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Formatted question description: https://leetcode.ca/all/813.html
813. Largest Sum of Averages (Medium)
We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve?
Note that our partition must use every number in A, and that scores are not necessarily integers.
Example: Input: A = [9,1,2,3,9] K = 3 Output: 20 Explanation: The best choice is to partition A into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20. We could have also partitioned A into [9, 1], [2], [3, 9], for example. That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.
Note:
1 <= A.length <= 100.1 <= A[i] <= 10000.1 <= K <= A.length.- Answers within
10^-6of the correct answer will be accepted as correct.
Related Topics:
Dynamic Programming
Solution 1. DP
Let dp[k][i] be the maximum score for the subproblem with k groups and A[0..i] subarray.
Let avg[i][j] be the average of numbers in subarray A[i..j].
dp[1][i] = avg[0][i]
dp[k][i] = max(dp[k][i], dp[k-1][j] + avg[j][i] | k-1 <= j <= i) where k >= 2
// OJ: https://leetcode.com/problems/largest-sum-of-averages/
// Time: O(N^2 * K)
// Space: O(NK + N^2)
class Solution {
public:
double largestSumOfAverages(vector<int>& A, int K) {
int N = A.size();
vector<vector<double>> avg(N + 1, vector<double>(N + 1));
for (int i = 0; i < N; ++i) {
int sum = 0;
for (int j = i; j < N; ++j) {
sum += A[j];
avg[i][j] = (double)sum / (j - i + 1);
}
}
vector<vector<double>> dp(K + 1, vector<double>(N));
for (int i = 0; i < N; ++i) dp[1][i] = avg[0][i];
for (int k = 2; k <= K; ++k) {
for (int i = 0; i < N; ++i) {
for (int j = k - 1; j <= i; ++j) {
dp[k][i] = max(dp[k][i], dp[k - 1][j - 1] + avg[j][i]);
}
}
}
return dp[K][N - 1];
}
};
Solution 2. DP
Count avg on the fly.
// OJ: https://leetcode.com/problems/largest-sum-of-averages/
// Time: O(N^2 * K)
// Space: O(NK)
class Solution {
public:
double largestSumOfAverages(vector<int>& A, int K) {
int N = A.size();
vector<vector<double>> dp(K + 1, vector<double>(N));
double sum = 0;
for (int i = 0; i < N; ++i) {
sum += A[i];
dp[1][i] = sum / (i + 1);
}
for (int k = 2; k <= K; ++k) {
for (int i = k - 1; i < N; ++i) {
double sum = 0;
for (int j = i; j >= k - 1; --j) {
sum += A[j];
dp[k][i] = max(dp[k][i], dp[k - 1][j - 1] + sum / (i - j + 1));
}
}
}
return dp[K][N - 1];
}
};
Solution 3. DP + Space Optimization
Since dp[k][i] is only dependent on dp[k-1][j-1] and j <= i, we can reduce the dp array to 1D array.
