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Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp=[0]*(len(arr)+1)
dp[1]=arr[0]
for i in range(1,len(arr)):
for j in range(max(0,i-k+1),i+1):
dp[i+1]=max(dp[i+1],dp[j]+max(arr[j:i+1])*(i-j+1))
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [[0 for _ in range(len(A))] for _ in range(2)]
dp2 = [0 for _ in range(len(A))]
for i in range(len(A)):
# we have 2 options - to partition or not to partition.
# if i - K + 1 < 0:
# maxCand = max(A[:i+1])
# size = i + 1
# else:
# maxCand = max(A[i - K + 1:i+1])
# size = K
# # if we choose not to partition
# dp[0][i] = size * maxCand + max(dp[0][i - K], dp[1][i - K])
# # if we choose to partition at this candidate
# dp[1][i] = A[i] + max(dp[0][i - 1], dp[1][i - 1])
startIndex = max(0, i - K + 1)
while startIndex <= i:
if startIndex - 1 >= 0:
dp2[i] = max(dp2[startIndex - 1] + max(A[startIndex:i+1])*(i + 1 - startIndex), dp2[i])
else:
dp2[i] = max(A[:i+1]) * (i + 1)
startIndex += 1
print(dp2)
return dp2[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
#
def rec(l,h):
if h-l<K:
dp[l][h] = (h-l+1)*max(A[l:h+1])
return dp[l][h]
m = []
if dp[l][h]>0:
return dp[l][h]
for i in range(0,K):
m += [rec(l,l+i) + rec(l+i+1,h)]
dp[l][h] = max(m)
return dp[l][h]
dp = [0]*len(A)
dp = [[0]*len(A) for i in range(len(A))]
rec(0,len(A)-1)
return dp[0][len(A)-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
size = len(arr)
mem = [0] * size
for i in range(size):
mem[i] = arr[i]
for j in range(0, k):
if i - j > 0:
candidate = max(arr[i - j: i + 1]) * (j + 1) + mem[i - j - 1]
mem[i] = max(mem[i], candidate)
elif i - j == 0:
candidate = max(arr[i - j: i + 1]) * (j + 1)
mem[i] = max(mem[i], candidate)
else:
break
return mem[size - 1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
A = arr
dp = [0] * (n := len(arr))
dp[0] = curr_max = arr[0]
for i in range(1, k):
curr_max = max(curr_max, A[i])
dp[i] = (i + 1) * curr_max
for i in range(k, n):
curr_max = A[i]
for j in range(k):
curr_max = max(curr_max, A[i - j])
dp[i] = max(dp[i], dp[i - j - 1] + curr_max * (j + 1))
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0]
for i in range(len(arr)):
max_sum = -float('inf')
for j in range(1, k+1):
if i - j + 1 < 0:
break
max_sum = max(max_sum, max(arr[i-j + 1: i+1]) * j + dp[i - j + 1])
dp.append(max_sum)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def getMax(self, arr, k, idx, cache):
# I am partitioning 1 element
if idx == len(arr) - 1:
return arr[idx]
# I need to get max sum after partitioning
maxSum = 0
# [1,2,9,30]
# I need to try partitioning from 1 -> K elements
for numInPartition in range(1, k + 1):
# I cannot choose elements if I would be choosing too many elements
if idx + numInPartition > len(arr):
break
startOfRecursiveIndex = idx + numInPartition
# select max value in first 'k' elements
maxVal = max(arr[idx:startOfRecursiveIndex])
# generates sum for this subset
partSum = maxVal * numInPartition
# I need to get the partition sum for the rest of the elements
rest = cache[startOfRecursiveIndex] if startOfRecursiveIndex in cache else self.getMax(arr, k, startOfRecursiveIndex, cache)
cache[startOfRecursiveIndex] = rest
maxSum = max(maxSum, partSum + rest)
return maxSum
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
'''
Input:
A: int[] - Array of numbers
K: int - Max length allowable for subarray
Output:
Return largest sum after partitioning.
Steps:
1. Make partition
2. Change all elements to max value
v
[1,2,9,30], 2
{1,2}{9,30} => {2,2}{30,30} => 4 + 60 = 64
{1}{2,9}{30} => {1}{9,9}{30} => 1 + 18 + 30 = 49
Example:
[1]
{1} = 1
[1,2]
{1}{2} => 1+2 = 3
{1,2} => {2,2} => 4
v
[1,2,9]
{1,2}{9} => {2,2} + {9} => 4 + 9 = 13
{1}{2,9} => {1},{9,9} = 1 + 18 = 19
{1,2,9}
{1}{2,9}{30} => {1}{9,9}{30} => 19 + 30 => 49
{1,2}{9,30} => {2,2}{30,30} => 4 + 60 => 64
v
[1,2,9,30]
{1,2,9} {30} => 30 + getMaxSum([1,2,9]) => 30 + 19 => 49
{1,2} {9,30} => 60 + getMaxSum([1,2]) => 60 + 4 = 64
{1}{2,9,30}
{1,2}{9,30}
{1,2,9}{30}
All about subproblems!!
Approach:
- Let's look into sliding window.
- We can frame this problem in terms of choices.
When considering an element, we can place it into its own partition OR with the preceeding partition.
Choice:
K-way choice
Do I group my current element in the last [1,k] elements?
Constraint:
Subset needs to be contiguous?
