Algorithm
Problem Name: 813. Largest Sum of Averages
Problem Link: https://leetcode.com/problems/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 <= 1001 <= nums[i] <= 1041 <= 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]
}
Input
nums = [9,1,2,3,9], k = 3
Output
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)
Input
nums = [9,1,2,3,9], k = 3
Output
20.00000