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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 util(self,i,n,A,K,d) :
if i >= n :
return 0
if i not in d :
count=0
ma=A[i]
temp=-1
while (i + count < n) and count<=K-1 :
if A[i+count] > ma :
ma=A[i+count]
nex=self.util(i+count+1,n,A,K,d)
if (count+1)*ma+nex > temp :
temp=(count+1)*ma+nex
count+=1
d[i]=temp
return d[i]
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
n=len(A)
d={}
return self.util(0,n,A,K,d)
|
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:
visited={}
def recurse(index):
if index>=len(A):
return 0
if index in visited:
return visited[index]
mxx=0
ans=0
for i in range(K):
if index+i<len(A):
mxx=max(mxx,A[i+index])
ans=max(ans,mxx*(i+1)+recurse(i+index+1))
visited[index]=ans
return ans
return recurse(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 = {}
def recur(index,arr,k):
if index >= len(arr):
return 0
if index in dp:
return dp[index]
maxi = 0
res = 0
for i in range(k):
if index+i < len(arr):
maxi = max(maxi, arr[index+i])
res = max(res, maxi * (i+1) + recur(index + i + 1, arr, k))
dp[index] = res
return res
return recur(0, 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:
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)
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 = {}
def recur(i,arr,k):
if i >= len(arr):
return 0
if i in dp:
return dp[i]
maxi,res = 0,0
for j in range(k):
if i+j < len(arr):
maxi = max(maxi, arr[i+j])
res = max(res, maxi * (j+1) + recur(i + j + 1, arr, k))
dp[i] = res
return res
return recur(0, 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, arr: List[int], k: int) -> int:
dp=[0]*len(arr)
dp[0]=arr[0]
max_value=arr[0]
max_sum=arr[0]
for i in range(1,k):
max_value=max(max_value,arr[i])
dp[i]=max_value*(i+1)
for i in range(k,len(arr)):
max_value=0
for j in range(k):
max_value=max(max_value,arr[i-j])
dp[i]=max(dp[i],dp[i-j-1]+max_value*(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
|
from collections import deque
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = deque([0 for i in range(K)])
for i in range(1, len(A)+1):
maxA, maxP = float('-inf'), float('-inf')
for j in range(1, min(K+1, i+1)):
maxA = max(maxA, A[i-j])
maxP = max(maxP, dp[-j] + j*maxA)
dp.popleft()
dp.append(maxP)
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
|
from collections import deque
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = deque([0 for i in range(K)])
for i in range(1, len(A)+1):
maxA, maxP = float('-inf'), float('-inf')
for j in range(1, min(K, i)+1):
maxA = max(maxA, A[i-j])
maxP = max(maxP, dp[-j] + j*maxA)
dp.popleft()
dp.append(maxP)
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:
inf = 0x3f3f3f3f
dp = [[(-inf, -inf) for _ in range(k+1)] for _ in range(2)]
last_max = 0
dp[0][0] = (0, 0)
for i in range(1, len(arr) + 1):
for j in range(1, k + 1):
if j == 1:
dp[i%2][j] = (last_max + arr[i-1], arr[i-1])
last_max = dp[i%2][j][0]
elif i >= j:
last_s, last_n = dp[(i-1)%2][j-1]
n = max(last_n, arr[i-1])
s = last_s - last_n * (j-1) + n * j
dp[i%2][j] = (s, n)
last_max = max(last_max, s)
# print(i, j, dp[i][j])
return last_max
|
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]
#Time Complexity- O(N*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:
N = len(A)
if K == 1 or N == 1:
return sum(A)
dp = [0] * (N + 1)
dp[0] = 0
for i in range(N):
mv = A[i]
dp[i+1] = dp[i] + A[i]
for j in range(1, min(i+1, K)):
mv = max(mv, A[i-j])
dp[i+1] = max(dp[i+1], dp[i-j] + (j+1)*mv)
# 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, A: List[int], K: int) -> int:
size = len(A)
if K == 1: return sum(A)
dp = [0] * (size+1)
for i in range(size):
cur_max = A[i]
block_size = 1
while block_size <= K and i - block_size + 1 >= 0:
cur_max = max(cur_max, A[i-block_size + 1])
dp[i+1] = max(dp[i+1], dp[i-block_size + 1] + block_size * cur_max)
block_size += 1
return dp[size]
|
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]*(n+1) for _ in range(n+1)]
def maxsum(i=0):
if i>=n:return 0
elif dp[i][n]!=-1:
return dp[i][n]
else:
mx=0
for j in range(i,min(i+k,n)):
mx=max(arr[j],mx)
dp[i][j]=mx*(j-i+1)
dp[i][n]=max(dp[i][n],dp[i][j]+maxsum(j+1))
return dp[i][n]
return maxsum()
|
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] * (n := len(arr))
dp[0] = curr_max = arr[0]
for i in range(1, k):
curr_max = max(curr_max, arr[i])
dp[i] = (i + 1) * curr_max
for i in range(k, n):
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] + (j + 1) * curr_max)
return dp[-1]
# @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
# 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
|
from collections import deque
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
dp = deque([0 for i in range(K)])
for i in range(1, len(A)+1):
maxA, maxP = float('-inf'), float('-inf')
for j in range(1, min(K+1, i+1)):
maxA = max(maxA, A[i-j])
maxP = max(maxP, dp[-j] + j*maxA)
dp.popleft()
dp.append(maxP)
return dp.pop()
|
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
|
'''
what would be the state ?
