Algorithm


Problem Name: 813. Largest Sum of Averages

You are given an integer array nums and an integer k. You can partition the array into at most k non-empty adjacent subarrays. The score of a partition is the sum of the averages of each subarray.

Note that the partition must use every integer in nums, and that the score is not necessarily an integer.

Return the maximum score you can achieve of all the possible partitions. Answers within 10-6 of the actual answer will be accepted.

 

Example 1:

Input: nums = [9,1,2,3,9], k = 3
Output: 20.00000
Explanation: 
The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.

Example 2:

Input: nums = [1,2,3,4,5,6,7], k = 4
Output: 20.50000

 

Constraints:

  • 1 <= nums.length <= 100
  • 1 <= nums[i] <= 104
  • 1 <= k <= nums.length

Code Examples

#1 Code Example with Javascript Programming

Code - Javascript Programming


const largestSumOfAverages = function(A, K) {
    const len = A.length
    const P = [0]
    for(let i = 0; i  <  len; i++) {
        P[i+1] = (P[i] || 0) + A[i]
    }
    const dp = []
    for(let j = 0; j  <  len; j++) {
        dp[j] = (P[len] - P[j]) / (len - j)
    }
    for(let m = 0; m  <  K - 1; m++) {
        for(let n = 0; n  <  len; n++) {
            for(let k = n + 1; k  <  len; k++) {
                dp[n] = Math.max(dp[n], (P[k] - P[n]) / (k - n) + dp[k])
            }
        }
    }
    return dp[0]
}
Copy The Code & Try With Live Editor

Input

x
+
cmd
nums = [9,1,2,3,9], k = 3

Output

x
+
cmd
20.00000

#2 Code Example with Python Programming

Code - Python Programming


class Solution:
    def largestSumOfAverages(self, A, K):
        memo = {}
        def search(n, k):
            if n < k: return 0
            if (n, k) not in memo:
                if k == 1: memo[n, k] = sum(A[:n]) / float(n)
                else:
                    cur = memo[n, k] = 0
                    for i in range(n - 1, 0, -1):
                        cur += A[i]
                        memo[n, k] = max(memo[n, k], search(i, k - 1) + cur / float(n - i))
            return memo[n, k]
        return search(len(A), K)
Copy The Code & Try With Live Editor

Input

x
+
cmd
nums = [9,1,2,3,9], k = 3

Output

x
+
cmd
20.00000
Advertisements

Demonstration


Previous
#812 Leetcode Largest Triangle Area Solution in C, C++, Java, JavaScript, Python, C# Leetcode
Next
#814 Leetcode Binary Tree Pruning Solution in C, C++, Java, JavaScript, Python, C# Leetcode