LeetCode 913: Cat and Mouse Solution
Master LeetCode problem 913 (Cat and Mouse), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
913. Cat and Mouse
Problem Explanation
Explanation:
The problem can be modeled as a graph problem where nodes represent the positions of the mouse and cat, and edges represent possible moves. We can use a depth-first search (DFS) approach with memoization to simulate all possible moves of the mouse and cat and determine the outcome of the game.
The idea is to create a memoization table that stores the results of all possible game states (mouse position, cat position, and turn). We start by exhaustively exploring all possible moves from the current state and recursively determine the outcome of the game based on the optimal moves of both players. We use a turn-based approach to keep track of whose turn it is.
To prevent cycles and infinite loops, we mark visited states in the memoization table. If we encounter a state that has already been visited, we know that the game is in a draw state.
We iterate through the graph and simulate all possible moves of the mouse and cat. If the game ends in a win for either player, we update the memoization table accordingly. If the game is a draw, we mark the state as a draw in the memoization table.
The time complexity of this approach is O(n^3) where n is the number of nodes in the graph, and the space complexity is also O(n^3) due to memoization.
:
Solution Code
class Solution {
public int catMouseGame(int[][] graph) {
int n = graph.length;
int[][][] memo = new int[n][n][2];
return dfs(graph, 1, 2, 0, memo);
}
private int dfs(int[][] graph, int mouse, int cat, int turn, int[][][] memo) {
if (memo[mouse][cat][turn] != 0) {
return memo[mouse][cat][turn];
}
if (mouse == 0) {
return memo[mouse][cat][turn] = 1;
}
if (mouse == cat) {
return memo[mouse][cat][turn] = 2;
}
if (turn == 0) {
int res = 2; // Assume cat wins
for (int next : graph[cat]) {
if (next == 0) continue;
int outcome = dfs(graph, mouse, next, 1, memo);
if (outcome == 1) return memo[mouse][cat][turn] = 1;
if (outcome == 0) res = 0;
}
return memo[mouse][cat][turn] = res;
} else {
int res = 1; // Assume mouse wins
for (int next : graph[mouse]) {
int outcome = dfs(graph, next, cat, 0, memo);
if (outcome == 2) return memo[mouse][cat][turn] = 2;
if (outcome == 0) res = 0;
}
return memo[mouse][cat][turn] = res;
}
}
}Try It Yourself
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