How do you solve Largest Sum of Averages on LeetCode?

Largest Sum of Averages (LeetCode #813) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n² * k) time complexity and O(n * k) space complexity. Key insight: Dynamic programming can significantly reduce the time complexity compared to brute force.

What is the brute force solution for Largest Sum of Averages?

The brute force approach for Largest Sum of Averages is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible partitions of the array into at most k subarrays. For each partition, we calculate the sum of the averages of the subarrays and keep track of the maximum score found. The optimal Optimal Solution approach improves this to O(n² * k).

What algorithm or data structure is used to solve Largest Sum of Averages?

Largest Sum of Averages uses the following concepts: Array, Dynamic Programming, Prefix Sum. The recommended patterns to study are: Dynamic Programming, Prefix Sum.

What are the interview tips for Largest Sum of Averages?

Explain your thought process clearly while coding. Discuss the trade-offs between brute force and optimal solutions. Practice writing clean and efficient code.

What are common mistakes when solving Largest Sum of Averages?

Not considering all possible partitions in brute force. Overlooking the need for prefix sums in the optimal solution.

#813

Largest Sum of Averages

Medium

ArrayDynamic ProgrammingPrefix SumDynamic ProgrammingPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n² * k)Space O(n * k)

The optimal solution uses dynamic programming to efficiently calculate the maximum score. We keep track of the best scores for each possible number of partitions and use previously computed results to build up to the final answer.

⚙️

Algorithm

3 steps

1Step 1: Create a DP array where dp[i][j] represents the maximum score using the first i elements with j partitions.
2Step 2: Use a prefix sum array to quickly calculate the average of any subarray.
3Step 3: Iterate through the array and update the DP table based on previous results.

solution.py14 lines

1# Full working Python code
2def largestSumOfAverages(nums, k):
3    n = len(nums)
4    prefix_sum = [0] * (n + 1)
5    for i in range(n):
6        prefix_sum[i + 1] = prefix_sum[i] + nums[i]
7    dp = [[0] * (k + 1) for _ in range(n + 1)]
8    for i in range(1, n + 1):
9        dp[i][1] = prefix_sum[i] / i
10    for j in range(2, k + 1):
11        for i in range(1, n + 1):
12            for m in range(1, i + 1):
13                dp[i][j] = max(dp[i][j], dp[m][j - 1] + (prefix_sum[i] - prefix_sum[m]) / (i - m))
14    return dp[n][k]

ℹ

Complexity note: The time complexity is O(n² * k) because we iterate through the array and for each element, we check all previous partitions, which results in a nested loop structure.

1Dynamic programming can significantly reduce the time complexity compared to brute force.
2Using prefix sums allows for efficient average calculations.

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