Asterisks – X Inner Layers

Replace characters in the X inner layers of an NxN matrix with asterisks

Understand the Problem

Problem Statement

Given an NxN character matrix and an integer X, replace all characters present in the X inner layers of the matrix with asterisks (*) and print the modified matrix.

A layer is defined as a concentric square boundary within the matrix. The outermost layer is layer 1, the next inner boundary is layer 2, and so on.

Constraints

2 ≤ N ≤ 50
1 ≤ X ≤ (N+1)/2
Each character in the matrix is a single printable character
The matrix is guaranteed to be square (NxN)

Examples

Example 1

Input

8
b z r p h u w f
c p g b p u a j
n h f a l b i g
j k x n n y k p
k v j f u b l r
z x m a t z x j
s o c u r q d j
x n t v a g y h
2

Output

b z r p h u w f
c p g b p u a j
n h * * * * i g
j k * * * * k p
k v * * * * l r
z x * * * * x j
s o c u r q d j
x n t v a g y h

Explanation

With X=2, the 2 inner layers form a 4x4 region in the center (rows 2-5, columns 2-5). All characters in this region are replaced with asterisks, while the outer 2 layers remain unchanged.

Example 2

Input

5
H Q G I B
W A U T W
F U O W C
U G R E W
R P K C M
1

Output

H Q G I B
W A U T W
F U * W C
U G R E W
R P K C M

Explanation

With X=1, only the innermost layer (center element) is replaced. In a 5x5 matrix, this is just the element at position [2][2], which becomes '*'.

Solution

#include <stdio.h>

int main() {
    int n;
    scanf("%d", &n);
    
    char matrix[n][n];
    // Read the matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            scanf(" %c", &matrix[i][j]);
        }
    }
    
    int x;
    scanf("%d", &x);
    
    // Calculate boundaries for the inner region
    int outerLayers = (n / 2) - x + (n % 2);
    int start = outerLayers;
    int end = n - outerLayers;
    
    // Replace characters in inner region with asterisks
    for (int i = start; i < end; i++) {
        for (int j = start; j < end; j++) {
            matrix[i][j] = '*';
        }
    }
    
    // Print the modified matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            printf("%c", matrix[i][j]);
            if (j < n - 1) printf(" ");
        }
        printf("\n");
    }
    
    return 0;
}

Time:O(N²) - We need to read the input matrix (O(N²)), then potentially modify up to O(N²) elements, and finally print the output (O(N²)). The dominant factor is the matrix operations.

Space:O(N²) - We store the N×N character matrix in memory. The input size itself is O(N²), making this the space complexity.

Approach:

C Solution Explanation:

Input Reading: Read matrix size N, then read N×N characters using nested loops and scanf
Boundary Calculation: Calculate the start and end indices of the inner region that needs modification
Matrix Modification: Iterate through the inner rectangular region and replace all characters with '*'
Output: Print the modified matrix with proper spacing between characters

The algorithm efficiently identifies the inner layers by calculating the preserved outer layers and working inward.

Visual Explanation

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