Skip to main content

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

Comments

Post a Comment

Popular posts from this blog

Segment Tree

Segment tree can be viewed as an abstract data structure which using some more space to trade for speed. For example, for a typical question with O(N^2) time complexity, the segment tree method can decrease it to O(N*log(N)).  To make it understandable, let us consider one example. Say we have an integer array of N size, and what we want is to query the maximum with a query range [idx1, idx2], where idx1 is the left indexes, and idx2 is the right indexes inclusive. If we only do this kind of query once, then we just need to scan through the array from idx1 to idx2 once, and record the maximum, done. The time complexity is O(N), which is decent enough in most cases even though it is not the optimal one (for example, with a segment tree built, the time complexity can decrease down to O(log(N))). However, how about we need to query the array N times? If we continue to use the naïve way above, then the time complexity is O(N^2), since for each query we need to scan the query range once. We
Post a Comment
Read more

Binary Search - Hard Level - Question 3

Binary Search - Hard Level - Question 3 878. Nth Magical Number A positive integer is magical if it is divisible by either a or b. Given the three integers n, a, and b, return the nth magical number. Since the answer may be very large, return it modulo 10^9 + 7. Analysis: Let us consider some examples first. Example 1, a = 4, b = 2. If b is dividable by a, then all the numbers which is dividable by a should be dividable by b as well. So the nth magical number should be n*b; Example 2, a = 3, b = 2. The multiples of 2 are: 2, 4, 6, 8, 10, 12, ... The multiple of 3 are: 3, 6, 9, 12, ... So the overlap is related to the minimum common multiple between a and b, and we need to remove the overlap which is double-counted. So now, we make some conclusions: 1. the upper bound of the nth magical number should be n*b, where a is the smaller one (or b <= a); 2. there are n*b/a magical numbers smaller than n*b; 3. there are n*b/(minimum common multiple) overlaps. Thus, the overall count is: n +
Post a Comment
Read more

Recursion - Example

Recursion - Example Leetcode 231  Power of Two Given an integer n, return true if it is a power of two. Otherwise, return false. An integer n is a power of two, if there exists an integer x such that n == 2^x Constraints: -2^31 <= n <= 2^31 - 1 Analysis: One way is to think about this question recursively: if n%2 == 1, then n must not be power of 2; if not, then we just need to consider whether (n/2) is a power of 2 or not. This is exactly the "same question with a smaller size"! It is trivial to figure out the base cases: if n == 0, return false; if n == 1, return true. See the code below: class Solution { public: bool isPowerOfTwo(int n) { // base cases if(n == 0) return false; if(n == 1) return true; // converging if(n%2 == 1) return false; return isPowerOfTwo(n/2); } }; If interested, there are some other ways to solve this problem. For example, using bit manipulation, we can have the following solution: class
Post a Comment
Read more