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Wipro NLTH Coding Sample Question- 2
PrepSter Wipro NLTH Coding Question – 2
Print a given matrix in spiral form
Ques: Given a 2D array, print it in spiral form. See the following examples.
NOTE:- Please comment down the code in other languages as well below .
Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Output: 1 2 3 4 8 12 16 15 14 13 9 5 6 7 11 10 Input: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Output: 1 2 3 4 5 6 12 18 17 16 15 14 13 7 8 9 10 11
C
#include <stdio.h> #define R 3 #define C 6 void spiralPrint (int m, int n, int a[R][C]) { int i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ while (k < m && l < n) { /* Print the first row from the remaining rows */ for (i = l; i < n; ++i) { printf ("%d ", a[k][i]); } k++; /* Print the last column from the remaining columns */ for (i = k; i < m; ++i) { printf ("%d ", a[i][n - 1]); } n--; /* Print the last row from the remaining rows */ if (k < m) { for (i = n - 1; i >= l; --i) { printf ("%d ", a[m - 1][i]); } m--; } /* Print the first column from the remaining columns */ if (l < n) { for (i = m - 1; i >= k; --i) { printf ("%d ", a[i][l]); } l++; } } } /* Driver program to test above functions */ int main () { int a[R][C] = { {1, 2, 3, 4, 5, 6}, {7, 8, 9, 10, 11, 12}, {13, 14, 15, 16, 17, 18} }; spiralPrint (R, C, a); return 0; }
C++
#include <bits/stdc++.h> using namespace std; #define R 3 #define C 6 void spiralPrint(int m, int n, int a[R][C]) { int i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ while (k < m && l < n) { /* Print the first row from the remaining rows */ for (i = l; i < n; ++i) { cout << a[k][i] << " "; } k++; /* Print the last column from the remaining columns */ for (i = k; i < m; ++i) { cout << a[i][n - 1] << " "; } n--; /* Print the last row from the remaining rows */ if (k < m) { for (i = n - 1; i >= l; --i) { cout << a[m - 1][i] << " "; } m--; } /* Print the first column from the remaining columns */ if (l < n) { for (i = m - 1; i >= k; --i) { cout << a[i][l] << " "; } l++; } } } /* Driver program to test above functions */ int main() { int a[R][C] = { { 1, 2, 3, 4, 5, 6 }, { 7, 8, 9, 10, 11, 12 }, { 13, 14, 15, 16, 17, 18 } }; spiralPrint(R, C, a); return 0; }
JAVA
// Java program to print a given matrix in spiral form import java.io.*; class PrepInsta { // Function print matrix in spiral form static void spiralPrint (int m, int n, int a[][]) { int i, k = 0, l = 0; /* k - starting row index m - ending row index l - starting column index n - ending column index i - iterator */ while (k < m && l < n) { // Print the first row from the remaining rows for (i = l; i < n; ++i) { System.out.print (a[k][i] + " "); } k++; // Print the last column from the remaining columns for (i = k; i < m; ++i) { System.out.print (a[i][n - 1] + " "); } n--; // Print the last row from the remaining rows */ if (k < m) { for (i = n - 1; i >= l; --i) { System.out.print (a[m - 1][i] + " "); } m--; } // Print the first column from the remaining columns */ if (l < n) { for (i = m - 1; i >= k; --i) { System.out.print (a[i][l] + " "); } l++; } } } // driver program public static void main (String[]args) { int R = 3; int C = 6; int a[][] = { {1, 2, 3, 4, 5, 6}, {7, 8, 9, 10, 11, 12}, {13, 14, 15, 16, 17, 18} }; spiralPrint (R, C, a); } }
Python
#Python3 program to print #given matrix in spiral form def spiralPrint (m, n, a): k = 0; l = 0 '' ' k - starting row index m - ending row index l - starting column index n - ending column index i - iterator ' '' while (k < m and l < n) : #Print the first row from #the remaining rows for i in range (l, n): print (a[k][i], end = " ") k += 1 #Print the last column from #the remaining columns for i in range (k, m): print (a[i][n - 1], end = " ") n -= 1 #Print the last row from #the remaining rows if (k < m) : for i in range (n - 1, (l - 1), -1): print (a[m - 1][i], end = " ") m -= 1 #Print the first column from #the remaining columns if (l < n) : for i in range (m - 1, k - 1, -1): print (a[i][l], end = " ") l += 1 #Driver Code a =[[1, 2, 3, 4, 5, 6], [7, 8, 9, 10, 11, 12],[13, 14, 15, 16, 17, 18]] R = 3; C = 6 spiralPrint (R, C, a)