i: current position
partition start
just try allof them is probably the answer
[1,15,7,9,2,5,10]
try all possible partitions and keep track of the max and sum so far
helper(start, i)
for j in range(i, k - i + start + 1): # these are the possible ends
helper(start, j)
if i = 2 and start = 1 and k = 3
from i=2 to 5 - 1 = 4
instead it should be 1, 2, 3, so only one more so i + 1 = i + k - i
if i = 2 and start = 1 and k = 4
it shld be 1, 2, 3, 4
i just want to know much left to get to k
k = end - start
k = j + i - start -> j = k - i + start
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
def helper(i, left, curr_max, sum_so_far):
print(\"helper i =\", i, \", left =\", left, \", curr_max =\", curr_max)
if i == len(arr):
self.res = max(self.res, sum_so_far + (curr_max * (k - left)))
return 0
# continue current partition
choice1 = float('-inf')
if left > 0:
choice1 = helper(i + 1, left - 1, max(curr_max, arr[i]), sum_so_far)
# start a new partition
choice2 = helper(i + 1, k, arr[i], sum_so_far + (curr_max * (k - left)))
print(\"choice1 =\", choice1, \", choice2 =\", choice2)
return max(choice1, choice2)
self.res = float('-inf')
helper(0, k, arr[0], 0)
return self.res
'''
'''
dp[i]: max sum for arr ending at i
'''
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], K: int) -> int:
n = len(arr)
dp = [0 for _ in range(n+1)]
for i in range(n): # i is considered the end of the partition
currMax = 0
for k in range(1, min(K, i + 1) + 1):
currMax = max(currMax, arr[i - k + 1])
dp[i] = max(dp[i], dp[i-k] + (currMax * 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 arr is None or len(arr) == 0:
return 0
dp = [0 for _ in range(len(arr))]
for i in range(len(arr)):
max_elem = arr[i]
j = 1
while j <= k and i - j + 1 >= 0:
max_elem = max(max_elem, arr[i - j + 1])
if i - j >= 0:
dp[i] = max(dp[i], max_elem * j + dp[i - j])
else:
dp[i] = max(dp[i], max_elem * 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, A: List[int], K: int) -> int:
n = len(A)
dp = [0] * (n+1)
for i, a in enumerate(A):
r = a
for j in range(1, K+1):
r = max(r, A[i-j+1] if i-j+1>=0 else float('-inf'))
# print(i, j, r, dp)
if i-j+1 >= 0:
dp[i+1] = max(dp[i+1], dp[i-j+1] + r*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:
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]+cacheMax[j][i]*(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:
n = len(arr)
self.dp = [-1 for i in range(n)]
def helper(pos:int)->int:
if pos >= n:
return 0
if self.dp[pos] != -1:
return self.dp[pos]
current_max = arr[pos]
ret = arr[pos] + helper(pos+1)
for i in range(1, k):
pos2 = pos + i
if pos2 >= n:
break
current_max = max(current_max, arr[pos2])
ret = max(ret, current_max * (i+1) + helper(pos2+1))
self.dp[pos] = ret
return ret
return helper(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)
# maxSum[i] = max(A[i] + maxSum[i+1], max(A[i], A[i+1]) + maxSum[i+2], max(A[i], A[i+1], A[i+2], maxSum[i+3]))
maxSum = [0] * n;
for i in range(-1, -K-1, -1):
maxSum[i] = max(A[i:])*(-i)
for i in range(n-K-1, -1, -1):
ms = 0
for k in range(1, K+1):
ms = max(ms, max(A[i:i+k]) * k + maxSum[i+k])
# print(i,k,ms)
maxSum[i] = ms
# print(maxSum)
return maxSum[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:
@lru_cache(None)
def helper(index, start, mx):
if index == len(arr):
return (index - start)* mx
if index - start + 1 <= k:
return max(helper(index + 1, start, max(arr[index], mx)), (index - start)*mx + helper(index + 1, index , arr[index]))
return (index - start)*mx + helper(index + 1, index , arr[index])
return helper(0, 0, 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, arr: List[int], k: int) -> int:
inf = 0x3f3f3f3f
dp = [[(-inf, -inf) for _ in range(k+1)] for _ in range(len(arr)+1)]
last_max = 0
dp[0][0] = (0, 0)
for i in range(1, len(arr) + 1):
for j in range(1, k + 1):
if j == 1:
dp[i][j] = (last_max + arr[i-1], arr[i-1])
last_max = dp[i][j][0]
elif i >= j:
last_s, last_n = dp[i-1][j-1]
n = max(last_n, arr[i-1])
s = last_s - last_n * (j-1) + n * j
dp[i][j] = (s, n)
last_max = max(last_max, s)
# print(i, j, dp[i][j])
return last_max
|
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)
for i in range(K):
dp[i] = max(A[:(i + 1)]) * (i + 1)
for i in range(K, len(A)):
dp[i] = max([dp[i - j] + max(A[(i - j + 1):(i + 1)]) * j for j in range(1, K + 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:
'''
For each element, it can either go to the previous group, or start a new group
