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Formatted question description: https://leetcode.ca/all/1219.html
1219. Path with Maximum Gold (Medium)
In a gold mine grid of size m * n, each cell in this mine has an integer representing the amount of gold in that cell, 0 if it is empty.
Return the maximum amount of gold you can collect under the conditions:
- Every time you are located in a cell you will collect all the gold in that cell.
- From your position you can walk one step to the left, right, up or down.
- You can't visit the same cell more than once.
- Never visit a cell with
0gold. - You can start and stop collecting gold from any position in the grid that has some gold.
Example 1:
Input: grid = [[0,6,0],[5,8,7],[0,9,0]] Output: 24 Explanation: [[0,6,0], [5,8,7], [0,9,0]] Path to get the maximum gold, 9 -> 8 -> 7.
Example 2:
Input: grid = [[1,0,7],[2,0,6],[3,4,5],[0,3,0],[9,0,20]] Output: 28 Explanation: [[1,0,7], [2,0,6], [3,4,5], [0,3,0], [9,0,20]] Path to get the maximum gold, 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7.
Constraints:
1 <= grid.length, grid[i].length <= 150 <= grid[i][j] <= 100- There are at most 25 cells containing gold.
Related Topics: Backtracking
Solution 1. DFS
// OJ: https://leetcode.com/problems/path-with-maximum-gold/
// Time: O(MN*4^(MN))
// Space: O(MN) because in the worst case the DFS will visit all cells.
class Solution {
int M, N, ans = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
void dfs(vector<vector<int>> &G, int i, int j, int cnt) {
int g = G[i][j];
G[i][j] = 0;
cnt += g;
ans = max(ans, cnt);
for (auto &[dx, dy] : dirs) {
int x = i + dx, y = j + dy;
if (x < 0 || y < 0 || x >= M || y >= N || G[x][y] == 0) continue;
dfs(G, x, y, cnt);
}
G[i][j] = g;
}
public:
int getMaximumGold(vector<vector<int>>& G) {
M = G.size(), N = G[0].size();
for (int i = 0; i < M; ++i) {
for (int j = 0; j < N; ++j) {
if (G[i][j] == 0) continue;
dfs(G, i, j, 0);
}
}
return ans;
}
};
-
class Solution { int max = 0; public int getMaximumGold(int[][] grid) { int rows = grid.length, columns = grid[0].length; boolean[][] visited = new boolean[rows][columns]; for (int i = 0; i < rows; i++) { for (int j = 0; j < columns; j++) { if (grid[i][j] != 0) backtrack(grid, i, j, 0, visited); } } return max; } public void backtrack(int[][] grid, int row, int column, int amount, boolean[][] visited) { int rows = grid.length, columns = grid[0].length; if (row < 0 || row >= rows || column < 0 || column >= columns || grid[row][column] == 0 || visited[row][column]) max = Math.max(max, amount); else { amount += grid[row][column]; visited[row][column] = true; int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} }; for (int[] direction : directions) backtrack(grid, row + direction[0], column + direction[1], amount, visited); amount -= grid[row][column]; visited[row][column] = false; } } } ############ class Solution { private int[][] grid; private int m; private int n; public int getMaximumGold(int[][] grid) { m = grid.length; n = grid[0].length; this.grid = grid; int ans = 0; for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { ans = Math.max(ans, dfs(i, j)); } } return ans; } private int dfs(int i, int j) { if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) { return 0; } int t = grid[i][j]; grid[i][j] = 0; int[] dirs = {-1, 0, 1, 0, -1}; int ans = 0; for (int k = 0; k < 4; ++k) { ans = Math.max(ans, t + dfs(i + dirs[k], j + dirs[k + 1])); } grid[i][j] = t; return ans; } } -
// OJ: https://leetcode.com/problems/path-with-maximum-gold/ // Time: O(MN*4^(MN)) // Space: O(MN) because in the worst case the DFS will visit all cells. class Solution { int M, N, ans = 0, dirs[4][2] = { {0,1},{0,-1},{1,0},{-1,0} }; void dfs(vector<vector<int>> &G, int i, int j, int cnt) { int g = G[i][j]; G[i][j] = 0; cnt += g; ans = max(ans, cnt); for (auto &[dx, dy] : dirs) { int x = i + dx, y = j + dy; if (x < 0 || y < 0 || x >= M || y >= N || G[x][y] == 0) continue; dfs(G, x, y, cnt); } G[i][j] = g; } public: int getMaximumGold(vector<vector<int>>& G) { M = G.size(), N = G[0].size(); for (int i = 0; i < M; ++i) { for (int j = 0; j < N; ++j) { if (G[i][j] == 0) continue; dfs(G, i, j, 0); } } return ans; } }; -
class Solution: def getMaximumGold(self, grid: List[List[int]]) -> int: def dfs(i, j): if not (0 <= i < m and 0 <= j < n and grid[i][j]): return 0 t = grid[i][j] grid[i][j] = 0 ans = t + max( dfs(i + a, j + b) for a, b in [[0, 1], [0, -1], [-1, 0], [1, 0]] ) grid[i][j] = t return ans m, n = len(grid), len(grid[0]) return max(dfs(i, j) for i in range(m) for j in range(n)) ############ # 1219. Path with Maximum Gold # https://leetcode.com/problems/path-with-maximum-gold/ class Solution: def getMaximumGold(self, grid: List[List[int]]): rows, cols = len(grid), len(grid[0]) res = 0 def dfs(x, y): tmp = grid[x][y] grid[x][y] = 0 total = 0 for dx,dy in ((x+1, y), (x-1, y), (x, y-1), (x, y+1)): if 0 <= dx < rows and 0 <= dy < cols and grid[dx][dy] != 0: total = max(total, grid[dx][dy] + dfs(dx,dy)) grid[x][y] = tmp return total for i in range(rows): for j in range(cols): if grid[i][j] > 0: res = max(res, grid[i][j] + dfs(i,j)) return res -
func getMaximumGold(grid [][]int) int { m, n := len(grid), len(grid[0]) var dfs func(i, j int) int dfs = func(i, j int) int { if i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0 { return 0 } t := grid[i][j] grid[i][j] = 0 ans := 0 dirs := []int{-1, 0, 1, 0, -1} for k := 0; k < 4; k++ { ans = max(ans, t+dfs(i+dirs[k], j+dirs[k+1])) } grid[i][j] = t return ans } ans := 0 for i := 0; i < m; i++ { for j := 0; j < n; j++ { ans = max(ans, dfs(i, j)) } } return ans } func max(a, b int) int { if a > b { return a } return b } -
/** * @param {number[][]} grid * @return {number} */ var getMaximumGold = function (grid) { const m = grid.length; const n = grid[0].length; function dfs(i, j) { if (i < 0 || i >= m || j < 0 || j >= n || grid[i][j] == 0) { return 0; } const t = grid[i][j]; grid[i][j] = 0; let ans = 0; const dirs = [-1, 0, 1, 0, -1]; for (let k = 0; k < 4; ++k) { ans = Math.max(ans, t + dfs(i + dirs[k], j + dirs[k + 1])); } grid[i][j] = t; return ans; } let ans = 0; for (let i = 0; i < m; ++i) { for (let j = 0; j < n; ++j) { ans = Math.max(ans, dfs(i, j)); } } return ans; };