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It's been a while since I have worked in C, so a language-specific review would not be so accurate. The only thing I can say in that aspect is that you should put curly bracket even in single-command body ifs as the code gets more readable and less error prone.

Taking a look at the problem and at your solution I think that you're overcomplicating it. You are allocating memory for each box and working with it when you actually don't need them. All you need is a square matrix - to be more precise, a lower triangular matrix - of ints (in this case).

The idea is the following:

1. Represent the boxes' status via the matrix \$A\$. If the box in the position \$(x, y)\$ is wet then \$A_{x,y} = 1\$, otherwise \$A_{x,y} = 0\$. So, initially, every item in the matrix is \$0\$.
2. When the box in position \$(x, y)\$ leaks you just set all the appropriate cells of \$A\$ to \$1\$ and count the number of cells whose value has changed from \$0\$ to \$1\$.
3. Return the value of the counter for each case.

In pseudo-code (I'm not so practical in C), it becomes something like the following:

int[,] A = new int[7,7];

void initialize_matrix(){
    for(int i = 0; i < 7, i++){
        for(int j = 0; j < 7; j++){
            A[i, j] = 0;
        }
    }
}

int count_wet_boxes(int x, int y){
    int counter = 0;

    for(int i = x; i >= 0; i--){
        for(int j = 0; (i + j) <= (x + y); j++){
            if(A[i, j] == 0){
                A[i, j] = 1;
                counter++;
            }
        }
    }

    return counter;
}

int main(){
    initialize_matrix();

    // the rest of the logic goes here
}

You can also try to ask in the Mathematics SE site if there is a formula you can use.

Let me know if anything's unclear.

It's been a while since I have worked in C, so a language-specific review would not be so accurate. The only thing I can say in that aspect is that you should put curly bracket even in single-command body ifs as the code gets more readable and less error prone.

Taking a look at the problem and at your solution I think that you're overcomplicating it. You are allocating memory for each box and working with it when you actually don't need them. All you need is a square matrix - to be more precise, a lower triangular matrix - of ints (in this case).

The idea is the following:

1. Represent the boxes' status via the matrix \$A\$. If the box in the position \$(x, y)\$ is wet then \$A_{x,y} = 1\$, otherwise \$A_{x,y} = 0\$. So, initially, every item in the matrix is \$0\$.
2. When the box in position \$(x, y)\$ leaks you just set all the appropriate cells of \$A\$ to \$1\$ and count the number of cells whose value has changed from \$0\$ to \$1\$.
3. Return the value of the counter for each case.

In pseudo-code (I'm not so practical in C), it becomes something like the following:

int[,] A = new int[7,7];

void initialize_matrix(){
    for(int i = 0; i < 7, i++){
        for(int j = 0; j < 7; j++){
            A[i, j] = 0;
        }
    }
}

int count_wet_boxes(int x, int y){
    int counter = 0;

    for(int i = x; i >= 0; i--){
        for(int j = 0; (i + j) <= (x + y); j++){
            if(A[i, j] == 0){
                A[i, j] = 1;
                counter++;
            }
        }
    }

    return counter;
}

int main(){
    initialize_matrix();

    // the rest of the logic goes here
}

You can also try to ask in the Mathematics SE site if there is a formula you can use.

Let me know if anything's unclear.

It's been a while since I haven't worked in C, so a language-specific review would not be so accurate. The only thing I can say in that aspect is that you should put curly bracket even in single-command body ifs as the code gets more readable and less error prone.

Taking a look at the problem and at your solution I think that you're overcomplicating it. You are allocating memory for each box and working with it when you actually don't need them. All you need is a square matrix - to be more precise, a lower triangular matrix - of ints (in this case).

The idea is the following:

1. Represent the boxes' status via the matrix \$A\$. If the box in the position \$(x, y)\$ is wet then \$A_{x,y} = 1\$, otherwise \$A_{x,y} = 0\$. So, initially, every item in the matrix is \$0\$.
2. When the box in position \$(x, y)\$ leaks you just set all the appropriate cells of \$A\$ to \$1\$ and count the number of cells whose value has changed from \$0\$ to \$1\$.
3. Return the value of the counter for each case.

In pseudo-code (I'm not so practical in C), it becomes something like the following:

int[,] A = new int[7,7];

void initialize_matrix(){
    for(int i = 0; i < 7, i++){
        for(int j = 0; j < 7; j++){
            A[i, j] = 0;
        }
    }
}

int count_wet_boxes(int x, int y){
    int counter = 0;

    for(int i = x; i >= 0; i--){
        for(int j = 0; (i + j) <= (x + y); j++){
            if(A[i, j] == 0){
                A[i, j] = 1;
                counter++;
            }
        }
    }

    return counter;
}

int main(){
    initialize_matrix();

    // the rest of the logic goes here
}

You can also try to ask in the Mathematics SE site if there is a formula you can use.

