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Distribute Repeating Integers

Hard

Asked by:

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Topics:

ArraysDynamic ProgrammingBit Manipulation

You are given an array of n integers, nums, where there are at most 50 unique values in the array. You are also given an array of m customer order quantities, quantity, where quantity[i] is the amount of integers the i^th customer ordered. Determine if it is possible to distribute nums such that:

The i^th customer gets exactly quantity[i] integers,
The integers the i^th customer gets are all equal, and
Every customer is satisfied.

Return true if it is possible to distribute nums according to the above conditions.

Example 1:

Input: nums = [1,2,3,4], quantity = [2]
Output: false
Explanation: The 0^th customer cannot be given two different integers.

Example 2:

Input: nums = [1,2,3,3], quantity = [2]
Output: true
Explanation: The 0^th customer is given [3,3]. The integers [1,2] are not used.

Example 3:

Input: nums = [1,1,2,2], quantity = [2,2]
Output: true
Explanation: The 0^th customer is given [1,1], and the 1st customer is given [2,2].

Constraints:

n == nums.length
1 <= n <= 10⁵
1 <= nums[i] <= 1000
m == quantity.length
1 <= m <= 10
1 <= quantity[i] <= 10⁵
There are at most 50 unique values in nums.

Try coding it in LeetCode

Distribute Repeating Integers

Hard

Asked by:

19 views

Topics:

ArraysDynamic ProgrammingBit Manipulation

The i^th customer gets exactly quantity[i] integers,
The integers the i^th customer gets are all equal, and
Every customer is satisfied.

Return true if it is possible to distribute nums according to the above conditions.

Example 1:

Input: nums = [1,2,3,4], quantity = [2]
Output: false
Explanation: The 0^th customer cannot be given two different integers.

Example 2:

Input: nums = [1,2,3,3], quantity = [2]
Output: true
Explanation: The 0^th customer is given [3,3]. The integers [1,2] are not used.

Example 3:

Input: nums = [1,1,2,2], quantity = [2,2]
Output: true
Explanation: The 0^th customer is given [1,1], and the 1st customer is given [2,2].

Constraints:

n == nums.length
1 <= n <= 10⁵
1 <= nums[i] <= 1000
m == quantity.length
1 <= m <= 10
1 <= quantity[i] <= 10⁵
There are at most 50 unique values in nums.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `quantity` and `nums` arrays? What is the range of values within those arrays?
Can `quantity` or `nums` contain zero or negative values?
If it's impossible to distribute the integers according to the `quantity` array, what should I return?
Is the order of distribution important? Does there have to be a specific mapping of quantities to numbers?
If multiple valid distributions exist, is any valid distribution acceptable, or is there a preferred or required order/arrangement?

Brute Force Solution

Approach

The brute force method to distribute repeating integers involves trying every possible way to assign the integers to groups. We systematically explore all arrangements, checking if any assignment satisfies the quantity requirements for each group.

Here's how the algorithm would work step-by-step:

Consider assigning the first integer to the first group, then consider assigning it to the second group, and so on, until all groups have been considered.
Next, consider assigning the second integer to the first group, then to the second, and so on.
Repeat this process for every integer, each time considering every group as a possible destination.
For each complete assignment of integers to groups, check if the number of integers assigned to each group matches the required quantity for that group.
If the quantity matches for all groups, then a valid distribution has been found.
If no valid distribution is found after trying all possibilities, then it is not possible to distribute the integers as required.

