10:I[13699,["/_next/static/chunks/6b91a202b724e571.js","/_next/static/chunks/6710f67d945945a1.js","/_next/static/chunks/cb0f98aea04386f6.js"],"default"] 11:I[25152,["/_next/static/chunks/6b91a202b724e571.js","/_next/static/chunks/6710f67d945945a1.js","/_next/static/chunks/cb0f98aea04386f6.js"],"default"] 12:T4ca,class Solution: def maximumAmount(self, coins: List[List[int]]) -> int: m, n = len(coins), len(coins[0]) dp = [[-math.inf] * 3 for _ in range(n + 1)] dp[1][0] = 0 dp[1][1] = 0 dp[1][2] = 0 for i in range(m): next_dp = [[-math.inf] * 3 for _ in range(n + 1)] for j in range(n): for k in range(3): # Base transition (no neutralization here) best_prev = max(dp[j + 1][k], next_dp[j][k]) if best_prev != -math.inf: next_dp[j + 1][k] = max(next_dp[j + 1][k], best_prev + coins[i][j]) # Transition with neutralization if k > 0: best_prev_power = max(dp[j + 1][k - 1], next_dp[j][k - 1]) if best_prev_power != -math.inf and coins[i][j] < 0: next_dp[j + 1][k] = max(next_dp[j + 1][k], best_prev_power) # Special setup for first cell handling on first row if i == 0: next_dp[0] = [-math.inf] * 3 dp = next_dp return max(dp[n])13:T51f,class Solution { public int maximumAmount(int[][] coins) { int m = coins.length, n = coins[0].length; int[][][] dp = new int[m][n][3]; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { Arrays.fill(dp[i][j], Integer.MIN_VALUE / 2); } } dp[0][0][0] = coins[0][0]; dp[0][0][1] = Math.max(coins[0][0], 0); dp[0][0][2] = Math.max(coins[0][0], 0); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { for (int k = 0; k < 3; k++) { if (i > 0) { dp[i][j][k] = Math.max(dp[i][j][k], dp[i - 1][j][k] + coins[i][j]); if (k > 0 && coins[i][j] < 0) { dp[i][j][k] = Math.max(dp[i][j][k], dp[i - 1][j][k - 1]); } } if (j > 0) { dp[i][j][k] = Math.max(dp[i][j][k], dp[i][j - 1][k] + coins[i][j]); if (k > 0 && coins[i][j] < 0) { dp[i][j][k] = Math.max(dp[i][j][k], dp[i][j - 1][k - 1]); } } } } } return dp[m - 1][n - 1][2]; } }14:T4ad,class Solution { public: int maximumAmount(vector>& coins) { int m = coins.size(), n = coins[0].size(); vector>> dp(m, vector>(n, vector(3, -1e18))); dp[0][0][0] = coins[0][0]; dp[0][0][1] = max(coins[0][0], 0); dp[0][0][2] = max(coins[0][0], 0); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { for (int k = 0; k < 3; k++) { if (i > 0) { dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j][k] + coins[i][j]); if (k > 0 && coins[i][j] < 0) { dp[i][j][k] = max(dp[i][j][k], dp[i - 1][j][k - 1]); } } if (j > 0) { dp[i][j][k] = max(dp[i][j][k], dp[i][j - 1][k] + coins[i][j]); if (k > 0 && coins[i][j] < 0) { dp[i][j][k] = max(dp[i][j][k], dp[i][j - 1][k - 1]); } } } } } return dp[m - 1][n - 1][2]; } };

Time Complexity

Space Complexity