This approach uses dynamic programming with a one-dimensional array to find the solution. We use an array dp where dp[i] stores the maximum sum we can get for the array arr from the 0th index to the ith index. For each position, we try to partition the last k elements and update the dp array accordingly, keeping track of the maximum value observed in those elements to account for possible transformations.
Time Complexity: O(n * k) since for each index, we consider up to k previous elements.
Space Complexity: O(n) for the dp array.
1def maxSumAfterPartitioning(arr, k):
2 n = len(arr)
3 dp = [0] * (n + 1)
4 for i in range(1, n + 1):
5 maxElem, maxSum = 0, 0
6 for j in range(1, min(k, i) + 1):
7 maxElem = max(maxElem, arr[i-j])
8 maxSum = max(maxSum, dp[i-j] + maxElem * j)
9 dp[i] = maxSum
10 return dp[n]
11
12arr = [1, 15, 7, 9, 2, 5, 10]
13k = 3
14print(maxSumAfterPartitioning(arr, k))This Python function operates using a dynamic programming approach encapsulated in a list named dp. It records the maximum sum feasible by iteratively determining the best partitions up to length k. During each iteration, potential partitions are evaluated, and the maximum combination is persisted in the dp array.
In this approach, a recursive function is used to solve the problem, combined with memoization to store previously computed results. The idea is to break the problem into subproblems by recursively partitioning the array from each position and recalculating sums. Alongside recursion, memoization saves time by avoiding recomputation of results for elements already processed.
Time Complexity: O(n * k), due to exponential recursive divisions curtailed by memoization.
Space Complexity: O(n) for memoization storage.
1#include <stdio.h>
2#include <limits.h>
3#include <string.h>
4#include <stdlib.h>
5
6int max(int a, int b) {
7 return a > b ? a : b;
8}
9
10int helper(int* arr, int n, int k, int* memo, int i) {
11 if (i >= n) return 0;
12 if (memo[i] != -1) return memo[i];
13 int maxElem = INT_MIN, maxSum = 0;
14 for (int j = i; j < i + k && j < n; ++j) {
15 maxElem = max(maxElem, arr[j]);
16 maxSum = max(maxSum, maxElem * (j-i+1) + helper(arr, n, k, memo, j+1));
17 }
18 return memo[i] = maxSum;
19}
20
21int maxSumAfterPartitioning(int* arr, int arrSize, int k) {
22 int* memo = (int*)malloc(arrSize * sizeof(int));
23 memset(memo, -1, arrSize * sizeof(int));
24 return helper(arr, arrSize, k, memo, 0);
25}
26
27int main() {
28 int arr[] = {1, 15, 7, 9, 2, 5, 10};
29 int k = 3;
30 int arrSize = sizeof(arr) / sizeof(arr[0]);
31 printf("%d\n", maxSumAfterPartitioning(arr, arrSize, k));
32 return 0;
33}In this C solution, a recursive function named helper is used to evaluate possible partitions lazily. The use of memoization, through an integer array memo, optimizes this by caching previously calculated subproblem results. This active arrangement ensures computative processes reuse stored values during recursion.