dp[i][j] = largest sum before i when last group's size is j, last group's largest value
'''
dp = [[(0, 0)] * k for _ in range(len(arr)+1)]
for i in range(1, len(arr)+1):
for j in range(min(i, k)):
if j == 0:
dp[i][j] = max((dp[i-1][m][0] + arr[i-1], arr[i-1]) for m in range(min(i, k)))
else:
new_max = max(dp[i-1][j-1][1], arr[i-1])
dp[i][j] = dp[i-1][j-1][0] - j * dp[i-1][j-1][1] + new_max * (j + 1), new_max
#print(dp)
return max(dp[-1])[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
for i in range(n-1, -1, -1):
curr_max = arr[i]
for j in range(0, k):
index = min(n-1, i+j)
curr_max = max(curr_max, arr[index])
if i+j+1 >= n:
dp[i] = (n-i)*curr_max
else:
dp[i] = max(dp[i], dp[i+j+1] + (j+1)*curr_max)
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:
if not arr:
return 0
local_max = {(idx, idx+1) :arr[idx] for idx in range(len(arr))}
for l in range(2, k+1):
for start in range(len(arr)-l+1):
local_max[(start, start+l)] = max(local_max[(start, start+l-1)], local_max [(start+1, start+l)])
f = [0]
for end in range(1, len(arr)+1):
_local = 0
for start in range(max(0, end-k), end):
_local = max(f[start]+local_max[(start, end)]*(end-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: List[int], k: int) -> int:
n=len(arr)
memo=[0]*n
for i in range(n):
if i<k:
memo[i]=(i+1)*max(arr[:i+1])
else:
memo[i]=max(memo[i-j-1]+(j+1)*max(arr[i-j:i+1]) for j in range(k))
return memo[-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:
# 动态规划
length = len(arr)
dp = [0]*(length+1)
maxval = [[0]*length for _ in range(length)]
for i in range(length):
for j in range(i, length):
if i == j:
maxval[i][j] = arr[i]
else:
maxval[i][j] = max(maxval[i][j-1], arr[j])
# print(maxval)
for i in range(1, length+1):
temp = 0
# print(\"------{}------\".format(i))
for t in range(1, k+1):
temp = max(temp, dp[i-t]+maxval[i-t][i-1]*t)
# print(temp)
dp[i] = temp
# 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] * len(A)
for i in range(K):
dp[i] = (i+1) * max(A[:i+1])
for i in range(K,len(A)):
maximum = 0
for j in range(1,K+1):
if i-j<0:
break
temp = dp[i-j] + max(A[i-j+1:i+1]) * j
if temp>maximum:
maximum = temp
dp[i] = maximum
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:
def dfs(i, j, max_num=0):
if i == -1:
return max_num * j
if opt[i][j] == -1:
a, b = 0, 0
if j < K:
a = dfs(i - 1, j + 1, max(max_num, A[i]))
b = j * max_num + dfs(i - 1, 1, A[i])
opt[i][j] = max(a, b)
return opt[i][j]
n = len(A)
opt = [[-1] * (K + 1) for _ in range(n)]
return dfs(n - 1, 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:
l = len(arr)
results = [0] * l
for idx, num in enumerate(arr):
if idx < k:
results[idx] = max(arr[:idx+1]) * (idx + 1)
else:
res = 0
for i in range(1, k+1):
start = idx -i
tmp = results[start] + max(arr[start+1:idx+1]) * i
res = max(res, tmp)
results[idx] = res
return results[-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:
max_sums = [0]*len(arr)
for i in range(len(arr)):
if i<=k-1:
max_sums[i] = (i+1)*max(arr[:i+1])
else:
res = 0
for j in range(1,k+1):
first = max_sums[i-j]
second = j*max(arr[i-j+1:i+1])
#print(i, j)
#print(arr[:i-j+1], arr[i-j+1:i+1])
#print(\"first, second \", first, second)
res = max(res, first+second)
max_sums[i] = res
#print(max_sums)
#print(max_sums)
return max_sums[-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:
maxx = arr[0]
dp = []
for i,_ in enumerate(arr[:k]):
maxx=max(arr[i], maxx)
dp.append(maxx*(i+1))
for i, _ in enumerate(arr[k::]):
i+=k
max_sum = 0
for j in range(k):
max_sum = max(
max_sum,dp[i-j-1] + (max(arr[i-j:i+1])*(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 maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
A=arr
K=k
n=len(A)
dp=[0]*n
for i in range(K):
dp[i]=max(A[:i+1])*(i+1)
for i in range(K,n):
for j in range(1,K+1):
dp[i]=max(dp[i],dp[i-j]+max(A[i-j+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, arr: List[int], k: int) -> int:
size = len(arr)
dp = [0]*size
for i in range(k):
dp[i] = max(arr[:i+1])*(i+1)
for i in range(k,size):
for j in range(1,k+1):
dp[i] = max(dp[i],dp[i-j]+max(arr[i-j+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 printm(self,arr):
[print(arr[i]) for i in range(len(arr))]
def maxSumAfterPartitioning(self, A,K):
size = len(A)
dp = [0]*size
for i in range(K):
dp[i] = max(A[:i+1])*(i+1)
# printm(dp)
for i in range(K,size):
for j in range(1,K+1):
dp[i] = max(dp[i],dp[i-j]+max(A[i-j+1:i+1])*j)
