1937. Maximum Number of Points with Cost LeetCode Solution
In this guide, you will get 1937. Maximum Number of Points with Cost LeetCode Solution with the best time and space complexity. The solution to Maximum Number of Points with Cost problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Maximum Number of Points with Cost
You are given an m x n integer matrix points (0-indexed). Starting with 0 points, you want to maximize the number of points you can get from the matrix.
To gain points, you must pick one cell in each row. Picking the cell at coordinates (r, c) will add points[r][c] to your score.
However, you will lose points if you pick a cell too far from the cell that you picked in the previous row. For every two adjacent rows r and r + 1 (where 0 <= r = 0.
-x for x < 0.
Example 1:
Input: points = [[1,2,3],[1,5,1],[3,1,1]]
Output: 9
Explanation:
The blue cells denote the optimal cells to pick, which have coordinates (0, 2), (1, 1), and (2, 0).
You add 3 + 5 + 3 = 11 to your score.
However, you must subtract abs(2 – 1) + abs(1 – 0) = 2 from your score.
Your final score is 11 – 2 = 9.
Input: points = [[1,5],[2,3],[4,2]]
Output: 11
Explanation:
The blue cells denote the optimal cells to pick, which have coordinates (0, 1), (1, 1), and (2, 0).
You add 5 + 3 + 4 = 12 to your score.
However, you must subtract abs(1 – 1) + abs(1 – 0) = 1 from your score.
Your final score is 12 – 1 = 11.
Constraints:
m == points.length
n == points[r].length
1 <= m, n <= 105
1 <= m * n <= 105
0 <= points[r][c] <= 105
Complexity Analysis
Time Complexity: O(mn)
Space Complexity: O(n)
1937. Maximum Number of Points with Cost LeetCode Solution in C++
class Solution {
public:
long long maxPoints(vector<vector<int>>& points) {
const int n = points[0].size();
// dp[j] := the maximum number of points you can have if points[i][j] is the
// most recent cell you picked
vector<long> dp(n);
for (const vector<int>& row : points) {
vector<long> leftToRight(n);
long runningMax = 0;
for (int j = 0; j < n; ++j) {
runningMax = max(runningMax - 1, dp[j]);
leftToRight[j] = runningMax;
}
vector<long> rightToLeft(n);
runningMax = 0;
for (int j = n - 1; j >= 0; --j) {
runningMax = max(runningMax - 1, dp[j]);
rightToLeft[j] = runningMax;
}
for (int j = 0; j < n; ++j)
dp[j] = max(leftToRight[j], rightToLeft[j]) + row[j];
}
return ranges::max(dp);
}
}; /* code provided by PROGIEZ */
1937. Maximum Number of Points with Cost LeetCode Solution in Java
class Solution {
public long maxPoints(int[][] points) {
final int n = points[0].length;
// dp[j] := the maximum number of points you can have if points[i][j] is the
// most recent cell you picked
long[] dp = new long[n];
for (int[] row : points) {
long[] leftToRight = new long[n];
long runningMax = 0;
for (int j = 0; j < n; ++j) {
runningMax = Math.max(runningMax - 1, dp[j]);
leftToRight[j] = runningMax;
}
long[] rightToLeft = new long[n];
runningMax = 0;
for (int j = n - 1; j >= 0; --j) {
runningMax = Math.max(runningMax - 1, dp[j]);
rightToLeft[j] = runningMax;
}
for (int j = 0; j < n; ++j)
dp[j] = Math.max(leftToRight[j], rightToLeft[j]) + row[j];
}
return Arrays.stream(dp).max().getAsLong();
}
} // code provided by PROGIEZ
1937. Maximum Number of Points with Cost LeetCode Solution in Python
class Solution:
def maxPoints(self, points: list[list[int]]) -> int:
n = len(points[0])
# dp[j] := the maximum number of points you can have if points[i][j] is the
# most recent cell you picked
dp = [0] * n
for row in points:
leftToRight = [0] * n
runningMax = 0
for j in range(n):
runningMax = max(runningMax - 1, dp[j])
leftToRight[j] = runningMax
rightToLeft = [0] * n
runningMax = 0
for j in range(n - 1, - 1, -1):
runningMax = max(runningMax - 1, dp[j])
rightToLeft[j] = runningMax
for j in range(n):
dp[j] = max(leftToRight[j], rightToLeft[j]) + row[j]
return max(dp) # code by PROGIEZ