Goal:
Maximum sum
1. When placing
Approach:
[0,0,0,0]
maxSum[i] = max(maxSum[i-1] + A[i] * 1, maxSum[i-2] + max(last two elements) * 2) and so on
'''
cache = {}
return self.getMax(A, K, 0, cache)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
mem = {}
def find(s, e):
if (s,e) in mem:
return mem[(s,e)]
if e-s == 1:
return arr[s]
elif e-s == 0:
return 0
else:
m = None
for i in range(s+1, min(s+k+1, len(arr)+1)):
subsum = max(arr[s:i])*(i-s) + find(i,e)
if m is None or subsum > m:
m = subsum
mem[(s,e)] = m
return m
return find(0, len(arr))
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
if not arr:
return 0
n = len(arr)
dp = [0] * n
for i in range(n):
for j in range(k):
if i - j < 0:
continue
_max = max(arr[i-j : i+1])
_sum = _max * (j+1) if i-j == 0 else dp[i-j-1] + _max*(j+1)
dp[i] = max(dp[i], _sum)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, nums: List[int], k: int) -> int:
n = len(nums)
dp = [nums[0]] + [0] * (n - 1)
for i in range(1, n):
for j in range(max(0, i - k + 1), i + 1):
dp[i] = max(dp[i], dp[j - 1] + max(nums[j:i + 1]) * (i + 1 - j))
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
from functools import lru_cache
@lru_cache(None)
def dp(i):
ans=0
for k in range(K):
if i-k>=0:
ans=max(ans,max(A[i-k:i+1])*(k+1)+dp(i-k-1))
return ans
return dp(len(A)-1)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
if not arr:
return 0
n = len(arr)
dp = [0]*n
for i in range(n):
for j in range(k):
if i-j < 0:
continue
_max = max(arr[i-j:i+1])
_sum = _max*(j+1) if i-j == 0 else dp[i-j-1] + _max*(j+1)
dp[i] = max(dp[i], _sum)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr, k):
if not arr:
return 0
dp = [0]*len(arr)
for i in range(len(arr)):
for j in range(k):
if i-j < 0 :
continue
max_val = max(arr[i-j : i+1])
if i-j == 0:
total = max_val * (j+1)
else:
total = dp[i-j-1] + max_val * (j+1)
dp[i] = max(dp[i], total)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
N = len(A)
dp = [0] * N
dp[0] = A[0]
for i in range(1, N):
for j in range(0, min(K, i+1)):
current_subarray = A[i-j:i+1]
if i >= K:
dp[i] = max(dp[i], dp[i-j-1] + max(current_subarray)*(j+1))
else:
dp[i] = max(dp[i], max(current_subarray)*(j+1))
return dp[-1]
# [1,15,7,9,2,5,10] k = 3
# No need to inspect all combination, focus on growth
# (1)
# [1] -> 1 - max(A[i])*1
# (1,15)
# [1,15] -> 30 - max(A[i-1], A[i])*2
# [1][15] -> 16 - DP[i-1] + A[i]
0,1
1,1
# (1,15,7)
# [1,15,7] -> 45 max(A[i]. A[i-1], A[-2])*3
# [1][15,7] -> 23 - DP[i-2] + max(A[i-1], A[i])*2
# [1,15][7] -> 37 - DP[i-1] + A[i]
# (1,15,7,9)
# [1,15,7][9] -> 54 DP[i-1] + A[i]
# [1,15][7,9] -> 32 DP[i-2] + max(A[i-1], A[i])*2
# [1][15,7,9] -> 46 DP[i-3] + max(A[i-2], A[i-1], A[i])*3
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * n
for i in range(n):
if i == 0:
dp[i] = A[0]
else:
# A[i-k+1, ..., i-1, i] is a subarray
# A[0, 1, 2, ..., i-k] is previous chuck with subarray length up to K
for k in range(1, K+1):
if i-k+1 < 0:
break
elif i-k+1 == 0:
dp[i] = max(dp[i], max(A[i-k+1:i+1]) * k)
else:
dp[i] = max(dp[i], dp[i-k] + max(A[i-k+1:i+1]) * k)
return dp[n-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
if not arr:
return 0
n = len(arr)
dp = [0]*n
for i in range(n):
for j in range(k):
if i-j < 0:
continue
_max = max(arr[i-j:i+1])
_sum = _max*(j+1) if i-j == 0 else dp[i-j-1] + _max*(j+1)
dp[i] = max(dp[i], _sum)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n + 1)
for i in range(n):
for k in range(1,K+1):
if i-k > -2: dp[i+1] = max(dp[i+1], max(A[max(0,i-k+1):i+1]) * k + dp[i+1-k])
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n + 1)
for i in range(n):
dp_ix = i+1
for k in range(1,K+1):
if i-k > -2:
dp[dp_ix] = max(dp[dp_ix], max(A[max(0,i-k+1):i+1]) * k + dp[dp_ix-k])
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [[0 for _ in range(len(A))] for _ in range(2)]
dp2 = [0 for _ in range(len(A))]
for i in range(len(A)):