# self.printm(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:
L = len(arr)
dp = [0] * L
for i in range(k):
dp[i] = max(arr[:i+1])*(i+1)
for i in range(k,L):
dp[i] = dp[i-k] + max(arr[i-k+1:i+1])*k
for j in range(1,k):
dp[i] = max(dp[i],dp[i-j] + max(arr[i-j+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
|
# dynamic programming
# time complexity: O(A*K*K)
# dp[i] = max sum of A[:i] after partitioning
class Solution:
def maxSumAfterPartitioning(self, A: List[int], K: int) -> int:
N = len(A)
dp = [float('-inf')] * (N+1)
for i in range(1,N+1):
if i <= K:
dp[i] = max(A[:i]) * i
else:
for j in range(1,K+1):
dp[i] = max(dp[i], dp[i-j]+max(A[i-j: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 = [0] * len(arr)
for i in range(k):
dp[i] = max(arr[:i+1]) * (i+1)
for i in range(k, len(arr)):
for j in range(1,k+1):
dp[i] = max(dp[i], dp[i-j] + j * max(arr[i-j+1:i+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:
M = [0 for i in range(len(arr))]
M[0] = arr[0]
for i in range(1, len(arr)):
if i < k:
M[i] = max(arr[:i+1]) * (i+1)
else:
for j in range(1, min(k+1, i+1)):
M[i] = max(M[i], M[i-j] + max(arr[i-j+1:i+1]) * j)
return M[-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:
rec = {}
def helper(idx):
if idx in rec:
return rec[idx]
if len(arr)-idx <= k:
temp = max(arr[idx:])*(len(arr)-idx)
rec[idx] = temp
return temp
pre_max = float('-inf')
temp = float('-inf')
for j in range(idx, min(idx+k, len(arr))):
pre_max = max(pre_max, arr[j])
temp = max(temp, pre_max*(j-idx+1)+helper(j+1))
rec[idx] = temp
return temp
return helper(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)
for i in range(k):
dp[i] = max(arr[:i+1])*(i+1)
#print(dp)
for i in range(k,len(arr)):
for j in range(1,k+1):
dp[i] = max(dp[i],dp[i-j]+max(arr[i-j+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:
def maxSum(i, mem):
if i in mem: return mem[i]
if len(A) - i <= K:
mem[i] = max(A[i:]) * (len(A) - i)
return mem[i]
ans = -float('inf')
for j in range(i+1, i+1+K):
if j > len(A): break
ans = max(ans, max(A[i:j]) * (j-i) + maxSum(j, mem))
mem[i] = ans
return mem[i]
mem = {}
return maxSum(0, mem)
|
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:
largest_sums = []
for i in range(len(A)):
if i < K:
largest_sums.append(max(A[:i+1]) * (i + 1))
continue
max_sum = 0
for j in range(K):
if i - (j+1) < 0:
continue
# print(j, i-j-1, largest_sums[i - j-1], max(A[i-j:i+1]))
cur_sum = largest_sums[i - j-1] + max(A[i-j:i+1]) * (j + 1)
if cur_sum > max_sum:
max_sum = cur_sum
# print(i, max_sum)
largest_sums.append(max_sum)
return largest_sums[-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 dfs(i, j, max_num=0):
if i == -1:
return max_num * j
if opt[i][j] == -1:
ans = 0
if j < K:
ans = max(ans, dfs(i - 1, j + 1, max(max_num, A[i])))
ans = max(ans, j * max_num + dfs(i - 1, 1, A[i]))
opt[i][j] = ans
return opt[i][j]
n = len(A)
opt = [[-1] * (K + 1) for _ in range(n)]
return dfs(n - 1, 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 for i in range(len(arr))]
for i in range(k):
dp[i] = (i+1)*max(arr[:i+1])
for i in range(k,len(arr)):
for j in range(k):
dp[i] = max(dp[i], dp[i-j-1]+(j+1)*max(arr[i-j:i+1]))
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:
n = len(A)
dp = [0] * n
for i in range(K):
dp[i] = max(A[:i+1]) * (i + 1)
for i in range(K, n):
for d in range(1, K + 1):
dp[i] = max(dp[i], dp[i - d] + max(A[i - d + 1: i + 1]) * d)
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:
mm = [[a for a in arr]]
for i in range(len(arr) - 1):
mm.append([0] * len(arr))
for i in range(1, k):
for j in range(i, len(arr)):
mm[i][j] = max(mm[i - 1][j - 1], arr[j])
dp = [0] * (len(arr) + 1)
for i in range(1, len(arr) + 1):
for j in range(max(k, i)):
dp[i] = max(dp[i], dp[i - j - 1] + mm[j][i - 1] * (j + 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, A: List[int], K: int) -> int:
aLen = len(A)
dp = {}
def helper(i0):
if i0 in dp:
return dp[i0]
if aLen - i0 <= K:
dp[i0] = max(A[i0:])*(aLen-i0)
return dp[i0]
subAns = 0
thisMax = A[i0]
for ki in range(1, K+1):
thisMax = max(thisMax, A[i0+ki-1])
subAns = max(subAns, thisMax*ki + helper(i0 +ki))
dp[i0] = subAns
return subAns
return helper(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)
maxes = [[None for _ in range(n)] for _ in range(n)]
for i in range(n):
maxes[i][i] = arr[i]
for i in range(n):
for j in range(i+1, n):
maxes[i][j] = max(arr[j], maxes[i][j-1])
dp = [None for _ in range(n+1)]
