How do you solve Largest Number After Digit Swaps by Parity on LeetCode?

Largest Number After Digit Swaps by Parity (LeetCode #2231) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Digits can only be swapped if they share the same parity.

What is the time complexity of Largest Number After Digit Swaps by Parity?

The optimal solution for Largest Number After Digit Swaps by Parity runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting of the even and odd digits, while the space complexity is O(n) for storing the digits.

What is the brute force solution for Largest Number After Digit Swaps by Parity?

The brute force approach for Largest Number After Digit Swaps by Parity is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible permutations of the digits of the number and checking which permutations can be formed by swapping digits of the same parity. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Largest Number After Digit Swaps by Parity?

Largest Number After Digit Swaps by Parity uses the following concepts: Sorting, Heap (Priority Queue). The recommended patterns to study are: Sorting, Two Pointers.

What are the interview tips for Largest Number After Digit Swaps by Parity?

Always clarify the constraints and requirements before diving into coding. Think about edge cases, such as numbers with all even or all odd digits. Discuss your thought process and reasoning while coding to demonstrate your understanding.

What are common mistakes when solving Largest Number After Digit Swaps by Parity?

Failing to separate digits by parity correctly. Not considering the impact of sorting on the final number.

#2231

Largest Number After Digit Swaps by Parity

Easy

SortingHeap (Priority Queue)SortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal solution involves separating the digits into even and odd groups, sorting them in descending order, and then reconstructing the number. This approach is efficient because it directly targets the problem without generating permutations.

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Algorithm

4 steps

1Step 1: Convert the number into a string to access each digit.
2Step 2: Separate the digits into two lists: one for even digits and one for odd digits.
3Step 3: Sort both lists in descending order.
4Step 4: Reconstruct the number by iterating through the original digits and replacing them with the largest available digit from the sorted lists based on parity.

solution.py16 lines

1# Full working Python code
2
3def largestNumber(num):
4    digits = list(str(num))
5    even_digits = sorted([d for d in digits if int(d) % 2 == 0], reverse=True)
6    odd_digits = sorted([d for d in digits if int(d) % 2 != 0], reverse=True)
7    even_index, odd_index = 0, 0
8    result = []
9    for d in digits:
10        if int(d) % 2 == 0:
11            result.append(even_digits[even_index])
12            even_index += 1
13        else:
14            result.append(odd_digits[odd_index])
15            odd_index += 1
16    return int(''.join(result))

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of the even and odd digits, while the space complexity is O(n) for storing the digits.

1Digits can only be swapped if they share the same parity.
2Sorting the digits allows us to maximize the number by placing larger digits in more significant positions.

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