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This approach uses dynamic programming where we keep track of the maximum points obtainable for each cell using two linear passes per row: one from left to right and the other from right to left. This ensures that we account for the cost of moving between rows efficiently.
Time Complexity: O(m * n) as we iterate through the cells twice per row. Space Complexity: O(n) due to the auxiliary arrays used.
1def maxPoints(points):
2 m, n = len(points), len(points[0])
3 prevRow = points[0][:]
4
5 for i in range(1, m):
6 leftToRight = prevRow[:]
7 for j in range(1, n):
8 leftToRight[j] = max(leftToRight[j], leftToRight[j - 1] - 1)
9
10 rightToLeft = prevRow[:]
11 for j in range(n - 2, -1, -1):
12 rightToLeft[j] = max(rightToLeft[j], rightToLeft[j + 1] - 1)
13
14 for j in range(n):
15 leftToRight[j] = points[i][j] + max(leftToRight[j], rightToLeft[j])
16
17 prevRow = leftToRight
18
19 return max(prevRow)
20
21points = [[1, 2, 3], [1, 5, 1], [3, 1, 1]]
22print("Maximum points:", maxPoints(points))
23Similar to the C-based implementations, this Python approach optimally determines maximum points by storing pre-computed row values. The leftToRight and rightToLeft adjustments ensure correct score after factoring in distance deductions.
This approach uses a dynamic array to store optimal points value and computes it via a prefix and suffix maximum accumulations, thereby reducing the transition complexity while still making sure every cell's deduction is feasible within two linear passes.
Time Complexity: O(m * n). Space Complexity: O(n), reducing allocation per row iteration.
1
public class Solution {
public static int MaxPoints(int[][] points) {
int m = points.Length;
int n = points[0].Length;
int[] prevRow = new int[n];
Array.Copy(points[0], prevRow, n);
for (int i = 1; i < m; i++) {
int[] leftMax = new int[n];
int[] rightMax = new int[n];
leftMax[0] = prevRow[0];
for (int j = 1; j < n; j++) {
leftMax[j] = Math.Max(leftMax[j - 1] - 1, prevRow[j]);
}
rightMax[n - 1] = prevRow[n - 1];
for (int j = n - 2; j >= 0; j--) {
rightMax[j] = Math.Max(rightMax[j + 1] - 1, prevRow[j]);
}
for (int j = 0; j < n; j++) {
prevRow[j] = points[i][j] + Math.Max(leftMax[j], rightMax[j]);
}
}
int maxPoints = 0;
foreach (var val in prevRow) {
if (val > maxPoints) maxPoints = val;
}
return maxPoints;
}
public static void Main() {
int[][] points = new int[][] {
new int[] {1, 2, 3},
new int[] {1, 5, 1},
new int[] {3, 1, 1}
};
Console.WriteLine("Maximum points: " + MaxPoints(points));
}
}
In this C# solution, careful attention is given to maintaining efficiency and state during transitions by deriving the potential from each end with optimal in-place adjustments.