LeetCode 1043: Partition Array for Maximum Sum Solution
Master LeetCode problem 1043 (Partition Array for Maximum Sum), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
1043. Partition Array for Maximum Sum
Problem Explanation
Explanation
To solve this problem, we can use dynamic programming. We will iterate over the elements of the array and maintain a dynamic programming table where dp[i] represents the maximum sum we can obtain up to index i in the array. At each index, we will consider all possible subarrays of length at most k ending at that index and calculate the maximum sum we can obtain by partitioning the array accordingly.
The dynamic programming formula is:
dp[i] = max(dp[i-j] + max_val * j) for j = 1 to k, where max_val is the maximum value in the subarray [i-j+1, i]
After completing the iteration, the answer will be stored in dp[arr.length - 1].
The time complexity of this solution is O(n*k) where n is the length of the input array and k is the maximum subarray length. The space complexity is O(n) where n is the length of the input array.
Solution Code
class Solution {
public int maxSumAfterPartitioning(int[] arr, int k) {
int n = arr.length;
int[] dp = new int[n];
for (int i = 0; i < n; i++) {
int maxVal = 0;
for (int j = 1; j <= k && i - j + 1 >= 0; j++) {
maxVal = Math.max(maxVal, arr[i - j + 1]);
dp[i] = Math.max(dp[i], (i - j >= 0 ? dp[i - j] : 0) + maxVal * j);
}
}
return dp[n - 1];
}
}Try It Yourself
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