What is the brute force solution for Count Number of Distinct Integers After Reverse Operations?

The brute force approach for Count Number of Distinct Integers After Reverse Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves reversing each number in the array and appending it to the end of the array. Then, we can check for distinct values by comparing each element, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Number of Distinct Integers After Reverse Operations?

Count Number of Distinct Integers After Reverse Operations uses the following concepts: Array, Hash Table, Math, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Number of Distinct Integers After Reverse Operations?

Always think about the data structures that can help with distinct counting. Clarify the problem requirements before jumping into coding. Test edge cases like single-element arrays or maximum constraints.

What are common mistakes when solving Count Number of Distinct Integers After Reverse Operations?

Not using a set, leading to incorrect distinct counts. Forgetting to handle leading zeros after reversal.

#2442

Count Number of Distinct Integers After Reverse Operations

Medium

ArrayHash TableMathCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a set to store the original numbers and their reversed counterparts. This allows us to efficiently track distinct integers without needing to create a combined array.

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Algorithm

4 steps

1Step 1: Initialize an empty set to hold distinct numbers.
2Step 2: For each number in the input array, add it to the set.
3Step 3: Reverse the number and add the reversed number to the set.
4Step 4: Return the size of the set as the count of distinct integers.

solution.py8 lines

1# Full working Python code
2nums = [1, 13, 10, 12, 31]
3distinct_set = set()
4for num in nums:
5    distinct_set.add(num)
6    reversed_num = int(str(num)[::-1])
7    distinct_set.add(reversed_num)
8print(len(distinct_set))

ℹ

Complexity note: The time complexity is O(n) because we process each number once, and adding to a set is average O(1). The space complexity is O(n) due to the storage of distinct numbers in the set.

1Using a set efficiently tracks distinct elements.
2Reversing digits can be done easily with string manipulation.

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