# we have 2 options - to partition or not to partition.
# if i - K + 1 < 0:
# maxCand = max(A[:i+1])
# size = i + 1
# else:
# maxCand = max(A[i - K + 1:i+1])
# size = K
# # if we choose not to partition
# dp[0][i] = size * maxCand + max(dp[0][i - K], dp[1][i - K])
# # if we choose to partition at this candidate
# dp[1][i] = A[i] + max(dp[0][i - 1], dp[1][i - 1])
startIndex = max(0, i - K + 1)
currMax = 0
while startIndex <= i:
currMax = max(currMax, A[startIndex])
if startIndex - 1 >= 0:
dp2[i] = max(dp2[startIndex - 1] + max(A[startIndex:i+1])*(i + 1 - startIndex), dp2[i])
else:
dp2[i] = max(A[:i+1]) * (i + 1)
startIndex += 1
print(dp2)
return dp2[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0] * len(arr)
for i in range(len(arr)):
print(arr[i])
max_so_far = arr[i] + (dp[i-1] if i > 0 else 0)
for j in range(1, min(i+1, k)):
max_in_range = max(arr[i-j:i+1])
max_so_far = max(max_so_far, dp[i-j-1] + max_in_range * (j+1))
dp[i] = max_so_far
return dp[len(arr)-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0] * (len(arr))
dp[0] = arr[0]
for i,val in enumerate(arr):
if i == 0: continue
cur_max = val + dp[i-1]
cur_k = 1
while(i-cur_k>=0 and cur_k < k):
temp = arr[i-cur_k:i+1]
m = max(temp)
if i-cur_k==0: cur_max = max(cur_max, m* (cur_k+1) )
else:
cur_max = max(cur_max,( m * (cur_k+1)) + dp[i-cur_k-1] )
cur_k += 1
dp[i] = cur_max
return dp[-1]
'''
#[1,4,1, 5, 7, 3, 6, 1, 9, 9, 3]
# 1 5
'''
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
DP = [0 for _ in range(len(arr)+1)]
for i in range(1,len(DP)):
loc = max(i-k,0)
run_max = 0
for j in range(loc,i):
run_max = max(run_max,DP[j]+max(arr[j:i])*(i-j))
DP[i] = run_max
return DP[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
if not K or K == 1:
return sum(A)
if not A:
return 0
dp = [0] * len(A)
for index, num in enumerate(A):
possible = []
for group in range(K):
if index - group >= 0:
if index - group - 1 >= 0:
previous = dp[index - group - 1]
else:
previous = 0
possible.append(
previous + max(A[index - group : index + 1]) * (group + 1)
)
dp[index] = max(possible)
# print(f\"{index=} {dp[index]=}\")
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
self.arr = arr
self.dp = {}
def helper(i,j,k):
if j <= i:
return 0
if (i,j) in self.dp:
return self.dp[(i,j)]
maxsum = -sys.maxsize
for length in range(1,k+1):
if i+length <= j:
currsum = (max(self.arr[i:i+length]) * length) + helper(i+length,j,k)
maxsum = max(maxsum,currsum)
self.dp[(i,j)] = maxsum
return self.dp[(i,j)]
return helper(0,len(arr),k)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp=[0 for _ in A]
for i,num in enumerate(A):
for j in range(K):
ans=0
if i-j>=0:
ans = ans + (j+1)*max(A[i-j:i+1])
if i-j-1>=0:
ans = ans + dp[i-j-1]
dp[i]=max(dp[i],ans)
# if i-p-1>=0:
# ans=dp[i-p-1]+(p+1)*max(A[i-p:i+1])
# print(A[i-p-1],(p+1)*max(A[i-p:i+1]))
# dp[i]=max(dp[i],ans)
#print(dp)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
d = [0] * (1 + len(arr) + k)
for ind in range(1, len(arr) + 1):
for sub_arr_st in range(min(ind, k)):
if arr[ind - sub_arr_st - 1] > arr[ind - 1]:
break
for sub_arr_len in range(sub_arr_st + 1, k + 1):
ind_x = ind - sub_arr_st + sub_arr_len - 1
d[ind_x] = max(d[ind_x], d[ind - sub_arr_st - 1] + arr[ind - 1] * sub_arr_len)
return d[len(arr)]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp=[0]*(len(arr)+1)
dp[0]=arr[0]
for i in range(1,len(arr)):
ans=-100000
for z in range(max(0,i-k+1) , i+1):
# print(z,i+1)
ans=max(dp[z-1]+max(arr[z:i+1])*(i-z+1) , ans)
# ans=max(dp[i-z-1]+max(arr[i-z-1:i+1])*(z+1),ans)
dp[i]=ans
return dp[-2]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
cacheMax=[[0]*len(arr) for i in range(len(arr))]
for i in range(len(arr)):
cacheMax[i][i]=arr[i]
for i in range(1,len(arr)):
for j in range(i):
cacheMax[j][i] = max(cacheMax[j][i-1],arr[i])
dp=[0]*(len(arr)+1)
dp[1]=arr[0]
for i in range(1,len(arr)):
for j in range(max(0,i-k+1),i+1):
dp[i+1]=max(dp[i+1],dp[j]+max(arr[j:i+1])*(i-j+1))
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
memo = {}
l_arr = len(arr)
for chain_len in range(0, k, 1):
for i in range(0, l_arr - chain_len, 1):
memo[(i, i + chain_len)] = max(arr[i : i + chain_len + 1]) * (chain_len + 1)
res_memo = {}
def dfs(idx):
if idx == l_arr:
return 0
if idx in res_memo:
return res_memo[idx]
res_memo[idx] = max( memo[(idx, idx + j)] + dfs(idx + j + 1) for j in range(k) if idx + j < l_arr )
return res_memo[idx]
return dfs(0)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
# use dp to solve this question
n = len(A)
dp = [0] * (n + 1)