Let me know if anything's unclear.

It's been a while since I have worked in C, so a language-specific review would not be so accurate. The only thing I can say in that aspect is that you should put curly bracket even in single-command body ifs as the code gets more readable and less error prone.

Taking a look at the problem and at your solution I think that you're overcomplicating it. You are allocating memory for each box and working with it when you actually don't need them. All you need is a square matrix - to be more precise, a lower triangular matrix - of ints (in this case).

The idea is the following:

1. Represent the boxes' status via the matrix \$A\$. If the box in the position \$(x, y)\$ is wet then \$A_{x,y} = 1\$, otherwise \$A_{x,y} = 0\$. So, initially, every item in the matrix is \$0\$.
2. When the box in position \$(x, y)\$ leaks you just set all the appropriate cells of \$A\$ to \$1\$ and count the number of cells whose value has changed from \$0\$ to \$1\$.
3. Return the value of the counter for each case.

In pseudo-code (I'm not so practical in C), it becomes something like the following:

int[,] A = new int[7,7];

void initialize_matrix(){
    for(int i = 0; i < 7, i++){
        for(int j = 0; j < 7; j++){
            A[i, j] = 0;
        }
    }
}

int count_wet_boxes(int x, int y){
    int counter = 0;

    for(int i = x; i >= 0; i--){
        for(int j = 0; (i + j) <= (x + y); j++){
            if(A[i, j] == 0){
                A[i, j] = 1;
                counter++;
            }
        }
    }

    return counter;
}

int main(){
    initialize_matrix();

    // the rest of the logic goes here
}

You can also try to ask in the Mathematics SE site if there is a formula you can use.

Let me know if anything's unclear.

It's been a while since I haven't worked in C, so a language-specific review would not be so accurate. The only thing I can say in that aspect is that you should put curly bracket even in single-command body ifs as the code gets more readable and less error prone.

Taking a look at the problem and at your solution I think that you're overcomplicating it. You are allocating memory for each box and working with it when you actually don't need them. All you need is a square matrix - to be more precise, a lower triangular matrix - of ints (in this case).

The idea is the following:

1. Represent the boxes' status via the matrix \$A\$. If the box in the position \$(x, y)\$ is wet then \$A_{x,y} = 1\$, otherwise \$A_{x,y} = 0\$. So, initially, every item in the matrix is \$0\$.
2. When the box in position \$(x, y)\$ leaks you just set all the appropriate cells of \$A\$ to \$1\$ and count the number of cells whose value has changed from \$0\$ to \$1\$.
3. Return the value of the counter for each case.

In pseudo-code (I'm not so practical in C), it becomes something like the following:

int[,] A = new int[7,7];

void initialize_matrix(){
    for(int i = 0; i < 7, i++){
        for(int j = 0; j < 7; j++){
            A[i, j] = 0;
        }
    }
}

int count_wet_boxes(int x, int y){
    int counter = 0;

    for(int i = x; i >= 0; i--){
        for(int j = 0; (i + j) <= (x + y); j++){
            if(A[i, j] == 0){
                A[i, j] = 1;
                counter++;
            }
        }
    }

    return counter;
}

int main(){
    initialize_matrix();

    // the rest of the logic goes here
}

Let me know if anything's unclear.

It's been a while since I haven't worked in C, so a language-specific review would not be so accurate. The only thing I can say in that aspect is that you should put curly bracket even in single-command body ifs as the code gets more readable and less error prone.

Taking a look at the problem and at your solution I think that you're overcomplicating it. You are allocating memory for each box and working with it when you actually don't need them. All you need is a square matrix - to be more precise, a lower triangular matrix - of ints (in this case).

The idea is the following:

1. Represent the boxes' status via the matrix \$A\$. If the box in the position \$(x, y)\$ is wet then \$A_{x,y} = 1\$, otherwise \$A_{x,y} = 0\$. So, initially, every item in the matrix is \$0\$.
2. When the box in position \$(x, y)\$ leaks you just set all the appropriate cells of \$A\$ to \$1\$ and count the number of cells whose value has changed from \$0\$ to \$1\$.
3. Return the value of the counter for each case.

In pseudo-code (I'm not so practical in C), it becomes something like the following:

int[,] A = new int[7,7];

void initialize_matrix(){
    for(int i = 0; i < 7, i++){
        for(int j = 0; j < 7; j++){
            A[i, j] = 0;
        }
    }
}

int count_wet_boxes(int x, int y){
    int counter = 0;

    for(int i = x; i >= 0; i--){
        for(int j = 0; (i + j) <= (x + y); j++){
            if(A[i, j] == 0){
                A[i, j] = 1;
                counter++;
            }
        }
    }

    return counter;
}

int main(){
    initialize_matrix();

    // the rest of the logic goes here
}

You can also try to ask in the Mathematics SE site if there is a formula you can use.

Let me know if anything's unclear.