dp[n] = 0
for i in range(n-1, -1, -1):
m = float('-inf')
for j in range(min(n-i,k)):
m = max(m, (j+1)*maxes[i][min(n-1, i+j)] + dp[min(n, i+j+1)])
dp[i] = m
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:
def helper(index, start, mx):
key = (index, start, mx)
if key in d:
return d[key]
if index == len(arr):
return (index - start)* mx
if index - start + 1 <= k:
d[key] = max(helper(index + 1, start, max(arr[index], mx)), (index - start)*mx + helper(index + 1, index , arr[index]))
return d[key]
d[key] = (index - start)*mx + helper(index + 1, index , arr[index])
return d[key]
d = dict()
return helper(0, 0, 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, A: List[int], K: int) -> int:
@lru_cache(None)
def checkSum(i, j, k):
if j - i <= k:
return max(A[i:j]) * (j - i)
else:
best = 0
for nk in range(1, K + 1):
best = max(best, checkSum(i, i + nk, K) + checkSum(i+nk, j, K))
return best
return checkSum(0, len(A), 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:
N = len(A)
arr = [[(0,0)]*(K+1) for _ in range(N)]
prev=0
for i in range(N):
arr[i][1] = (A[i]+prev,A[i])
prev+=A[i]
for j in range(2,min(i+2,K+1)):
mx = max(arr[i-1][j-1][1],A[i])
s = arr[i-1][j-1][0] - arr[i-1][j-1][1]*(j-1) + mx*j
arr[i][j] = (s,mx)
prev = max(prev,s)
# print(prev,arr[i][1:])
return max(arr[-1][k][0] for k in range(1,K+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 A:
return 0
N = len(A)
if K == N:
return N*max(A)
dp = [0 for i in range(N)]
for i in range(K):
dp[i] = max(A[:i+1])*(i+1)
for i in range(K, N):
a = max(A[i-K+1:i+1])
# print(a, dp[i-K], dp[i-1])
# dp[i] = max(dp[i-K]+K*a, dp[i-1]+A[i])
dp[i] = max([dp[i-j]+j*max(A[i-j+1:i+1]) for j in range(1, K+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:
@lru_cache(None)
def help(A,K):
n = len(A)
if K==1:
return sum(A)
if K>=n:
return n*max(A)
cMax = [A[0]]
for i in range(1,K):
cMax.append(max(cMax[-1],A[i]))
return max((i+1)*cMax[i] + help(A[i+1:],K) for i in range(K))
return help(tuple(A),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, arr: List[int], k: int) -> int:
#dpdp
grid = [0]*len(arr)
for i in range(len(arr)-1, -1, -1):
if len(arr) - i <= k:
grid[i] = max(arr[i:len(arr)])*(len(arr)-i)
else:
maxi = 0
current_max = 0
for t in range(k):
current_max = max(current_max, arr[i+t])
maxi = max(maxi, current_max*(t+1) + grid[i+t+1])
grid[i] = maxi
return grid[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
if k==1:
return sum(arr)
self.memo={}
def makepart(arr,n):
if n in self.memo:
return self.memo[n]
if not arr:
return 0
l=min(k,len(arr))
maxele=0
total=0
for i in range(l):
maxele=max(maxele,arr[i])
temp=(maxele*(i+1))+makepart(arr[i+1:],n+i+1)
total=max(total,temp)
self.memo[n]=total
return total
return makepart(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, A: List[int], k: int) -> int:
def dfs(i, j, max_num=0):
if i == -1:
return max_num * j
if (i, j, max_num) not in opt:
ans = 0
if j < k:
ans = max(ans, dfs(i - 1, j + 1, max(max_num, A[i])))
ans = max(ans, j * max_num + dfs(i - 1, 1, A[i]))
opt[(i, j, max_num)] = ans
return opt[(i, j, max_num)]
n = len(A)
opt = dict()
return dfs(n - 1, 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, k: int) -> int:
n = len(arr)
mask = [[0] * n for i in range(n)]
for i in range(n):
for j in range(i, n):
mask[i][j] = max(arr[j], mask[i][j - 1] if j > i else 0)
for i in range(n):
for j in range(i, n):
mask[i][j] = (j + 1 - i) * mask[i][j]
dp = [0]*n
for i in range(n):
for h in range(1,k+1):
dp[i]= max((dp[i-h] if i-h>=0 else 0)+(mask[i-h+1][i] if i-h+1<n else 0), 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, k: int) -> int:
n = len(arr)
mask = [[0] * n for i in range(n)]
for i in range(n):
for j in range(i, n):
mask[i][j] = max(arr[j], mask[i][j - 1] if j > i else 0)
for i in range(n):
for j in range(i, n):
mask[i][j] = (j + 1 - i) * mask[i][j]
dp = [0]*n
for i in range(n):
for h in range(1,k+1):
dp[i]= max((dp[i-h] if i-h>=0 else 0)+(mask[i-h+1][i] if i-h+1<n else 0), 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, A: List[int], K: int) -> int:
self.answer = {}
def max_starting_i (index_A):
array = A[index_A:]
if index_A in self.answer:
return self.answer[index_A]
#print (array)
if len(array) == 0:
self.answer[index_A] = 0
return 0
if len(array) <= K:
self.answer[index_A] = max(array) * len(array)
return self.answer[index_A]
max_start_here = 0
for i in range(0,K):
#print (i)
