**Max Sum Partitioning - Amazon Top Interview Questions**

### Problem Statement :

You are given a list of integers nums and an integer k. You must partition nums into contiguous groups of size at most k each, and then set each element in nums to the maximum value of its group. Return the maximum possible sum of the resulting list after partitioning. Constraints n ≤ 1,000 where n is the length of nums 1 ≤ k ≤ n Example 1 Input nums = [1, 6, 3, 2, 2, 5, 1] k = 3 Output 35 Explanation We can partition the list into [1, 6, 3] + [2] + [2, 5, 1] to get [6, 6, 6, 2, 5, 5, 5].

### Solution :

` ````
Solution in C++ :
int solve(vector<int>& nums, int k) {
int n = nums.size();
vector<int> dp(n + 1, 0);
for (int i = 1; i <= n; i++) {
for (int j = i, curr_max = 0; j > i - k and j; j--) {
curr_max = curr_max > nums[j - 1] ? curr_max : nums[j - 1];
dp[i] = max(dp[i], dp[j - 1] + curr_max * (i - j + 1));
}
}
return dp[n];
}
```

` ````
Solution in Python :
class Solution:
def solve(self, nums, k):
n = len(nums)
@lru_cache(None)
def dp(i) -> int:
if i >= n:
return 0
max_elem = 0
max_sum = 0
for j in range(i, min(n, i + k)):
max_elem = max(max_elem, nums[j])
max_fill = (j - i + 1) * max_elem
rtr = max_fill + dp(j + 1)
max_sum = max(max_sum, rtr)
return max_sum
return dp(0)
```

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