# dp[i] is the answer for array A[0] ... A[i-1]
# for j = 1 .. k,
# dp[i] is the max of dp[i-j] + max(A[i-1] ... A[i-j]) * j
# j = 1
# dp[i-1] + max(A[i-1]) * 1
# dp[i-2] + max(A[i-1], A[i-2]) * 2
# dp[i-3] + max(A[i-1], A[i-2], A[i-3]) * 3
# ...
# dp[i-k] + max(A[i-1], ..., A[i-k]) * 3
for i in range(1, n+1):
for j in range(1, K+1):
if i-j>=0:
dp[i] = max(dp[i], dp[i-j] + max(A[i-j:i]) * j) # from i-j to i-1, with i-1-i+j+1 = j elements
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [0 for i in range(len(A)+1)]
for i in range(1, len(A)+1):
dp[i] = max([dp[i-j] + j * max(A[i-j:i]) for j in range(1, K+1) if j <= i])
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
res = [0] * (len(arr) + 1)
for i in range(1, len(arr) + 1):
for j in range(1, k + 1):
if i - j > -1:
res[i] = max(res[i], res[i - j] + max(arr[i - j : i]) * j)
return res[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
ar_len = len(arr)
dp= [0] * (ar_len+1)
for i in range(1,len(dp)):
tmp_max = []
for j in range(1,k+1):
if i-j >=0:
max_t = dp[i-j] + max(arr[i-j:i]) * j
tmp_max.append(max_t)
dp[i] = max(tmp_max)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
if k == n: return n * max(arr)
dp = collections.deque([0] * k)
for i in range(n - 1, -1, -1):
m = 0
result = 0
for j in range(min(k, n - i)):
m = max(m, arr[i + j])
result = max(result, m * (j + 1) + dp[j])
dp.appendleft(result)
dp.pop()
return dp[0]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
N = len(A)
dp = [0] * N
dp[0] = A[0]
#print('DP:', dp)
#print()
for i in range(1, N):
for j in range(0, min(K, i+1)):
current_subarray = A[i-j:i+1]
#print( max(current_subarray)*(j+1), current_subarray, i, j)
if i >= K:
dp[i] = max(dp[i], dp[i-j-1] + max(current_subarray)*(j+1))
else:
dp[i] = max(dp[i], max(current_subarray)*(j+1))
#print('DP:', dp)
#print()
return dp[-1]
# [1,15,7,9,2,5,10] k = 3
# No need to inspect all combination, focus on growth
# (1)
# [1] -> 1 - max(A[i])*1
# (1,15)
# [1,15] -> 30 - max(A[i-1], A[i])*2
# [1][15] -> 16 - DP[i-1] + A[i]
0,1
1,1
# (1,15,7)
# [1,15,7] -> 45 max(A[i]. A[i-1], A[-2])*3
# [1][15,7] -> 23 - DP[i-2] + max(A[i-1], A[i])*2
# [1,15][7] -> 37 - DP[i-1] + A[i]
# (1,15,7,9)
# [1,15,7][9] -> 54 DP[i-1] + A[i]
# [1,15][7,9] -> 32 DP[i-2] + max(A[i-1], A[i])*2
# [1][15,7,9] -> 46 DP[i-3] + max(A[i-2], A[i-1], A[i])*3
# N = len(A)
# dp = [0] * (N + 1)
# for i in range(N):
# curMax = 0
# for k in range(1, min(K, i + 1) + 1):
# curMax = max(curMax, A[i - k + 1])
# dp[i] = max(dp[i], dp[i - k] + curMax * k)
# print(dp)
# print()
# return dp[N - 1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, nums: List[int], k: int) -> int:
'''
v
[1,30,45,_,_,_,_,????]
[1,15,7, 9,2,5,10], 2
dp[i] = max( (windowSize * maxInWindow) + dp[i - windowSize] ) for valid windowSize from 1 to k
return dp[-1]
30 + [7] => 37
1 + [15,7] => 31
0 + [1,15,7] => 45 <<<<<
[1,15]
'''
dp = [0] * len(nums)
for i in range(len(nums)):
# generate all window sizes
# [1,2,3]
# if k = 3
for windowSize in range(1, k + 1):
startIndex = i - windowSize + 1
# Bounding Function to make sure that I don't run out of bounds
if startIndex < 0:
break
# v
# v
# # [1,15,7,9,2,5,10]
maxVal = max(nums[startIndex: i + 1])
currVal = windowSize * maxVal
currSum = currVal
if startIndex > 0:
currSum += dp[startIndex - 1]
dp[i] = max(dp[i], currSum)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
memo = {n:0}
def helper(arr, i):
if i in memo:
return memo[i]
res = 0
for j in range(i, min(n,i+k)):
res = max(res, helper(arr, j+1)+max(arr[i:j+1])*(j+1-i))
memo[i] = res
return res
return helper(arr, 0)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, nums: List[int], k: int) -> int:
'''
v
[1,30,45,_,_,_,_,????]
[1,15,7, 9,2,5,10], 2
dp[i] = max( (windowSize * maxInWindow) + dp[i - windowSize] ) for valid windowSize from 1 to k
return dp[-1]
30 + [7] => 37
1 + [15,7] => 31
0 + [1,15,7] => 45 <<<<<
[1,15]
'''
dp = [0] * len(nums)
for i in range(len(nums)):
# generate all window sizes
# [1,2,3]
# if k = 3
for windowSize in range(1, k + 1):
startIndex = i - windowSize + 1
# Bounding Function to make sure that I don't run out of bounds
if startIndex < 0:
break
# v
# v
# # [1,15,7,9,2,5,10]
maxVal = max(nums[startIndex: i + 1])
currVal = windowSize * maxVal
currSum = currVal
if startIndex > 0:
currSum += dp[startIndex - 1]
dp[i] = max(dp[i], currSum)
print(dp)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [0 for i in range(len(A)+1)]
for i in range(1, len(A)+1):
dp[i] = max([dp[i-j] + j * max(A[i-j:i]) for j in range(1, K+1) if j <= i])
print(dp)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0] * len(arr)
for i in range(len(arr)):
max_so_far = arr[i] + (dp[i-1] if i > 0 else 0)
for j in range(1, min(i+1, k)):
max_in_range = max(arr[i-j:i+1])
max_so_far = max(max_so_far, dp[i-j-1] + max_in_range * (j+1))
dp[i] = max_so_far
return dp[len(arr)-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
result = [0]*len(arr)
result[0] = arr[0]
max_val = result[0]
for i in range(1,k):
