Cat and Mouse II

Cat and Mouse II - Problem

Array Math Dynamic Programming Graph Topological Sort Memoization Matrix Game Theory Hard

A game is played by a cat and a mouse named Cat and Mouse.

The environment is represented by a grid of size rows x cols, where each element is a wall, floor, player (Cat, Mouse), or food.

Players are represented by the characters 'C' (Cat), 'M' (Mouse).
Floors are represented by the character '.' and can be walked on.
Walls are represented by the character '#' and cannot be walked on.
Food is represented by the character 'F' and can be walked on.
There is only one of each character 'C', 'M', and 'F' in grid.

Mouse and Cat play according to the following rules:

Mouse moves first, then they take turns to move.
During each turn, Cat and Mouse can jump in one of the four directions (left, right, up, down). They cannot jump over the wall nor outside of the grid.
catJump, mouseJump are the maximum lengths Cat and Mouse can jump at a time, respectively. Cat and Mouse can jump less than the maximum length.
Staying in the same position is allowed.
Mouse can jump over Cat.

The game can end in 4 ways:

If Cat occupies the same position as Mouse, Cat wins.
If Cat reaches the food first, Cat wins.
If Mouse reaches the food first, Mouse wins.
If Mouse cannot get to the food within 1000 turns, Cat wins.

Given a rows x cols matrix grid and two integers catJump and mouseJump, return true if Mouse can win the game if both Cat and Mouse play optimally, otherwise return false.

Input & Output

Example 1 — Basic Grid

$ Input: grid = [["M",".","F"],[".","#","."],[".",".","C"]], catJump = 2, mouseJump = 1

› Output: true

💡 Note: Mouse can reach food in 2 moves while cat needs more moves due to wall blocking and lower jump capacity

Example 2 — Cat Advantage

$ Input: grid = [["M",".",".",".","C"],[".",".",".",".","."],[".",".","F",".","."]], catJump = 3, mouseJump = 1

› Output: false

💡 Note: Cat has higher jump capacity and can intercept mouse before it reaches food

Example 3 — Wall Protection

$ Input: grid = [["C","#","M"],["#","#","#"],["F",".","."]], catJump = 1, mouseJump = 2

› Output: true

💡 Note: Wall separates cat from mouse, giving mouse safe path to food with higher jump

Constraints

rows == grid.length
cols = grid[i].length
1 ≤ rows, cols ≤ 8
grid[i][j] consist only of characters 'C', 'M', 'F', '.', '#'
There is only one of each character 'C', 'M', and 'F' in grid
1 ≤ catJump, mouseJump ≤ 8

Visualization

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Asked in

G Google 12 M Microsoft 8 A Amazon 6

This is a classic minimax game theory problem. The key insight is modeling it as alternating turns where mouse tries to maximize winning probability while cat tries to minimize it. Best approach uses memoization to cache game states. Time: O(rows×cols×rows×cols×turns), Space: O(rows×cols×rows×cols×turns)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force Game Simulation

⏱️ Time: O(4^(catJump+mouseJump)) Space: O(rows × cols × turns)

Try every possible move for both players at each turn, exploring the complete game tree until reaching a terminal state. This approach considers all possible paths the game could take.

Memoization (Dynamic Programming)

⏱️ Time: O(rows × cols × rows × cols × turns) Space: O(rows × cols × rows × cols × turns)

Same as brute force but memoize results for each unique game state (mouse position, cat position, turn, moves). This dramatically reduces repeated work.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

typedef struct {
    int student_id;
    char name[100];
    int age;
    int grade;
} Student;

Student* solution(Student* students, int size) {
    // In C, grades are already integers after parsing
    // No conversion needed
    return students;
}

int parseJson(const char* json, Student* students) {
    int count = 0;
    const char* pos = json + 1; // Skip opening [
    
    while (*pos && *pos != ']') {
        if (*pos == '{') {
            pos++; // Skip {
            Student* student = &students[count];
            
            while (*pos && *pos != '}') {
                // Skip whitespace
                while (*pos && isspace(*pos)) pos++;
                if (*pos == '"') {
                    pos++; // Skip opening quote
                    
                    // Read key
                    char key[50] = {0};
                    int keyLen = 0;
                    while (*pos && *pos != '"') {
                        key[keyLen++] = *pos++;
                    }
                    pos++; // Skip closing quote
                    
                    // Skip to value
                    while (*pos && *pos != ':') pos++;
                    pos++; // Skip :
                    while (*pos && isspace(*pos)) pos++;
                    
                    if (strcmp(key, "student_id") == 0) {
                        student->student_id = atoi(pos);
                        while (*pos && isdigit(*pos)) pos++;
                    } else if (strcmp(key, "name") == 0) {
                        pos++; // Skip opening quote
                        int nameLen = 0;
                        while (*pos && *pos != '"') {
                            student->name[nameLen++] = *pos++;
                        }
                        student->name[nameLen] = '\0';
                        pos++; // Skip closing quote
                    } else if (strcmp(key, "age") == 0) {
                        student->age = atoi(pos);
                        while (*pos && isdigit(*pos)) pos++;
                    } else if (strcmp(key, "grade") == 0) {
                        student->grade = (int)atof(pos);
                        while (*pos && (*pos == '.' || isdigit(*pos))) pos++;
                    }
                }
                
                if (*pos == ',') pos++;
            }
            
            if (*pos == '}') pos++;
            count++;
        }
        
        if (*pos == ',') pos++;
        while (*pos && isspace(*pos)) pos++;
    }
    
    return count;
}

void printJson(Student* students, int count) {
    printf("[");
    for (int i = 0; i < count; i++) {
        printf("{\"student_id\":%d,\"name\":\"%s\",\"age\":%d,\"grade\":%d}", 
               students[i].student_id, students[i].name, students[i].age, students[i].grade);
        if (i < count - 1) printf(",");
    }
    printf("]\n");
}

int main() {
    char input[10000];
    fgets(input, sizeof(input), stdin);
    
    Student students[1000];
    int count = parseJson(input, students);
    
    Student* result = solution(students, count);
    printJson(result, count);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

✓ Linear Space

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Medium Frequency

~45 min Avg. Time

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