# Distribution of candies according to ages of students

Given two integer arrays and where stores the ages of different students and an element of stores the number of candies that packet has (complete array represent the number of packets). The candies can be distributed among students such that:

- Every student must get only one pack of candies.
- All students of the same age must get equal number of candies.
- A student which is older must get more candies than all the student who are younger than him.

The task is to determine whether it is possible to distribute candies in the described manner. If possible then print else 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

**Approach:**

- Make 2 frequency arrays, one which will store the number of students with a particular age and one which will store the number of packets with a particular amount of candies.
- Then traverse the frequency array for ages starting from the youngest age and for every age in ascending try to find the candy packets that are greater than or equal to the number of students for the selected age (starting from the least number of candies a packet has)
- If the above case fails then the answer is else repeat the above steps until all the student have got the candies and print in the end.

Below is the implementation of the above approach:

`# Python3 implementation of the approach ` ` ` `# Function to check The validity of distribution ` `def` `check_distribution(n, k, age, candy): ` ` ` ` ` `# Stroring the max age of all students + 1 ` ` ` `mxage ` `=` `max` `(age)` `+` `1` ` ` ` ` `# Stroring the max candy + 1 ` ` ` `mxcandy ` `=` `max` `(candy)` `+` `1` ` ` `fr1 ` `=` `[` `0` `] ` `*` `mxage ` ` ` `fr2 ` `=` `[` `0` `] ` `*` `mxcandy ` ` ` ` ` `# creating the frequency array of the ` ` ` `# age of students ` ` ` `for` `j ` `in` `range` `(n): ` ` ` `fr1[age[j]] ` `+` `=` `1` ` ` ` ` `# Creating the frequency array of the ` ` ` `# packets of candies ` ` ` `for` `j ` `in` `range` `(k): ` ` ` `fr2[candy[j]] ` `+` `=` `1` ` ` ` ` `# pointer to tell whether we have reached ` ` ` `# the end of candy frequency array ` ` ` `k ` `=` `0` ` ` ` ` `# Flag to tell if distribution is possible or not ` ` ` `Tf ` `=` `True` ` ` `for` `j ` `in` `range` `(mxage): ` ` ` `if` `(fr1[j] ` `=` `=` `0` `): ` ` ` `continue` ` ` ` ` `# Flag to tell if we can choose some ` ` ` `# candy packets for the students with age j ` ` ` `flag ` `=` `False` ` ` `while` `(k < mxcandy): ` ` ` ` ` `# If the quantity of packets is greater ` ` ` `# than or equal to the number of students ` ` ` `# of age j, then we can choose these ` ` ` `# packets for the students ` ` ` `if` `(fr1[j] <` `=` `fr2[k]): ` ` ` `flag ` `=` `True` ` ` `break` ` ` `k ` `+` `=` `1` ` ` ` ` `# Start searching from k + 1 in next operation ` ` ` `k ` `=` `k ` `+` `1` ` ` ` ` `# If we cannot choose any packets ` ` ` `# 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) ` |

*chevron_right*

*filter_none*

**Output:**

YES

## Recommended Posts:

- Maximum number of candies that can be bought
- Find the minimum and maximum amount to buy all N candies
- Exploring Data Distribution | Set 2
- Exploring Data Distribution | Set 1
- NLP | Storing Frequency Distribution in Redis
- Inverse Gamma Distribution in Python
- NLP | Storing Conditional Frequency Distribution in Redis
- Maximum students to pass after giving bonus to everybody and not exceeding 100 marks
- Fractional Knapsack Queries
- Minimum Possible sum of digits in a positive multiple of N
- Multitape Nondeterministic Turing Machine simulator
- Python | sympy RGS method
- Python | sympy Partition() method
- Python | sympy Relational() method

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.