How do you solve Maximize Score After N Operations on LeetCode?

Maximize Score After N Operations (LeetCode #1799) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute force approach involves generating all possible pairs of elements from the array and calculating the score for each combination. This method is straightforward but inefficient as it explores all combinations exhaustively.

What is the time complexity of Maximize Score After N Operations?

The optimal solution for Maximize Score After N Operations runs in undefined time and uses undefined space. undefined

#1799

Maximize Score After N Operations

Hard

ArrayMathDynamic ProgrammingBacktrackingBit ManipulationNumber TheoryBitmask

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute force approach involves generating all possible pairs of elements from the array and calculating the score for each combination. This method is straightforward but inefficient as it explores all combinations exhaustively.

⚙️

Algorithm

3 steps

1Step 1: Generate all combinations of pairs from the nums array.
2Step 2: For each combination, calculate the score using the formula i * gcd(x, y) where i is the operation number.
3Step 3: Keep track of the maximum score obtained through all combinations.

solution.py20 lines

1# Full working Python code
2import itertools
3import math
4def maxScore(nums):
5    n = len(nums) // 2
6    def backtrack(used, score, op):
7        if op > n:
8            return score
9        max_score = 0
10        for i in range(len(nums)):
11            if used[i]: continue
12            for j in range(i + 1, len(nums)):
13                if used[j]: continue
14                used[i] = used[j] = True
15                current_score = op * math.gcd(nums[i], nums[j])
16                max_score = max(max_score, backtrack(used, score + current_score, op + 1))
17                used[i] = used[j] = False
18        return max_score
19    return backtrack([False] * len(nums), 0, 1)
20

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