max_split_here = (i+1)*max(array[:i+1]) + max_starting_i(index_A+i+1)
max_start_here = max(max_start_here, max_split_here)
self.answer[index_A] = max_start_here
return max_start_here
return max_starting_i(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:
#dpdp
grid = [0]*len(arr)
for i in range(len(arr)-1, -1, -1):
if len(arr) - i <= k:
grid[i] = max(arr[i:len(arr)])*(len(arr)-i)
else:
maxi = 0
for t in range(k):
maxi = max(maxi, max(arr[i:i+t+1])*(t+1) + grid[i+t+1])
grid[i] = maxi
return grid[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, l: List[int], k: int) -> int:
def _max_sum(l, start, k, memo):
if (len(l)-start) <= k:
return (len(l)-start)*max(l[start:])
if start in memo:
return memo[start]
max_sum = l[start]
for i in range(1, k+1):
curr_sum = max(l[start:start+i])*i + _max_sum(l, start+i, k, memo)
max_sum = max(max_sum, curr_sum)
memo[start] = max_sum
return max_sum
if not l:
return 0
return _max_sum(l, 0, k, dict())
|
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:
# fMax = [A[0]]
# for num in A[1:]:
# fMax.append(max(fMax[-1], num))
aLen = len(A)
# bMax = [num for num in A]
# for i in range(aLen-2, -1, -1):
# bMax[i] = max(bMax[i], bMax[i+1])
# ans = 0
# for i in range(aLen-1):
# ans = max(ans, (i+1)*fMax[i] + (aLen-i-1)*bMax[i+1])
# return ans
dp = {}
def helper(i0):
if i0 in dp:
return dp[i0]
if aLen - i0 <= K:
dp[i0] = max(A[i0:])*(aLen-i0)
return dp[i0]
subAns = 0
thisMax = A[i0]
for ki in range(1, K+1):
thisMax = max(thisMax, A[i0+ki-1])
subAns = max(subAns, thisMax*ki + helper(i0 +ki))
dp[i0] = subAns
return subAns
return helper(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:
self.dp = [-1] * (len(arr)+k)
return self.solve(arr, 0, k)
def solve(self, arr, start, k):
if start >= len(arr):
return 0
if start + k >= len(arr):
end = min(len(arr), start+k)
return max(arr[start:end]) * (end-start)
if self.dp[start] != -1:
return self.dp[start]
result = float('-inf')
for p in range(1, k+1):
temp = max(arr[start:start+p]) * p + self.solve(arr, start+p, k)
result = max(temp, result)
self.dp[start] = result
return result
|
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 partition(self, arr, k, index, memo):
if len(arr) == 0:
return 0
if index in memo:
return memo[index]
maxNum = arr[0]
maxSum = maxNum
for i in range(1, min(k+1, len(arr)+1)):
maxNum = max(maxNum, arr[i-1])
subarr = arr[:i]
sumArr = self.partition(arr[i:], k, index + i, memo)
# i is the size of the array you're examining
sumArr += maxNum * (i)
maxSum = max(sumArr, maxSum)
memo[index] = maxSum
return maxSum
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
memo = {}
num = self.partition(arr, k, 0, memo)
print(memo)
return num
|
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.cache = {}
def helper(arr, k, start):
if start in self.cache:
return self.cache[start]
_arr = arr[start:]
if len(_arr) == 0:
return 0
if len(_arr) <= k:
return max(_arr) * len(_arr)
ans = -1
for i in range(1, k + 1):
ans = max(ans, i * max(_arr[:i]) + helper(arr, k, start + i))
self.cache[start] = ans
return ans
return helper(arr, k, 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+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:i])
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
|
import typing
class Solution:
def maxSumAfterPartitioning(self, arr: typing.List[int], k: int) -> int:
storage = {}
def getSumRec(arr:list,i:int,k:int)->int:
if i in storage:
return storage[i]
if not i<len(arr):
return 0
if len(arr) - i < k:
return max(arr[i:])*len(arr[i:])
listOfSums = []
for offset in range(k):
listOfSums.append( max(arr[i:i+offset+1] )*(offset+1) + getSumRec(arr,i+offset+1,k))
storage[i] = max(listOfSums)
return storage[i]
return getSumRec(arr,0,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:
best_a = [0] * len(A)
best_a[0] = A[0]
for i in range(1, len(A)):
max_v = 0
for j in range(i, max(-1, i-K), -1):
sa = A[j:i+1]
v = best_a[j-1] + max(sa) * len(sa)
if v > max_v:
max_v = v
#print(best_a,max_v,i,j,sa)
best_a[i] = max_v
return best_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:
dp = [0] * (len(arr)+1)
for i in range(0, len(arr)):
m = 0
for j in range(i, i-k, -1):
if j < 0:
break
t = max(arr[j:i+1]) * (i - j + 1) + dp[j]
if t > m:
m = t
dp[i+1] = m
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:
best_a = [0] * len(A)
best_a[0] = A[0]
for i in range(1,len(A)):
maxv = 0
for j in range(i,max(-1,i - K),-1):
sa = A[j:i + 1]
