Given a square matrix mat[][] and a positive integer K. The task is to print the Kth boundary of mat[][].
Examples:
Input: mat[][] = {{1, 2, 3, 4, 5}, K = 1
{6, 7, 8, 9, 10}
{11, 12, 13, 14, 15}
{16, 17, 18, 19, 20}
{21, 22, 23, 24, 25}}
Output: 1 2 3 4 5
6 10
11 15
16 20
21 22 23 24 25
Explanation: The first boundary of mat[][] is above.Input: mat[][] = {{1, 2, 3}, K = 2
{4, 5, 6}
{7, 8, 9}}
Output: 5
Approach: This problem is implementation-based. Traverse the matrix and check for every element if that element lies on the Kth boundary or not. If yes then print the element else print space character. Follow the steps below to solve the given problem.
- for i in from 0 to N
- for j in from 0 to N
- if((i == K – 1 or i == N – K) and (j >= K – 1 and j <= N – K))
- print mat[i][j]
- else if (j == K – 1 or j == N – K) and (i >= K – 1 and i <= N – K):
- print mat[i][j]
- if((i == K – 1 or i == N – K) and (j >= K – 1 and j <= N – K))
- for j in from 0 to N
- This will give the required Kth border of mat[][]
Below is the implementation of the above approach.
C++
// C++ Program to implement// the above approach #include <bits/stdc++.h>using namespace std;// Function to print Kth border of a matrixvoid printKthBorder(vector<vector<int>> mat, int N, int K){ for (int i = 0; i < N; i++) { cout << endl; for (int j = 0; j < N; j++) { // To keep track of which element to skip int flag = 0; if ((i == K - 1 || i == N - K) && (j >= K - 1 && j <= N - K)) { // Print the element cout << mat[i][j] << " "; flag = 1; } else if ((j == K - 1 || j == N - K) && (i >= K - 1 && i <= N - K)) { // Print the element cout << mat[i][j] << " "; flag = 1; } if (flag == 0) cout << " "; } }}// Driver codeint main() { int N = 5; int K = 1; vector<vector<int>> mat = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}}; printKthBorder(mat, N, K);}// This code is contributed by Samim Hossain Mondal. |
Java
// Java Program to implement// the above approach import java.util.*;public class GFG{// Function to print Kth border of a matrixstatic void printKthBorder(int [][]mat, int N, int K){ for (int i = 0; i < N; i++) { System.out.println(); for (int j = 0; j < N; j++) { // To keep track of which element to skip int flag = 0; if ((i == K - 1 || i == N - K) && (j >= K - 1 && j <= N - K)) { // Print the element System.out.print(mat[i][j] + " "); flag = 1; } else if ((j == K - 1 || j == N - K) && (i >= K - 1 && i <= N - K)) { // Print the element System.out.print(mat[i][j] + " "); flag = 1; } if (flag == 0) System.out.print(" "); } }}// Driver codepublic static void main(String args[]) { int N = 5; int K = 1; int [][]mat = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}}; printKthBorder(mat, N, K);}}// This code is contributed by Samim Hossain Mondal. |
Python3
# Python program for above approach# Function to print Kth border of a matrixdef printKthBorder(mat, N, K): for i in range(N): print() for j in range(N): # To keep track of which element to skip flag = 0 if((i == K-1 or i == N-K) \ and (j >= K-1 and j <= N-K)): # Print the element print(mat[i][j], end =" ") flag = 1 elif (j == K-1 or j == N-K) \ and (i >= K-1 and i <= N-K): # Print the element print(mat[i][j], end =" ") flag = 1 if flag == 0: print(end =" ")# Driver codeN = 5K = 1mat = [[1, 2, 3, 4, 5], \ [6, 7, 8, 9, 10], \ [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], \ [21, 22, 23, 24, 25]]printKthBorder(mat, N, K) |
C#
// C# Program to implement// the above approach using System;class GFG{ // Function to print Kth border of a matrixstatic void printKthBorder(int [,]mat, int N, int K){ for (int i = 0; i < N; i++) { Console.WriteLine(); for (int j = 0; j < N; j++) { // To keep track of which element to skip int flag = 0; if ((i == K - 1 || i == N - K) && (j >= K - 1 && j <= N - K)) { // Print the element Console.Write(mat[i, j] + " "); flag = 1; } else if ((j == K - 1 || j == N - K) && (i >= K - 1 && i <= N - K)) { // Print the element Console.Write(mat[i, j] + " "); flag = 1; } if (flag == 0) Console.Write(" "); } }}// Driver codepublic static void Main() { int N = 5; int K = 1; int [,]mat = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}}; printKthBorder(mat, N, K);}}// This code is contributed by Samim Hossain Mondal. |
Javascript
<script> // JavaScript Program to implement // the above approach // Function to print Kth border of a matrix function printKthBorder(mat, N, K) { for (let i = 0; i < N; i++) { document.write('<br>') for (let j = 0; j < N; j++) { // To keep track of which element to skip flag = 0 if ((i == K - 1 || i == N - K) && (j >= K - 1 && j <= N - K)) { // Print the element document.write(mat[i][j] + " ") flag = 1 } else if ((j == K - 1 || j == N - K) && (i >= K - 1 && i <= N - K)) { // Print the element document.write(mat[i][j] + " ") flag = 1 } if (flag == 0) document.write(" "); } } } // Driver code N = 5 K = 1 mat = [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20], [21, 22, 23, 24, 25]] printKthBorder(mat, N, K) // This code is contributed by Potta Lokesh </script> |
Output
1 2 3 4 5 6 10 11 15 16 20 21 22 23 24 25
Time Complexity: O(N^2)
Space Complexity: O(1)
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