How do you solve Distribute Repeating Integers on LeetCode?

Distribute Repeating Integers (LeetCode #1655) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(n) space complexity. Key insight: Count the frequencies of each number to understand how many can be assigned.

What is the brute force solution for Distribute Repeating Integers?

The brute force approach for Distribute Repeating Integers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to assign each customer their required quantity of integers from the available numbers, checking all possible combinations. This method is straightforward but can be inefficient due to its exhaustive nature. The optimal Optimal Solution approach improves this to O(n * m).

What are the interview tips for Distribute Repeating Integers?

Always clarify the problem requirements before jumping into code. Discuss your thought process and the approach you plan to take. Be ready to optimize your initial solution if time allows.

What are common mistakes when solving Distribute Repeating Integers?

Not counting the frequencies correctly. Failing to sort the quantity array before attempting to distribute.

#1655

Distribute Repeating Integers

Hard

ArrayHash TableDynamic ProgrammingBacktrackingBit ManipulationCountingBitmaskHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n * m)
Space	O(1)	O(n)

💡

Intuition

Time O(n * m)Space O(n)

The optimal solution uses a backtracking approach to efficiently explore the distribution of integers. By counting the frequencies of each integer and trying to satisfy each customer's order one by one, we can prune unnecessary checks and find a valid distribution faster.

⚙️

Algorithm

3 steps

1Step 1: Count the frequency of each unique integer in nums.
2Step 2: Sort the quantity array in descending order to prioritize larger orders first.
3Step 3: Use a backtracking function to attempt to assign integers to each customer, checking if the current assignment can satisfy the quantity requirements.

solution.py19 lines

1# Full working Python code
2from collections import Counter
3
4def canDistribute(nums, quantity):
5    freq = Counter(nums)
6    unique_numbers = sorted(freq.values(), reverse=True)
7    quantity.sort(reverse=True)
8    return backtrack(unique_numbers, quantity, 0)
9
10def backtrack(unique_numbers, quantity, index):
11    if index == len(quantity):
12        return True
13    for i in range(len(unique_numbers)):
14        if unique_numbers[i] >= quantity[index]:
15            unique_numbers[i] -= quantity[index]
16            if backtrack(unique_numbers, quantity, index + 1):
17                return True
18            unique_numbers[i] += quantity[index]
19    return False

ℹ

Complexity note: The time complexity is driven by the backtracking approach, where n is the number of unique integers and m is the number of customers. Each customer may require a different number of checks, but we avoid unnecessary checks through pruning.

1Count the frequencies of each number to understand how many can be assigned.
2Sort the quantity array to prioritize larger orders, which helps in backtracking.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.