max_val = max(max_val,arr[i])
result[i] = max_val*(i+1)
#print(result)
for i in range(k,len(arr)):
max_val = arr[i]
for j in range(1,k+1):
max_val = max(max_val,arr[i-j+1])
result[i] = max(result[i],result[i-j]+max_val*j)
#print(i,j,max_val,result[i])
#print(result)
return result[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
A = arr
K = k
n = len(A)
dp = [0] * n
curMax = 0
for i in range(n):
if i < K:
curMax = max(curMax, A[i])
dp[i] = curMax * (i + 1)
else:
curMax = 0
for j in range(1, K + 1):
curMax = max(A[i - j + 1], curMax)
dp[i] = max(dp[i], dp[i - j] + curMax * j)
# print(dp)
return dp[n - 1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
if not arr:
return 0
n = len(arr)
dp = [0] * n
curr_max = arr[0]
for i in range(k):
curr_max = max(curr_max, arr[i])
dp[i] = curr_max * (i + 1)
for i in range(k, n):
dp[i] = arr[i]
curr_max = arr[i]
for j in range(k):
curr_max = max(curr_max, arr[i-j])
dp[i] = max(dp[i], dp[i-j-1] + curr_max * (j + 1))
return dp[-1]
# @lru_cache(None)
# def dp(idx):
# if idx < 0:
# return 0
# left = max(0, idx - k + 1)
# curr_max = arr[idx]
# res = arr[idx]
# for i in range(idx, left - 1, -1):
# curr_max = max(curr_max, arr[i])
# res = max(res, dp(i-1) + curr_max * (idx - i + 1))
# return res
# return dp(len(arr)-1)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * (n + 1)
for i in range(1, n+1):
max_val = float('-inf')
for j in range(1, min(i, k)+1):
max_val = max(max_val, arr[i-j])
dp[i] = max(dp[i], dp[i-j] + j*max_val)
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp=[0]*len(arr)
dp[0]=arr[0]
max1=arr[0]
for i in range(1,k):
max1=max(max1,arr[i])
dp[i]=max1*(i+1)
for i in range(k,len(arr)):
max1=arr[i]
for j in range(1,k+1):
dp[i]=max(dp[i],max1*j+dp[i-j])
max1=max(arr[i-j],max1)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
f = [0]
for end in range(1, len(arr)+1):
_local = 0
_cur_max = 0
for start in range(end-1, max(0, end-k)-1, -1):
_cur_max = max(_cur_max, arr[start])
_local = max(_cur_max *(end - start)+f[start], _local)
f.append(_local)
return f[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr, k):
n = len(arr)
dp = [0]*(n+1)
for i in range(n):
cmax = 0
for j in range(1, min(k, i+1) + 1):
cmax = max(cmax, arr[i-j+1])
dp[i] = max(dp[i], dp[i-j] + cmax*j)
return dp[-2]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * (n + 1)
for i in range(n):
curMax = 0
for j in range(1, min(k, i + 1) + 1):
curMax = max(curMax, arr[i - j + 1])
dp[i] = max(dp[i], dp[i - j] + curMax * j)
return dp[n - 1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
best_sums = [0] * (len(arr)+1)
for i in range(len(arr)):
curr_max = 0
for j in range(1,min(k+1,i+2)):
curr_max = max(curr_max, arr[i-j+1])
best_sums[i] = max(best_sums[i], best_sums[i-j] + j*curr_max)
return best_sums[-2]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
DP=[0]*len(arr)
N=len(arr)
DP[0]=arr[0]
m=arr[0]
for i in range(k):
m=max(m,arr[i])
DP[i]=m*(i+1)
# print(DP)
for i in range(k,N):
mm=arr[i]
DP[i]=DP[i-1]+arr[i]
# print(DP[i],i)
# print(i,DP,mm,arr[i])
for j in range(1,k):
mm=max(mm,arr[i-j])
# print(i,mm,arr[i-j],DP[i-j])
DP[i]=max(DP[i],DP[i-j-1]+mm*(j+1))
#DP[i]=max(DP[i-1]+arr)
# print(DP)
return DP[-1]
# [5,7,7,7,7]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n=len(arr)
dp=[0]*(n+1)
for i in range(n):
cur=0
for j in range(1,min(k,i+1)+1):
cur=max(cur,arr[i-j+1])
dp[i]=max(dp[i],cur*j+dp[i-j])
return dp[-2]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * (n + 1)
for i in range(n):
cur_max = 0
for j in range(1, min(k, i + 1) + 1):
cur_max = max(cur_max, arr[i - j + 1])
dp[i] = max(dp[i], dp[i - j] + cur_max * j)
return dp[n - 1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
@lru_cache(None)
def dp(idx):
if idx < 0:
return 0
left = max(idx - k + 1, 0)
curr_max = arr[idx]
res = arr[idx]
for i in range(idx, left - 1, -1):
curr_max = max(curr_max, arr[i])
res = max(res, dp(i-1) + (idx - i + 1) * curr_max)
return res
# print(dp(3))
return dp(len(arr) - 1)
# A = arr
# dp = [0] * (n := len(arr))
# dp[0] = curr_max = arr[0]
# for i in range(1, k):
# curr_max = max(curr_max, A[i])
# dp[i] = (i + 1) * curr_max
# for i in range(k, n):
# curr_max = A[i]
# for j in range(k):
# curr_max = max(curr_max, A[i-j])
# dp[i] = max(dp[i], dp[i-j-1] + curr_max * (j + 1))
# return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
N = len(arr)
dp = [0] * (N+1)
for i in range(N):
cur_max = 0
for j in range(1, min(k, i+1)+1):
cur_max = max(cur_max, arr[i-j+1]) # 从当前元素开始
dp[i] = max(dp[i], dp[i-j]+cur_max*j)
return dp[N-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
# O(NK) coming sooooon
N = len(arr)
dp = [0 for _ in range(N)]
dp[0] = arr[0]
max_so_far = arr[0]
for i in range(1, k):
max_so_far = max(max_so_far, arr[i])
dp[i] = (i+1) * max_so_far
# now the actual thing
for i in range(k, N):
max_so_far = -sys.maxsize
for j in range(i, i-k, -1):
max_so_far = max(max_so_far, arr[j])
dp[i] = max(dp[i], dp[j-1] + (i-j+1) * max_so_far)
return dp[N-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n + 1)