v = best_a[j - 1] + max(sa) * len(sa)
if v > maxv:
maxv = v
best_a[i] = maxv
return best_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:
n=len(arr)
dp_sum=[0]*n
dp_sum[0]=arr[0]
max_so_far=arr[0]
for i in range(1,k):
max_so_far = max(max_so_far, arr[i])
dp_sum[i] = (i+1)*max_so_far
for i in range(k,n):
partition_max = 0
for back in range(k):
partition_max = max(partition_max, arr[i-back])
dp_sum[i] = max(dp_sum[i], dp_sum[i-back-1] + (back+1)*partition_max)
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, arr: List[int], k: int) -> int:
opt = [arr[0]]
for i in range(1, len(arr)):
opt.append(0) # At index i
subseqLenMax = min(k, i+1) # To avoid counting past index 0
# For the length of each possible subsequence which includes (and ends at) index i:
for subseqLen in range(1, subseqLenMax+1):
# Find the max possible sum of the subsequence
subseqSum = subseqLen * max(arr[i-subseqLen+1 : i+1])
# Add to the optimal solution from before
if subseqLen < i+1:
prevOpt = opt[i-subseqLen]
else: # Entire arr
prevOpt = 0
optTemp = prevOpt + subseqSum
# print('for opt', i, ' sublen =', subseqLen, ' subseqSum = ',subseqSum, ' optTemp = ', optTemp)
# Compare to the other possible subsequence sums
if optTemp > opt[i]:
opt[i] = optTemp
# print(opt)
return opt[len(opt)-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 K == 1:
return sum(A)
# dp[i]: ans considering A[0:i]
dp = [0] * (len(A)+1)
for i in range(1, len(A)+1):
cand = []
for j in range(1, K+1):
# consider to A[i-j:i]
if j > i:
break
temp = dp[i-j] + max(A[i-j:i]) * j
cand.append(temp)
dp[i] = max(cand)
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:
n = len(A)
dp = [0] * (n + 1)
for i in range(1, n + 1):
for w in range(1, K + 1):
if i - w < 0: break
dp[i] = max(dp[i], dp[i - w] + max(A[i - w:i]) * 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:
dp = [0 for i in range(len(A)+1)]
dp[0] = 0
dp[1] = A[0]
for i in range(1, len(A)):
result = 0
for j in range(max(0, i-K+1), i+1):
result = max(result, dp[j] + max(A[j:i+1])*(i-j+1))
dp[i+1] = result
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):
for j in range(1, min(k+1,i+1)):
dp[i]= max(dp[i], dp[i-j]+ max(arr[i-j:i])*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:
dp = [0]* (len(arr)+1)
dp[1] = arr[0]
for i in range(2, len(dp)):
for j in range(1, k+1):
if i-j < 0:
continue
#print(i, j, i-j, i)
dp[i] = max(dp[i], dp[i-j] + j * max(arr[i-j:i]))
print(dp)
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, arr: List[int], k: int) -> int:
n = len(arr)
if not arr:
return 0
if n == 1:
return arr[0]
if n < k:
return sum(arr)
dp = [0] * (n)
for i in range(n):
for j in range(i, max(-1, i-k), -1):
dp[i] = max(dp[i], max(arr[j:i+1]) * (i-j+1) + (dp[j-1] if j-1 >= 0 else 0))
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)
T = [0] * (N+1)
for i in range(1, N+1):
for j in range(1, k+1):
if j > i: break
#if i == 2: print(max(arr[(i-j): i]), arr[(i-j): i], j)
T[i] = max(T[i], T[i-j] + max(arr[(i-j): i]) * j)
#print(T)
return T[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:
M = [0 for i in range(len(arr))]
M[0] = arr[0]
#if len(arr) > 1:
# M[1] = max(arr[0],arr[1]) * 2
#if len(arr) > 2:
# M[2] = max(arr[0],arr[1],arr[2]) * 3
for i in range(1, len(arr)):
if i < k:
M[i] = max(arr[:i+1]) * (i+1)
for j in range(1, min(k+1, i+1)):
M[i] = max(M[i], M[i-j] + max(arr[i-j+1:i+1]) * j)
#print(M)
return M[-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 k == 1:
return sum(arr)
if k == len(arr):
max_val = max(arr)
return max_val * len(arr)
sums = [-1 for index in range(k)]
maxs = [-1 for index in range(k)]
max_sum = 0
sums[0] = arr[0]
maxs[0] = arr[0]
for idx in range(1, len(arr)):
val = arr[idx]
max_sum = max(sums)
for ki in range(k-1, 0, -1):
max_val = maxs[ki-1]
if not max_val < 0:
if val <= max_val:
maxs[ki] = max_val
sums[ki] = sums[ki-1] + max_val
else:
maxs[ki] = val
sums[ki] = sums[ki-1] - max_val * ki + val * (ki + 1)
sums[0] = max_sum + val
maxs[0] = val
return max(sums)
|
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
currMax = 0
for i in range(n):
if i < k:
currMax = max(currMax,arr[i])
dp[i] = currMax * (i+1)
else:
currMax = 0
for j in range(1,k+1):
currMax = max(currMax,arr[i-j+1])
dp[i] = max(dp[i], dp[i-j]+currMax*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:
'''Dynamic program using a list named dp, where dp[i] indicates the largest sum of the array arr[:i].'''
