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.


n ≤ 1,000 where n is the length of nums

1 ≤ k ≤ n

Example 1


nums = [1, 6, 3, 2, 2, 5, 1]
k = 3




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)

        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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