for i in range(1, n + 1):
max_val = float('-inf')
for w in range(1, K + 1):
if i - w < 0: break
max_val = max(max_val, A[i - w])
dp[i] = max(dp[i], dp[i - w] + max_val * w)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n + 1)
for i in range(1, n + 1):
cur = 0
for j in range(1, K + 1):
if i >= j:
cur = max(cur, A[i - j])
dp[i] = max(dp[i], dp[i - j] + cur * j)
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * (n+1)
for i in range(1,n+1):
j = i - 1
mx = -float('inf')
while i - j <= k and j >= 0:
mx = max(mx,arr[j])
dp[i] = max(dp[i],dp[j]+mx * (i - j))
j -= 1
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n+1)
for i in range(1,n+1):
j = i - 1
mx = -float('inf')
while i - j <= K and j >= 0:
mx = max(mx,A[j])
dp[i] = max(dp[i],dp[j]+mx * (i - j))
j -= 1
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0] * (len(arr)+1)
for i in range(1, len(dp)):
maxone = 0
for j in range(i, i-k, -1):
if j >= 1:
maxone = max(maxone, arr[j-1])
dp[i] = max(dp[i], dp[j-1]+maxone * (i-j+1))
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A, K):
N = len(A)
dp = [0] * (N + 1)
for i in range(N):
curMax = 0
for k in range(1, min(K, i + 1) + 1):
curMax = max(curMax, A[i - k + 1])
dp[i] = max(dp[i], dp[i - k] + curMax * k)
return dp[N - 1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
N = len(A)
DP = [0]*(N+1)
for i in range(N):
curMax = 0
for k in range(1, min(K, i+1) + 1):
curMax = max(curMax, A[i-k+1])
DP[i] = max(DP[i], DP[i-k] + curMax * k)
return DP[N-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * (n+1)
for i in range(n):
cur_max = 0
for d in range(1, min(k, i+1)+1):
cur_max = max(cur_max, arr[i-d+1])
dp[i+1] = max(dp[i+1], dp[i+1-d] + cur_max * d)
# print(dp)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n=len(arr)
dp=[0]*(n+1)
for i in range(n):
m=-1
x=-1
for j in range(1,k+1):
if (i-j+1)<0:
break
x=max(x,arr[i-j+1])
m=max(m,dp[i-j+1]+x*j)
dp[i+1]=m
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0 for _ in arr]
rolling_max = -1
for i in range(0, k):
rolling_max = max(rolling_max, arr[i])
dp[i] = rolling_max * (i + 1)
for i in range(k, len(arr)):
rolling_max = arr[i]
for j in range(1, k + 1):
rolling_max = max(rolling_max, arr[i - j + 1])
dp[i] = max(rolling_max * j + dp[i - j], dp[i])
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0]
for i in range(1, len(arr) + 1) :
max_arr_num = 0
now_max = 0
for j in range(1, k + 1) :
idx = i - j
if idx >= 0 :
max_arr_num = max(max_arr_num, arr[idx])
now_max = max(now_max, dp[idx] + max_arr_num * j)
dp.append(now_max)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n+1)
for i in range(1,n+1):
m = float('-inf')
for j in range(1,K+1):
if i - j >= 0:
m = max(A[i-j],m)
dp[i] = max(dp[i],dp[i-j] + m*j)
return dp[-1]
# n = len(A)
# dp = [0] * (n+1)
# for i in range(1,n+1):
# m = float('-inf')
# for j in range(1,K+1):
# if i - j >= 0:
# m = max(A[i-j],m)
# dp[i] = max(dp[i],dp[i-j]+m*j)
# return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n+1)
res = 0
for i in range(n):
max_val = 0
for k in range(1, min(K, i+1)+1):
max_val = max(max_val, A[i-k+1])
dp[i] = max(dp[i], dp[i-k] + max_val*k)
return dp[n-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0] * (len(arr))
dp[0] = arr[0]
for i,val in enumerate(arr):
if i == 0: continue
cur_max = val + dp[i-1]
cur_k = 1
m = val
while(i-cur_k>=0 and cur_k < k):
# temp = arr[i-cur_k:i+1]
m = max(m,arr[i-cur_k])
if i-cur_k==0: cur_max = max(cur_max, m* (cur_k+1) )
else:
cur_max = max(cur_max,( m * (cur_k+1)) + dp[i-cur_k-1] )
cur_k += 1
dp[i] = cur_max
return dp[-1]
'''
#[1,4,1, 5, 7, 3, 6, 1, 9, 9, 3]
# 1 5
'''
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
d = [0] * (len(arr) + 1)
for i in range(len(arr)):
cur_max = 0
for kk in range(1, k + 1):
if i - kk + 1 < 0:
continue
cur_max = max(cur_max, arr[i-kk+1])
# print(i, kk, \",\", i - kk + 1, cur_max, arr[i-kk+1], ':', d[i], d[i-kk])
d[i] = max(d[i], d[i-kk] + cur_max*kk)
print(d)
return d[-2]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
N = len(A)
dp = [0]*(N+1)
for i in range(N):
curr_max = 0
for j in range(1, K+1):
if i+j>N: break
curr_max = max(curr_max, A[i+j-1])
dp[i+j] = max(dp[i+j], dp[i]+curr_max*j)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [float('-inf') for i in range(len(A)+1)]
dp[-1] = 0
dp[-2] = A[-1]
for j in reversed(list(range(len(A)-1))):
cur_max = float('-inf')
for k in range(K):
if j+k == len(A):
break
cur_max = max(cur_max, A[j+k])
dp[j] = max(dp[j], (k+1)*cur_max + dp[j+k+1])
return dp[0]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp=[0]*len(arr)
dp[0]=arr[0]
for i in range(1, len(arr)):
maxval=arr[i]
for j in range(i, i-k, -1):
if j<0:
break
maxval=max(maxval, arr[j])
temp=dp[j-1] if j-1>=0 else 0
temp+=maxval*(i-j+1)
dp[i]=max(dp[i], temp)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], k: int) -> int:
N = len(A)
dp = [0] * (N+1)
for i in range(N):
mx = 0
#print('i={}'.format(i))
for j in range(i, max(-1, i-k), -1):
mx = max(mx, A[j])
#print(' j={}, mx={}'.format(j, mx))