// OJ: https://leetcode.com/problems/largest-sum-of-averages/
// Time: O(N^2 * K)
// Space: O(N)
class Solution {
public:
double largestSumOfAverages(vector<int>& A, int K) {
int N = A.size();
vector<double> dp(N);
double sum = 0;
for (int i = 0; i < N; ++i) {
sum += A[i];
dp[i] = sum / (i + 1);
}
for (int k = 2; k <= K; ++k) {
for (int i = N - 1; i >= k - 1; --i) {
double sum = 0;
dp[i] = 0;
for (int j = i; j >= k - 1; --j) {
sum += A[j];
dp[i] = max(dp[i], dp[j - 1] + sum / (i - j + 1));
}
}
}
return dp[N - 1];
}
};
-
class Solution { public double largestSumOfAverages(int[] A, int K) { int length = A.length; double[] prefixSums = new double[length + 1]; for (int i = 1; i <= length; i++) prefixSums[i] = prefixSums[i - 1] + A[i - 1]; double[][] dp = new double[length + 1][K + 1]; for (int i = 1; i <= length; i++) { dp[i][1] = prefixSums[i] / i; int maxGroups = Math.min(i, K); for (int k = 2; k <= maxGroups; k++) { for (int j = 1; j < i; j++) dp[i][k] = Math.max(dp[i][k], dp[j][k - 1] + (prefixSums[i] - prefixSums[j]) / (i - j)); } } return dp[length][K]; } } ############ class Solution { private Double[][] f; private int[] s; private int n; public double largestSumOfAverages(int[] nums, int k) { n = nums.length; s = new int[n + 1]; f = new Double[n + 1][k + 1]; for (int i = 0; i < n; ++i) { s[i + 1] = s[i] + nums[i]; } return dfs(0, k); } private double dfs(int i, int k) { if (i == n) { return 0; } if (k == 1) { return (s[n] - s[i]) * 1.0 / (n - i); } if (f[i][k] != null) { return f[i][k]; } double ans = 0; for (int j = i; j < n; ++j) { double t = (s[j + 1] - s[i]) * 1.0 / (j - i + 1) + dfs(j + 1, k - 1); ans = Math.max(ans, t); } return f[i][k] = ans; } } -
// OJ: https://leetcode.com/problems/largest-sum-of-averages/ // Time: O(N^2 * K) // Space: O(NK + N^2) class Solution { public: double largestSumOfAverages(vector<int>& A, int K) { int N = A.size(); vector<vector<double>> avg(N + 1, vector<double>(N + 1)); for (int i = 0; i < N; ++i) { int sum = 0; for (int j = i; j < N; ++j) { sum += A[j]; avg[i][j] = (double)sum / (j - i + 1); } } vector<vector<double>> dp(K + 1, vector<double>(N)); for (int i = 0; i < N; ++i) dp[1][i] = avg[0][i]; for (int k = 2; k <= K; ++k) { for (int i = 0; i < N; ++i) { for (int j = k - 1; j <= i; ++j) { dp[k][i] = max(dp[k][i], dp[k - 1][j - 1] + avg[j][i]); } } } return dp[K][N - 1]; } }; -
class Solution: def largestSumOfAverages(self, nums: List[int], k: int) -> float: @cache def dfs(i, k): if i == n: return 0 if k == 1: return (s[-1] - s[i]) / (n - i) ans = 0 for j in range(i, n): t = (s[j + 1] - s[i]) / (j - i + 1) + dfs(j + 1, k - 1) ans = max(ans, t) return ans n = len(nums) s = list(accumulate(nums, initial=0)) return dfs(0, k) ############ class Solution: def largestSumOfAverages(self, A, K): """ :type A: List[int] :type K: int :rtype: float """ n = len(A) m_ = [[0 for i in range(n + 1)] for j in range(K + 1)] sum_ = [0] * (n + 1) for i in range(1, n + 1): sum_[i] = sum_[i - 1] + A[i - 1] return self.LSA(A, sum_, m_, n, K) # Largest sum of averages for first n elements in A partioned into K groups def LSA(self, A, sum_, m_, n, k): if m_[k][n] > 0: return m_[k][n] if k == 1: return sum_[n] / n for i in range(k - 1, n): m_[k][n] = max(m_[k][n], self.LSA(A, sum_, m_, i, k - 1) + (sum_[n] - sum_[i]) / (n - i)) return m_[k][n] -
func largestSumOfAverages(nums []int, k int) float64 { n := len(nums) s := make([]int, n+1) f := [110][110]float64{} for i, v := range nums { s[i+1] = s[i] + v } var dfs func(i, k int) float64 dfs = func(i, k int) float64 { if i == n { return 0 } if k == 1 { return float64(s[n]-s[i]) / float64(n-i) } if f[i][k] > 0 { return f[i][k] } var ans float64 for j := i; j < n; j++ { t := float64(s[j+1]-s[i])/float64(j-i+1) + dfs(j+1, k-1) ans = math.Max(ans, t) } f[i][k] = ans return ans } return dfs(0, k) }