In a N x N grid representing a field of cherries, each cell is one of three possible integers.
Your task is to collect maximum number of cherries possible by following the rules below:
Example 1:
Input: grid = [[0, 1, -1], [1, 0, -1], [1, 1, 1]] Output: 5 Explanation: The player started at (0, 0) and went down, down, right right to reach (2, 2). 4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]]. Then, the player went left, up, up, left to return home, picking up one more cherry. The total number of cherries picked up is 5, and this is the maximum possible.
Note:
grid is an N by N 2D array, with 1 <= N <= 50.grid[i][j] is an integer in the set {-1, 0, 1}.Intuition
\nLet\'s find the most cherries we can pick up with one path, pick them up, then find the most cherries we can pick up with a second path on the remaining field.
\nThough a counter example might be hard to think of, this approach fails to find the best answer to this case:
\n11100\n00101\n10100\n00100\n00111\n
Algorithm
\nWe can use dynamic programming to find the most number of cherries dp[i][j] that can be picked up from any location (i, j) to the bottom right corner. This is a classic question very similar to Minimum Path Sum, refer to the link if you are not familiar with this type of question.
After, we can find an first path that maximizes the number of cherries taken by using our completed dp as an oracle for deciding where to move. We\'ll choose the move that allows us to pick up more cherries (based on comparing dp[i+1][j] and dp[i][j+1]).
After taking the cherries from that path (and removing it from the grid), we\'ll take the cherries again.
\n\nComplexity Analysis
\nTime Complexity: , where is the length of grid. Our dynamic programming consists of two for-loops of length N.
Space Complexity: , the size of dp.
Intuition
\nInstead of walking from end to beginning, let\'s reverse the second leg of the path, so we are only considering two paths from the beginning to the end.
\nNotice after t steps, each position (r, c) we could be, is on the line r + c = t. So if we have two people at positions (r1, c1) and (r2, c2), then r2 = r1 + c1 - c2. That means the variables r1, c1, c2 uniquely determine 2 people who have walked the same r1 + c1 number of steps. This sets us up for dynamic programming quite nicely.
Algorithm
\nLet dp[r1][c1][c2] be the most number of cherries obtained by two people starting at (r1, c1) and (r2, c2) and walking towards (N-1, N-1) picking up cherries, where r2 = r1+c1-c2.
If grid[r1][c1] and grid[r2][c2] are not thorns, then the value of dp[r1][c1][c2] is (grid[r1][c1] + grid[r2][c2]), plus the maximum of dp[r1+1][c1][c2], dp[r1][c1+1][c2], dp[r1+1][c1][c2+1], dp[r1][c1+1][c2+1] as appropriate. We should also be careful to not double count in case (r1, c1) == (r2, c2).
Complexity Analysis
\nTime Complexity: , where is the length of grid. Our dynamic programming has states.
Space Complexity: , the size of memo.
Intuition
\nLike in Approach #2, we have the idea of dynamic programming.
\nSay r1 + c1 = t is the t-th layer. Since our recursion only references the next layer, we only need to keep two layers in memory at a time.
Algorithm
\nAt time t, let dp[c1][c2] be the most cherries that we can pick up for two people going from (0, 0) to (r1, c1) and (0, 0) to (r2, c2), where r1 = t-c1, r2 = t-c2. Our dynamic program proceeds similarly to Approach #2.
Complexity Analysis
\nTime Complexity: , where is the length of grid. We have three for-loops of size .
Space Complexity: , the sizes of dp and dp2.