p = dp[i+1]
dp[i+1] = max(dp[i+1], dp[j] + mx * (i-j+1))
#print(' p={}, c={}'.format(p, dp[i+1]))
#print('c={}'.format(dp[i+1]))
return dp[N]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [float('-inf') for i in range(len(A)+1)]
dp[-1] = 0
dp[-2] = A[-1]
for j in reversed(range(len(A)-1)):
cur_max = float('-inf')
for k in range(K):
if j+k == len(A):
break
cur_max = max(cur_max, A[j+k])
dp[j] = max(dp[j], (k+1)*cur_max + dp[j+k+1])
return dp[0]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [0] * (len(A) + 1)
for i in range(1, len(A) + 1):
tmp = 0
tmp_max = 0
for j in range(1, min(i,K) + 1):
tmp_max = max(A[i-j], tmp_max)
tmp = max(tmp, dp[i-j] + tmp_max * j)
dp[i] = tmp
return dp[len(A)]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A, K):
# https://leetcode.com/problems/partition-array-for-maximum-sum/discuss/290863/JavaC%2B%2BPython-DP
# Use dp (record[i+1]) to record the maximum sum can be obtained from A[0]-A[i]
# For each position, we loop back 1 to k elements and find the maximum within each range
# And replace them all with the maximum and try to find the maximum sum based on record[i-k+1]
n = len(A)
record = [0]*(n+1)
for i in range(n):
curMax = 0
for k in range(1,min(K,i+1)+1):
curMax = max(curMax,A[i-k+1])
record[i+1] = max(record[i+1],record[i-k+1]+curMax*k)
return record[n]
# N = len(A)
# dp = [0] * (N + 1)
# for i in range(N):
# curMax = 0
# for k in range(1, min(K, i + 1) + 1):
# curMax = max(curMax, A[i - k + 1])
# dp[i+1] = max(dp[i+1], dp[i+1 - k] + curMax * k)
# return dp[N]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = [0] * (len(A) + 1)
for i in range(len(A)):
value = A[i]
for j in range(1, K + 1):
if i + j - 1 >= len(A):
break
value = max(A[i + j - 1], value)
dp[i + j] = max(dp[i] + value * j, dp[i + j])
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * (n + 1)
for i in range(1, n + 1):
m = -float('inf')
d = 0
while d < k and i - d > 0:
m = max(m, arr[i - 1 - d])
dp[i] = max(dp[i], dp[i - d - 1] + m * (d + 1))
d += 1
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
'''
So this can be a subproblem of different lengths of k and subarrays
so f([1,15,7,9,2,5,10]) = 1 + f([15,7,9,2,5,10]]) or 15 * 2 + f([7,9,2,5,10]]) or
15 * 3 + f([7,9,2,5,10]])
if not arr, then we return
'''
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
N = len(arr)
memo = {}
def maxSum2(i):
if i >= N:
return 0
if i in memo: return memo[i]
val = -1
sol = -1
# traverse k times or until end of array. + 1 to offset 0 index so we can multiply val * j straight up
for j in range(1, min(k+1, N - i + 1)):
val = max(val, arr[i + j -1])
sol = max(sol, val * j + maxSum2(i + j))
memo[i] = sol
return sol
return maxSum2(0)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * (n + 1)
for i in range(n):
currMax = arr[i]
size = 1
while size <= k and i - size + 1 >= 0:
currMax = max(currMax, arr[i - size + 1])
dp[i + 1] = max(dp[i + 1], dp[i + 1 - size] + (currMax * size))
size += 1
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
memo = {}
def max_sum(idx):
if idx >= len(arr):
return 0
if idx in memo:
return memo[idx]
subarr_max = float('-inf')
options = []
for end in range(idx, min(idx + k, len(arr))):
subarr_max = max(subarr_max, arr[end])
subarr_sum = subarr_max * (end - idx + 1)
options.append(subarr_sum + max_sum(end + 1))
memo[idx] = max(options)
return memo[idx]
return max_sum(0)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
memo = {}
def dfs(i):
if i >= n:
return 0
if i in memo:
return memo[i]
res = float('-inf')
cur_max = float('-inf')
for j in range(i, min(i + k, n)):
cur_max = max(cur_max, arr[j])
res = max(res, cur_max * (j - i + 1) + dfs(j + 1))
memo[i] = res
return res
return dfs(0)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
N = len(arr)
maxSoFar = [0]*N
for i in range(N):
curMax = 0
for prev in range(1,min(i+1,k)+1):
curMax = max(curMax,arr[i-prev+1])
lastPartition = maxSoFar[i-prev] if i >= prev else 0
maxSoFar[i] = max(maxSoFar[i], lastPartition + curMax*prev)
return maxSoFar[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
current_max = 0
dp = [0] * n
for i in range(n):
if i < k:
current_max = max(current_max, arr[i])
dp[i] = (i+1) * current_max
else:
current_max = 0
for j in range(1, k+1):
current_max = max(current_max, arr[i - j + 1])
dp[i] = max(dp[i], dp[i-j] + current_max * j)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A, K):
N = len(A)
dp = [0] * (N + 1)
for i in range(N):
curMax = 0
for k in range(1, min(K, i + 1) + 1):
curMax = max(curMax, A[i - k + 1])
dp[i+1] = max(dp[i+1], dp[i+1 - k] + curMax * k)
print(dp)
return dp[N]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0] * n
for i in range(n):
currMax = arr[i]
size = 1
while size <= k and i - size + 1 >= 0:
currMax = max(currMax, arr[i - size + 1])
dp[i] = max(dp[i], (dp[i - size] if i - size >= 0 else 0) + (currMax * size))
size += 1
# print(dp)
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
d = [0] * (len(arr))
for i in range(len(arr)):
cur_max = 0
