1

 

# If we can’t select any packages

# then the answer is NO

if (flag = = False ):

Tf = False

break

  if ( Tf):

print ( "YES" )

  else :

print ( "NO" )

 
# Driver code

age = [ 5 , 15 , 10 ]

candy = [ 2 , 2 , 2 , 3 , 3 , 4 ]

n = len (age)

k = len ( candy)

check_distribution (n, k, age, candy)

Exit :

 YES 

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