Code Implementation

def distribute_repeating_integers_brute_force(integer_counts, group_sizes):
    number_of_groups = len(group_sizes)
    number_of_integers = len(integer_counts)

    def can_distribute(index, current_group_assignments):
        # Base case: all integers have been assigned.
        if index == number_of_integers:
            group_counts = [0] * number_of_groups
            for integer_index in range(number_of_integers):
                group_counts[current_group_assignments[integer_index]] += integer_counts[integer_index]

            # Check if each group's size requirement is met.
            for group_index in range(number_of_groups):
                if group_counts[group_index] != group_sizes[group_index]:
                    return False
            return True

        # Try assigning the current integer to each group.
        for group_index in range(number_of_groups):
            current_group_assignments[index] = group_index

            # Recursively check if a valid distribution can be found.
            if can_distribute(index + 1, current_group_assignments):
                return True

        return False

    # Initialize an array to track which group each integer is assigned to.
    group_assignments = [0] * number_of_integers
    # Start the recursive process
    return can_distribute(0, group_assignments)

Big(O) Analysis

Time Complexity

O(m^n) – Let m be the number of customer groups and n be the number of distinct integers. The algorithm explores all possible assignments of each distinct integer to the customer groups. For each of the n distinct integers, there are m possible customer groups it can be assigned to. Therefore, the total number of possible assignments is m multiplied by itself n times, which is m^n. Checking each assignment to see if it's valid takes additional time, but the dominating factor is the exponential growth of possible assignments, leading to a time complexity of O(m^n).

Space Complexity

O(K^N) – The brute force method explores all possible assignments of N integers to K groups. Each level of recursion represents assigning an integer to one of the K groups. Therefore, the recursion tree can have a depth of N, and at each level, there are K possible choices, leading to a maximum of K^N recursive calls. The space complexity is primarily determined by the maximum depth of the call stack which is N. Each recursive call requires constant space for local variables; however, the primary driver of space complexity is the data structure needed to represent all partial and complete assignments of integers to groups which implicitly exists via the function call stack and local variables. Hence the overall space complexity is approximated as O(K^N) due to exploring all combinations.

Optimal Solution

Approach

The best way to distribute repeating numbers efficiently involves figuring out if we can fulfill each customer's order. We use a clever trick to check orders starting from the largest to the smallest to see if they can be fulfilled while avoiding unnecessary combinations.

Here's how the algorithm would work step-by-step:

First, count how many times each unique number appears.
Then, figure out how much each customer wants of a particular number and arrange these demands from highest to lowest.
Try to match each customer's demand with a number that appears enough times, starting with the customer that wants the most.
If you can't fulfill a customer's demand with the available numbers, then you know it's impossible to satisfy all customers.
Keep going until all customers are satisfied or you find it's impossible to satisfy them all.

Code Implementation

def canDistribute(nums, quantity):
    number_counts = {}
    for number in nums:
        number_counts[number] = number_counts.get(number, 0) + 1

    available_counts = list(number_counts.values())
    quantity.sort(reverse=True)

    def backtrack(customer_index):
        if customer_index == len(quantity):
            return True

        # Try to fulfill the current customer's demand with each count.
        for i in range(len(available_counts)):
            if available_counts[i] >= quantity[customer_index]:

                available_counts[i] -= quantity[customer_index]

                # Recursively check if remaining customers can be satisfied.
                if backtrack(customer_index + 1):
                    return True

                available_counts[i] += quantity[customer_index]

        return False

    # Need to start the backtracking from the first customer.
    return backtrack(0)

Big(O) Analysis

Time Complexity

O(2^m + n log n) – The complexity is driven by two major parts. First, we count the frequency of each number, which takes O(n) time where n is the length of the input array nums. Then we sort the customer array, which takes O(m log m) time, where m is the length of the customer array. The dominating factor usually comes from the recursive attempt to assign counts to customers, which has a worst-case time complexity of O(2^m) due to exploring all possible subsets of number counts for each customer. Combining the sorting and recursion leads to a total time complexity of approximately O(2^m + n log n), assuming customer array is already provided in sorted order.

Space Complexity

O(N + K) – The algorithm first counts the frequency of each unique number in the input array of size N, which requires a hash map (or similar data structure) potentially storing up to N unique numbers. Next, the customer demands, represented by an array of size K, need to be sorted. While the plain English explanation doesn't specify the sorting algorithm used, in the worst case, sorting an array of size K might take O(K) auxiliary space depending on the algorithm implementation (e.g., merge sort). Therefore, the overall auxiliary space is dominated by the frequency map and potentially the space for sorting customer demands, resulting in O(N + K) space complexity where N is the size of the input array of numbers and K is the number of customers (i.e., the length of the demands array).

