1043. Partition Array for Maximum Sum

Problem Statement

Given an integer array arr, 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. Test cases are generated so that the answer fits in a 32-bit integer.

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

Intuition

Parition using front and solve for the right part
Just like Palindrome partitioning 2

Links

https://leetcode.com/problems/partition-array-for-maximum-sum/description/

Video Links

https://www.youtube.com/watch?v=PhWWJmaKfMc&ab_channel=takeUforward

Approach 1:

Memoization

C++

class Solution {
public:
    int find_ans(vector<int>& arr, int k, int index, int n, vector<int> &dp){
        if(index == arr.size())
            return 0;
        
        if(dp[index] != -1)
            return dp[index];

        int maxi = INT_MIN;
        int len = 0;
        int max_sum = INT_MIN;

        for(int i=index; i< min(n,index+k); i++){
            len++;
            maxi = max(maxi, arr[i]);

            int sum = len*maxi + find_ans(arr, k, i+1, n, dp);
            max_sum = max(max_sum, sum);
        }

        return dp[index] = max_sum;
    }

    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        int n = arr.size();
        vector<int> dp(n, -1);

        return find_ans(arr, k, 0, n, dp);
    }
};

Approach 2:

Tabulation

C++

class Solution {
public:
    int maxSumAfterPartitioning(vector<int>& arr, int k) {
        int n = arr.size();
        vector<int> dp(n+1, 0);

        for(int index=n-1; index>=0; index--){
            int maxi = INT_MIN;
            int len = 0;
            int max_sum = INT_MIN;

            for(int i=index; i< min(n,index+k); i++){
                len++;
                maxi = max(maxi, arr[i]);

                int sum = len*maxi + dp[i+1];
                max_sum = max(max_sum, sum);
            }

            dp[index] = max_sum;
        }

        return dp[0];
    }
};

Approach 3:

C++

Approach 4:

C++