How do you solve Maximum Sum Obtained of Any Permutation on LeetCode?

Maximum Sum Obtained of Any Permutation (LeetCode #1589) 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: Higher frequency indices should be paired with larger values to maximize the sum.

What is the time complexity of Maximum Sum Obtained of Any Permutation?

The optimal solution for Maximum Sum Obtained of Any Permutation 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 frequency and nums arrays. The space complexity is O(n) for the frequency array.

What is the brute force solution for Maximum Sum Obtained of Any Permutation?

The brute force approach for Maximum Sum Obtained of Any Permutation is Brute Force, with O(n!) time complexity and O(n) space complexity. The brute force approach involves generating all possible permutations of the array `nums` and calculating the total sum for each permutation based on the given requests. This method is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Maximum Sum Obtained of Any Permutation?

Maximum Sum Obtained of Any Permutation uses the following concepts: Array, Greedy, Sorting, Prefix Sum. The recommended patterns to study are: Greedy, Sorting, Prefix Sum.

What are the interview tips for Maximum Sum Obtained of Any Permutation?

Always think about how to optimize your solution instead of brute-forcing. Understand the problem constraints and how they affect your approach. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Maximum Sum Obtained of Any Permutation?

Failing to account for the frequency of index access in requests. Not sorting the arrays before calculating the final sum.

#1589

Maximum Sum Obtained of Any Permutation

Medium

ArrayGreedySortingPrefix SumGreedySortingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

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

The optimal approach focuses on understanding the frequency of each index in the requests. By sorting `nums` in descending order and assigning larger values to indices that are accessed more frequently, we maximize the total sum efficiently.

⚙️

Algorithm

5 steps

1Step 1: Create a frequency array to count how many times each index is included in the requests.
2Step 2: For each request, increment the frequency for the start index and decrement after the end index to mark the range.
3Step 3: Compute the prefix sum of the frequency array to get the actual counts for each index.
4Step 4: Sort the frequency array and the `nums` array in descending order.
5Step 5: Multiply the sorted values together and sum them up to get the maximum total sum.

solution.py19 lines

1# Full working Python code
2import numpy as np
3
4MOD = 10**9 + 7
5
6def maxSum(nums, requests):
7    n = len(nums)
8    freq = [0] * (n + 1)
9    for start, end in requests:
10        freq[start] += 1
11        if end + 1 < n:
12            freq[end + 1] -= 1
13    for i in range(1, n):
14        freq[i] += freq[i - 1]
15    freq = freq[:-1]  # Remove the last extra element
16    nums.sort(reverse=True)
17    freq.sort(reverse=True)
18    max_sum = sum(f * n for f, n in zip(freq, nums)) % MOD
19    return max_sum

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of the frequency and nums arrays. The space complexity is O(n) for the frequency array.

1Higher frequency indices should be paired with larger values to maximize the sum.
2Using a frequency array allows us to efficiently calculate the contribution of each index.

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