Edge Cases

Case	How to Handle
Empty nums or quantity array	Return true immediately if quantity is empty; return false if nums is empty but quantity is not.
nums array has only one element	Return true if the quantity array contains only one element and that element is less than or equal to the number of occurences of the element in nums, otherwise return false.
quantity array contains only one element	Check if the counts of any number in nums are greater than or equal to the one element of the quantity array; if so, return true, otherwise return false.
The sum of the quantity array is greater than the length of the nums array	Return false, since it's impossible to distribute more quantities than numbers available.
Large input size (nums and quantity arrays)	Ensure the solution employs efficient data structures (e.g., hash maps) and algorithms (e.g., backtracking with memoization) to avoid time limit exceeded errors.
Cases where no distribution is possible (e.g., quantity demands exceed available counts)	The backtracking algorithm should explore all possibilities and return false only if every distribution attempt fails.
Integer overflow when calculating counts of elements or during summation	Use appropriate data types (e.g., long) to store counts or sums to prevent integer overflows.
All elements in nums are the same and a quantity is larger than the count of this element	Return false if a quantity is larger than the count of the only element in nums after grouping similar nums.

Empty nums or quantity array

How to Handle:

Return true immediately if quantity is empty; return false if nums is empty but quantity is not.

nums array has only one element

How to Handle:

Return true if the quantity array contains only one element and that element is less than or equal to the number of occurences of the element in nums, otherwise return false.

quantity array contains only one element

How to Handle:

Check if the counts of any number in nums are greater than or equal to the one element of the quantity array; if so, return true, otherwise return false.

The sum of the quantity array is greater than the length of the nums array

How to Handle:

Return false, since it's impossible to distribute more quantities than numbers available.

Large input size (nums and quantity arrays)

How to Handle:

Ensure the solution employs efficient data structures (e.g., hash maps) and algorithms (e.g., backtracking with memoization) to avoid time limit exceeded errors.

Cases where no distribution is possible (e.g., quantity demands exceed available counts)

How to Handle:

The backtracking algorithm should explore all possibilities and return false only if every distribution attempt fails.

Integer overflow when calculating counts of elements or during summation

How to Handle:

Use appropriate data types (e.g., long) to store counts or sums to prevent integer overflows.

All elements in nums are the same and a quantity is larger than the count of this element

How to Handle:

Return false if a quantity is larger than the count of the only element in nums after grouping similar nums.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the constraints on the size of the `quantity` and `nums` arrays? What is the range of values within those arrays?
Can `quantity` or `nums` contain zero or negative values?
If it's impossible to distribute the integers according to the `quantity` array, what should I return?
Is the order of distribution important? Does there have to be a specific mapping of quantities to numbers?
If multiple valid distributions exist, is any valid distribution acceptable, or is there a preferred or required order/arrangement?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider assigning the first integer to the first group, then consider assigning it to the second group, and so on, until all groups have been considered.
Next, consider assigning the second integer to the first group, then to the second, and so on.
Repeat this process for every integer, each time considering every group as a possible destination.
For each complete assignment of integers to groups, check if the number of integers assigned to each group matches the required quantity for that group.
If the quantity matches for all groups, then a valid distribution has been found.
If no valid distribution is found after trying all possibilities, then it is not possible to distribute the integers as required.