dp = [0] * len(arr)
for i in range(len(arr)):
maxsum = 0
for j in range(k):
if i - j >= 1:
maxsum = max(maxsum, (j + 1) * max(arr[i - j: i + 1]) + dp[i - j - 1])
elif i - j >= 0:
maxsum = max(maxsum, (j + 1) * max(arr[i - j: i + 1]))
dp[i] = maxsum
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 = [float('-inf') for _ in range(n + 1)]
dp[0] = 0
for i in range(1, n + 1):
j = i - 1
cnt = k
while j >= 0 and cnt >= 1:
maximum = max(arr[j:i])
dp[i] = max(dp[i], dp[j] + maximum*(i-j))
j -= 1
cnt -= 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:
n = len(A)
dp = [0] * (n+1)
for i, a in enumerate(A, 1):
for j in range(1, K+1):
# print(i, j, dp)
if i-j >= 0:
dp[i] = max(dp[i], dp[i-j] + max(A[i-j:i])*j)
else:
break
# 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(i) = max array sum using elements up to index i
# dp(i) = for j = 1 to k, find max of dp(i-j) + j * max(arr[i-j:i])
dp = [0 for i in range(len(arr))]
dp[0] = arr[0]
for i in range(1, len(arr)):
ans = 0
for j in range(1, k+1):
if i-j >= 0:
ans = max(ans, dp[i-j] + j * max(arr[i-j+1:i+1]))
elif i-j == -1:
ans = max(ans, j * max(arr[i-j+1: i+1]))
dp[i] = ans
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:
#don't really understand question. how did they partition exactly????
#dynamic programming!
#let d[i] be the answer for A[0]...A[i-1]
'''
S: d[i] be the answer for A[0]...A[i] answer = max sum of given array after partition.
R: want maximum of elements within K away from i. from hint 2:
d[i] = d[i-j] + max{A[i-j],..., A[i]}*j, for j between 1 and k.
#times j because the max of that partition becomes the corresponding max for all the positions in transformed A.
B = d[0]
'''
#base case:
#reference: https://leetcode.com/problems/partition-array-for-maximum-sum/discuss/783353/dp-with-python-and-some-comments-of-idea
dp = [0]*(len(A)+1)
for i in range(1, len(A)+1):
temp = []
for j in range(1, K+1): #go through all possible values of j
if i - j >= 0:
k = dp[i-j] + max(A[i-j:i])*j
temp.append(k)
# print(k)
# dp[i] = max(k, dp[i])
#storing stuff in the same index so replacing it every time.
dp[i] = max(temp)
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, arr: List[int], k: int) -> int:
n = len(arr)
dp = [0]*(n+1)
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(arr[i-j:i])*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, A: List[int], K: int) -> int:
dp = [0]
for i in range(len(A)):
curr = 0
for j in range(i, max(-1, i - K), -1):
curr = max(max(A[j: i + 1]) * (i - j + 1) + dp[j], curr)
dp.append(curr)
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)
# 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(arr[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, arr: List[int], k: int) -> int:
n=len(arr)
dp=[0]*(n+1)
for i in range(n):
m=-1
for j in range(1,k+1):
if (i-j+1)<0:
break
m=max(m,dp[i-j+1]+max(arr[i-j+1:i+1])*(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 range(k + 1)]
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 % (k + 1)] = max(rolling_max * j + dp[(i - j) % (k + 1)], dp[i % (k + 1)])
return dp[(len(arr) - 1) % (k + 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
|
# 10:28 -> misunderstood the problem
# dp
class Solution:
def maxSumAfterPartitioning(self, arr: List[int], k: int) -> int:
dp = [0 for num in arr]
for i in range(len(arr)):
prev_sum = 0 if i - 1 < 0 else dp[i - 1]
max_at_i = arr[i] + prev_sum
possible_ks = i if i < k - 1 else k - 1
for j in range(1, possible_ks + 1):
current_window = arr[i - j: i + 1]
current_max = max(current_window)
current_window_sum = len(current_window) * current_max
prev_window_sum = 0 if i - j - 1 < 0 else dp[i - j - 1]
total_sum = current_window_sum + prev_window_sum
if total_sum > max_at_i:
max_at_i = total_sum
dp[i] = max_at_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:
L=len(arr)
record=dict()
record[0]=0
for idx in range(1,L+1):
r=-float('inf')
for gap in range(1,k+1):
if idx-gap<0:
continue
else:
r=max(r,max(arr[idx-gap:idx])*gap+record[idx-gap])
record[idx]=r
return record[L]
|
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, a in enumerate(A, 1):
for j in range(1, K+1):
# print(i, j, dp)
if i-j >= 0:
dp[i] = max(dp[i], dp[i-j] + max(A[i-j:i])*j)
# 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) + 1)
dp[0] = 0
for i in range(len(arr)):
dp[i+1] = dp[i] + arr[i]
for j in range(k, 1, -1):
if i-j+1 >= 0:
dp[i+1] = max(dp[i+1], dp[i-j+1] + max(arr[i-j+1:i+1]) * j)
# 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:
hashmap = {}
def partitionMatrix(lst, currScore):
key = tuple(lst)
if key in hashmap:
return hashmap[key] + currScore
if lst == []:
hashmap[key] = 0
return currScore
for i in range(1, k + 1):
if len(lst) == i:
hashmap[key] = (max(lst) * i)
return currScore + (max(lst) * i)
best = currScore
for i in range(1, k + 1):
subScore = max(lst[:i]) * i
best = max(best, partitionMatrix(lst[i:], currScore + subScore))
hashmap[key] = best - currScore
return best
return partitionMatrix(arr, 0)
|
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