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Dynamic Programming - Hard Level - Question 1

Dynamic Programming - Hard Level - Question 1 


Leetcode 1937 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 < m - 1), picking cells at coordinates (r, c1) and (r + 1, c2) will subtract abs(c1 - c2) from your score.

Return the maximum number of points you can achieve.

abs(x) is defined as:

x for x >= 0.

-x for x < 0.

Constraints:

m == points.length

n == points[r].length

1 <= m, n <= 10^5

1 <= m * n <= 10^5

0 <= points[r][c] <= 10^5



Analysis:

Based on the description, we can define dp[i][j] as the maximum number of points with cost choosing the points[i][j].

Except the last row, for elements in all the other rows have this transition formula:

dp[i][j] = max(dp[i+1][k] - abs(j - k)) + points[i][j], where k is from 0 to n-1.

So the time complexity is O(m*n*n), which cannot pass the big-data test case. Since this is the base for the optimization, the code is given below:


class Solution {
public:
    long long maxPoints(vector<vector<int>>& points) {
        long long res = 0, val = -1e6;
        int m = points.size(), n = points.front().size();
        vector<vector<long long>> dp(m, vector<long long>(n, 0));
        for(int i=0; i<n; ++i) {
            dp[m-1][i] = points[m-1][i];
            res = max(res, dp[m-1][i]);
        }
        for(int i=m-2; i>=0; --i) {
            for(int j=0; j<n; ++j) {
                long long t = val;
                for(int k=0; k<n; ++k) {
                    t = max(t, dp[i+1][k] - abs(j - k));
                }
                dp[i][j] = t + points[i][j];
                res = max(res, dp[i][j]);
            }
        }
        return res;
    }
};

 

Then the question becomes how to optimize the most inside loop from O(n) to constant time.

The line involved is 

t = max(t, dp[i+1][k] - abs(j-k))

So the first step can remove the abs function,

t = max(t, dp[i+1][k] - j + k), when j>=k;

t = max(t, dp[i+1][k] +j - k), when j<k;

We can re-write them as,

t = max(t, dp[i+1][k] + k) - j, when j>=k;

t = max(t, dp[i+1][k] - k) + j, when j<k;

So what we need actually is the maximum value of (dp[i][k] + k) on the left-side of j (smaller than) and the maximum value of (dp[i][k] - k) on the right-side of j(larger than).

The optimization becomes to maintain a maximum from the left and from the right, respectively, which is trivial to do. 

The overall time complexity is O(m*n) now.


See the code below:

class Solution {
public:
    long long maxPoints(vector<vector<int>>& points) {
        long long res = 0, val = -1e6;
        int m = points.size(), n = points.front().size();
        vector<vector<long long>> dp(m, vector<long long>(n, 0));
        for(int i=0; i<n; ++i) {
            dp[m-1][i] = points[m-1][i];
            res = max(res, dp[m-1][i]);
        }
        for(int i=m-2; i>=0; --i) {
            vector<long long> left(n, val), right(n, val);
            left[0] = dp[i+1][0];
            for(int j=1; j<n; ++j) left[j] = max(left[j-1], dp[i+1][j] + j);
            right[n-1] = dp[i+1][n-1] - (n-1);
            for(int j=n-2; j>=0; --j) right[j] = max(right[j+1], dp[i+1][j] - j);
            for(int j=0; j<n; ++j) {
                dp[i][j] = max(dp[i][j], points[i][j] + left[j] - j);
                dp[i][j] = max(dp[i][j], points[i][j] + right[j] + j);
                res = max(res, dp[i][j]);
            }
        }
        return res;
    }
};


Upper Layer

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