875013. Maximum score from at most K jumps
Medium
Maximum score from at most K jumps
Summary
The problem is to find the maximum score that can be obtained by making at most K jumps in a given array. The score for each jump is calculated based on the value at the starting and ending indices of the jump. This problem involves dynamic programming, arrays, and map data structures.
Detailed Explanation
The approach to solve this problem is to use dynamic programming with memoization. We create a 2D DP table dp where dp[i][j] represents the maximum score that can be obtained by making at most j jumps starting from index i.
Here's the step-by-step breakdown of the solution:
- Initialize the DP table with zeros:
dp = [[0 for _ in range(K + 1)] for _ in range(n + 1)] - Fill the DP table using a bottom-up approach:
- For each index i and jump count j, calculate the maximum score that can be obtained by making at most j jumps starting from index i.
- If j is 0, set
dp[i][j] = arr[i]because we cannot make any more jumps. - Otherwise, iterate over all possible ending indices for the current jump:
- For each ending index k, calculate the score for the current jump:
score = dp[k][j - 1] + (arr[k] - arr[i]) - Update the DP table:
dp[i][j] = max(dp[i][j], score)
- For each ending index k, calculate the score for the current jump:
- The maximum score is stored in
dp[0][K], which represents the maximum score that can be obtained by making at most K jumps starting from index 0.
Time complexity: O(n * K) Space complexity: O(n * K)
Optimized Solutions
Java
public int maxScore(int[] arr, int K) {
int n = arr.length;
int[][] dp = new int[n][K + 1];
for (int i = 0; i < n; i++) {
dp[i][0] = arr[i];
}
for (int j = 1; j <= K; j++) {
for (int i = j - 1; i >= 0; i--) {
int score = dp[i + j][j - 1] + (arr[i + j] - arr[i]);
dp[i][j] = Math.max(dp[i][j], score);
}
}
return dp[0][K];
}