LeetCode 813: Largest Sum of Averages Solution

Master LeetCode problem 813 (Largest Sum of Averages), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

813. Largest Sum of Averages

Medium

Array Dynamic Programming Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We define a 2D array dp where dp[i][j] represents the maximum score we can achieve by partitioning the subarray nums[0..i] into j+1 groups.

The recurrence relation for calculating dp[i][j] is as follows:

Initialize dp[i][0] as the average of nums[0..i].
For each j from 1 to k-1, we iterate over all possible partition points m (where m ranges from j-1 to i-1) to find the maximum score by considering the score of the partition nums[m+1..i] and the maximum score of partitioning nums[0..m] into j-1 groups.

The final answer will be stored in dp[n-1][k-1], where n is the length of the nums array.

Solution Code

class Solution {
    public double largestSumOfAverages(int[] nums, int k) {
        int n = nums.length;
        double[][] dp = new double[n][k];
        
        double sum = 0;
        for (int i = 0; i < n; i++) {
            sum += nums[i];
            dp[i][0] = sum / (i + 1);
        }
        
        for (int j = 1; j < k; j++) {
            for (int i = j; i < n; i++) {
                double maxScore = 0;
                double currSum = 0;
                for (int m = i; m >= j; m--) {
                    currSum += nums[m];
                    maxScore = Math.max(maxScore, dp[m - 1][j - 1] + currSum / (i - m + 1));
                }
                dp[i][j] = maxScore;
            }
        }
        
        return dp[n - 1][k - 1];
    }
}

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