1043. Partition Array for Maximum Sum

1043. Partition Array for Maximum Sum (Medium)

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

Companies:
Amazon

Related Topics:
Array, Dynamic Programming

Solution 1. Bottom-up DP

Let dp[i + 1] be the answer to the subproblem on A[0..i].

At each dp[i + 1], we can let A[t..i] be the last subarray 0 <= t <= i && i - t + 1 <= K.

dp[i+1] = max( dp[t] + max(t, i) * (i - t + 1) | 0 <= t <= i && i - t + 1 <= K )
dp[0] = 0

// OJ: https://leetcode.com/problems/partition-array-for-maximum-sum/
// Author: github.com/lzl124631x
// Time: O(NK)
// Space: O(N)
class Solution {
public:
    int maxSumAfterPartitioning(vector<int>& A, int K) {
        int dp[501] = {}, N = A.size();
        for (int i = 0; i < N; ++i) {
            int mx = 0;
            for (int t = i, last = max(0, i + 1 - K); t >= last; --t) {
                mx = max(mx, A[t]);
                dp[i + 1] = max(dp[i + 1], dp[t] + mx * (i - t + 1));
            }
        }
        return dp[N];
    }
};

Solution 2. Top-down DP

Same idea as Solution 1, just that the dp[i] is the answer to the subproblem on A[i..(N-1)].

For dp[i], we can take A[i..j] as the first subarray i <= j < N && j - i + 1 <= K.

// OJ: https://leetcode.com/problems/partition-array-for-maximum-sum/
// Author: github.com/lzl124631x
// Time: O(NK)
// Space: O(N)
class Solution {
    int m[501] = {}, N, K;
    int dp(vector<int> &A, int i) {
        if (i == N) return 0;
        if (m[i] != -1) return m[i];
        int mx = 0;
        for (int j = i; j < N && j - i + 1 <= K; ++j) {
            mx = max(mx, A[j]);
            m[i] = max(m[i], mx * (j - i + 1) + dp(A, j + 1));
        }
        return m[i];
    }
public:
    int maxSumAfterPartitioning(vector<int>& A, int K) {
        N = A.size();
        this->K = K;
        memset(m, -1, sizeof(m));
        return dp(A, 0);
    }
};

Solution 3. DP with Space Optimization

Since dp[i + 1] is only dependent on the previous K dp values, we can reduce the space complexity to O(K).

// OJ: https://leetcode.com/problems/partition-array-for-maximum-sum/
// Author: github.com/lzl124631x
// Time: O(NK)
// Space: O(K)
class Solution {
public:
    int maxSumAfterPartitioning(vector<int>& A, int K) {
        int N = A.size();
        vector<int> dp(K);
        for (int i = 0; i < N; ++i) {
            int mx = 0, val = 0;
            for (int t = i, last = max(0, i + 1 - K); t >= last; --t) {
                mx = max(mx, A[t]);
                val = max(val, dp[t % K] + mx * (i - t + 1));
            }
            dp[(i + 1) % K] = val;
        }
        return dp[N % K];
    }
};