Code Implementation

def distribute_repeating_integers_brute_force(integer_counts, group_sizes):
    number_of_groups = len(group_sizes)
    number_of_integers = len(integer_counts)

    def can_distribute(index, current_group_assignments):
        # Base case: all integers have been assigned.
        if index == number_of_integers:
            group_counts = [0] * number_of_groups
            for integer_index in range(number_of_integers):
                group_counts[current_group_assignments[integer_index]] += integer_counts[integer_index]

            # Check if each group's size requirement is met.
            for group_index in range(number_of_groups):
                if group_counts[group_index] != group_sizes[group_index]:
                    return False
            return True

        # Try assigning the current integer to each group.
        for group_index in range(number_of_groups):
            current_group_assignments[index] = group_index

            # Recursively check if a valid distribution can be found.
            if can_distribute(index + 1, current_group_assignments):
                return True

        return False

    # Initialize an array to track which group each integer is assigned to.
    group_assignments = [0] * number_of_integers
    # Start the recursive process
    return can_distribute(0, group_assignments)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, count how many times each unique number appears.
Then, figure out how much each customer wants of a particular number and arrange these demands from highest to lowest.
Try to match each customer's demand with a number that appears enough times, starting with the customer that wants the most.
If you can't fulfill a customer's demand with the available numbers, then you know it's impossible to satisfy all customers.
Keep going until all customers are satisfied or you find it's impossible to satisfy them all.

Code Implementation

def canDistribute(nums, quantity):
    number_counts = {}
    for number in nums:
        number_counts[number] = number_counts.get(number, 0) + 1

    available_counts = list(number_counts.values())
    quantity.sort(reverse=True)

    def backtrack(customer_index):
        if customer_index == len(quantity):
            return True

        # Try to fulfill the current customer's demand with each count.
        for i in range(len(available_counts)):
            if available_counts[i] >= quantity[customer_index]:

                available_counts[i] -= quantity[customer_index]

                # Recursively check if remaining customers can be satisfied.
                if backtrack(customer_index + 1):
                    return True

                available_counts[i] += quantity[customer_index]

        return False

    # Need to start the backtracking from the first customer.
    return backtrack(0)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty nums or quantity array	Return true immediately if quantity is empty; return false if nums is empty but quantity is not.
nums array has only one element	Return true if the quantity array contains only one element and that element is less than or equal to the number of occurences of the element in nums, otherwise return false.
quantity array contains only one element	Check if the counts of any number in nums are greater than or equal to the one element of the quantity array; if so, return true, otherwise return false.
The sum of the quantity array is greater than the length of the nums array	Return false, since it's impossible to distribute more quantities than numbers available.
Large input size (nums and quantity arrays)	Ensure the solution employs efficient data structures (e.g., hash maps) and algorithms (e.g., backtracking with memoization) to avoid time limit exceeded errors.
Cases where no distribution is possible (e.g., quantity demands exceed available counts)	The backtracking algorithm should explore all possibilities and return false only if every distribution attempt fails.
Integer overflow when calculating counts of elements or during summation	Use appropriate data types (e.g., long) to store counts or sums to prevent integer overflows.
All elements in nums are the same and a quantity is larger than the count of this element	Return false if a quantity is larger than the count of the only element in nums after grouping similar nums.

Empty nums or quantity array

How to Handle:

Return true immediately if quantity is empty; return false if nums is empty but quantity is not.

nums array has only one element

How to Handle:

Return true if the quantity array contains only one element and that element is less than or equal to the number of occurences of the element in nums, otherwise return false.

quantity array contains only one element

How to Handle:

Check if the counts of any number in nums are greater than or equal to the one element of the quantity array; if so, return true, otherwise return false.

The sum of the quantity array is greater than the length of the nums array

How to Handle:

Return false, since it's impossible to distribute more quantities than numbers available.

Large input size (nums and quantity arrays)

How to Handle:

Ensure the solution employs efficient data structures (e.g., hash maps) and algorithms (e.g., backtracking with memoization) to avoid time limit exceeded errors.

Cases where no distribution is possible (e.g., quantity demands exceed available counts)

How to Handle:

The backtracking algorithm should explore all possibilities and return false only if every distribution attempt fails.

Integer overflow when calculating counts of elements or during summation

How to Handle:

Use appropriate data types (e.g., long) to store counts or sums to prevent integer overflows.

All elements in nums are the same and a quantity is larger than the count of this element

How to Handle:

Return false if a quantity is larger than the count of the only element in nums after grouping similar nums.