for kk in range(1, k + 1):
if i - kk + 1 < 0:
continue
cur_max = max(cur_max, arr[i-kk+1])
# print(i, kk, \",\", i - kk + 1, cur_max, arr[i-kk+1], ':', d[i], d[i-kk])
if i - kk < 0:
prev = 0
else:
prev = d[i-kk]
d[i] = max(d[i], prev + cur_max*kk)
print(d)
return d[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0 for i in range(len(arr)+1)]
for i in range(len(arr)):
val = arr[i]
for j in range(1, k+1):
if i + j -1 >= len(arr):
break
val = max(arr[i+j-1], val)
dp[i+j] = max(dp[i] + val *j, dp[i+j])
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = (len(arr) + 1) * [0]
for i in range(0, len(arr)):
sub_max = 0
for j in range(k):
if j <= i:
sub_max = max(sub_max, arr[i-j])
dp[i + 1] = max(dp[i+1], sub_max * (j + 1) + dp[i-j])
return dp[len(arr)]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
def dfs(start, memo):
if start in memo:
return memo[start]
N = len(A)
if start >= N:
return 0
maxSum = 0
maxEle = 0
for i in range(start, min(N, start + K)):
maxEle = max(maxEle, A[i])
maxSum = max(maxSum, maxEle * (i - start + 1) + dfs(i+1, memo))
memo[start] = maxSum
return maxSum
return dfs(0, {})
# max_heap = []
# counter = collections.defaultdict(int)
# N = len(A)
# result = []
# for i in range(K):
# heappush(max_heap, -A[i])
# counter[A[i]] += 1
# for i, n in enumerate(A):
# if i+k < N:
# result.append(-heap[0][0])
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
N = len(arr)
dp = [0] * (N + 1)
for i in range(N):
curMax = 0
for j in range(1, min(k, i + 1) + 1):
curMax = max(curMax, arr[i - j + 1])
dp[i] = max(dp[i], dp[i - j] + curMax * j)
return dp[N - 1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [-1] * (len(arr) + 1)
dp[0] = 0
for i in range(len(arr)):
cur_max = -1
for j in range(k):
if (i - j) < 0:
dp[i + 1] = max(dp[i + 1], arr[i])
break
cur_max = max(cur_max, arr[i-j])
dp[i + 1] = max(dp[i + 1], dp[i-j] + cur_max * (j + 1))
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
if arr == [] or k == 0:
return 0
if len(arr) == 1 and k == 1:
return arr[0]
l = len(arr)
res = [0]*(l)
for i in range(0, l):
sub_max = 0
for j in range(k):
if j <= i:
sub_max = max(sub_max, arr[i-j])
res[i] = max(res[i], sub_max*(j + 1) + (res[i-j-1] if i-j-1 >=0 else 0))
return res[l-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0]*(n+1)
dp[0] = 0
for i in range(1,n+1):
mx=0
for j in range(k):
if(i-j-1>=0):
mx = max(mx, arr[i-j-1])
dp[i] = max(dp[i], mx*(j+1) + dp[i-j-1])
return dp[n]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = len(arr) * [0]
dp[0] = arr[0]
for i in range(1, len(arr)):
sub_max = 0
for j in range(k):
if j <= i:
sub_max = max(sub_max, arr[i-j])
dp[i] = max(dp[i], sub_max * (j + 1) + (dp[i-j-1] if i-j-1 >= 0 else 0))
return dp[len(arr)-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
# walk through list: at every element, look backwards up to K spaces
# compare combination of 1, 2 .. K repeats of current segment max
# vs total sum, and retain the largest (dp_sum memoization)
# dummy index dp_sum[0] for index bound checking
# O(AK) time, O(A) extra space
dp_sum = [0 for _ in range(len(A)+1)]
for i in range(len(A)):
seg_max = A[i]
for j in range(1, K+1):
if i-j+1 < 0: break
seg_max = max(seg_max, A[i-j+1])
tmp = dp_sum[i-j+1] + seg_max * j
dp_sum[i+1] = max(dp_sum[i+1], tmp)
return dp_sum[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, A: List[int], k: int) -> int:
L=len(A)
vis={}
def pos(n):
if n>=L:
return 0
if n in vis:
return vis[n]
currmax=A[n]
ans=A[n]+pos(n+1)
for i in range(1,k):
n1=n+i
if n1>=L:
break
currmax=max(currmax,A[n1])
ans=max(ans,currmax*(i+1)+pos(n1+1))
vis[n]=ans
return vis[n]
return pos(0)
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
if not arr:
return 0
n = len(arr)
dp = [0 for _ in range(n)]
for i in range(n):
temp_max = arr[i]
j = 1
while i-j+1>=0 and j<=k:
temp_max = max(temp_max, arr[i-j+1])
if i>=j:
dp[i] = max(dp[i], dp[i-j]+(temp_max*j))
else:
dp[i] = max(dp[i], temp_max*j)
j+=1
return dp[-1]
|
Given an integer array arr, you should partition the array into (contiguous) subarrays of length at most k. After partitioning, each subarray has their values changed to become the maximum value of that subarray.
Return the largest sum of the given array after partitioning.
Example 1:
Input: arr = [1,15,7,9,2,5,10], k = 3
Output: 84
Explanation: arr becomes [15,15,15,9,10,10,10]
Example 2:
Input: arr = [1,4,1,5,7,3,6,1,9,9,3], k = 4
Output: 83
Example 3:
Input: arr = [1], k = 1
Output: 1
Constraints:
1 <= arr.length <= 500
0 <= arr[i] <= 109
1 <= k <= arr.length
|
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
n = len(arr)
dp = [-1 for _ in range(n)]
def max_sum(i):
# print(i)
if i >= n:
return 0
if i == (n - 1):
return arr[i]
if dp[i] != -1:
return dp[i]
ans = float('-inf')
maxi = arr[i]
for p in range(1, k + 1):
v = maxi*p + max_sum(i + p)
# print(i, p, v)
ans = max(v, ans)
if (i + p) < n:
maxi = max(maxi, arr[i + p])
else:
break
dp[i] = ans
return ans
v = max_sum(0)
return v
|
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