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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.
1import java.util.*;
2
3public class MaxPoints {
4 public static int maxPoints(int[][] points) {
5 int m = points.length;
6 int n = points[0].length;
7
8 int[] prevRow = new int[n];
9 System.arraycopy(points[0], 0, prevRow, 0, n);
10
11 for (int i = 1; i < m; i++) {
12 int[] leftToRight = new int[n];
13 int[] rightToLeft = new int[n];
14 int[] currentRow = new int[n];
15
16 leftToRight[0] = prevRow[0];
17 for (int j = 1; j < n; j++) {
18 leftToRight[j] = Math.max(leftToRight[j - 1] - 1, prevRow[j]);
19 }
20
21 rightToLeft[n - 1] = prevRow[n - 1];
22 for (int j = n - 2; j >= 0; j--) {
23 rightToLeft[j] = Math.max(rightToLeft[j + 1] - 1, prevRow[j]);
24 }
25
26 for (int j = 0; j < n; j++) {
27 currentRow[j] = points[i][j] + Math.max(leftToRight[j], rightToLeft[j]);
28 }
29
30 prevRow = currentRow;
31 }
32
33 return Arrays.stream(prevRow).max().getAsInt();
34 }
35
36 public static void main(String[] args) {
37 int[][] points = {{1, 2, 3}, {1, 5, 1}, {3, 1, 1}};
38 System.out.println("Maximum points: " + maxPoints(points));
39 }
40}
41
Using three arrays, this Java solution calculates the max points with linear iterations over each row where transitions between rows are handled by comparing values in leftToRight
and rightToLeft
.
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#include <algorithm>
using namespace std;
int maxPoints(vector<vector<int>>& points) {
int m = points.size();
int n = points[0].size();
vector<int> prevRow(points[0].begin(), points[0].end());
for (int i = 1; i < m; ++i) {
vector<int> leftMax(n);
vector<int> rightMax(n);
vector<int> currentRow(n);
leftMax[0] = prevRow[0];
for (int j = 1; j < n; ++j) {
leftMax[j] = max(leftMax[j - 1] - 1, prevRow[j]);
}
rightMax[n - 1] = prevRow[n - 1];
for (int j = n - 2; j >= 0; --j) {
rightMax[j] = max(rightMax[j + 1] - 1, prevRow[j]);
}
for (int j = 0; j < n; ++j) {
currentRow[j] = points[i][j] + max(leftMax[j], rightMax[j]);
}
prevRow = currentRow;
}
return *max_element(prevRow.begin(), prevRow.end());
}
int main() {
vector<vector<int>> points = {{1, 2, 3}, {1, 5, 1}, {3, 1, 1}};
int result = maxPoints(points);
printf("Maximum points: %d\n", result);
return 0;
}
The primary aim in this C++ code is to optimize the array management and expedite the element-wise summation for determining ideal points. By using dynamic arrays, computation resources are conserved and enhanced.