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1043. Partition Array for Maximum Sum 👍

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  int maxSumAfterPartitioning(vector<int>& arr, int k) {
    const int n = arr.size();
    vector<int> dp(n + 1);

    for (int i = 1; i <= n; ++i) {
      int mx = INT_MIN;
      for (int j = 1; j <= min(i, k); ++j) {
        mx = max(mx, arr[i - j]);
        dp[i] = max(dp[i], dp[i - j] + mx * j);
      }
    }

    return dp[n];
  }
};

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class Solution {
  public int maxSumAfterPartitioning(int[] arr, int k) {
    final int n = arr.length;
    int[] dp = new int[n + 1];

    for (int i = 1; i <= n; ++i) {
      int mx = Integer.MIN_VALUE;
      for (int j = 1; j <= Math.max(i, k); ++j) {
        mx = Math.max(mx, arr[i - j]);
        dp[i] = Math.max(dp[i], dp[i - j] + mx * j);
      }
    }

    return dp[n];
  }
}

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class Solution:
  def maxSumAfterPartitioning(self, arr: list[int], k: int) -> int:
    n = len(arr)
    dp = [0] * (n + 1)

    for i in range(1, n + 1):
      mx = -math.inf
      for j in range(1, min(i, k) + 1):
        mx = max(mx, arr[i - j])
        dp[i] = max(dp[i], dp[i - j] + mx * j)

    return dp[n]