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C++ Program to Print Spiral Pattern of Numbers
Displaying numbers in different formats comes under fundamental coding problems to learn different concepts of coding like conditional statements and looping statements. There are different programs where we have printed special characters like stars to make triangles or squares. In this article, we will print numbers in spiral form like a square in C++.
We will take the number of lines n as input, then starting from the top-left corner we will move towards the right, then down, then left and then up, then again towards the right, and so on.
Spiral Pattern with Numbers
1 2 3 4 5 6 7 24 25 26 27 28 29 8 23 40 41 42 43 30 9 22 39 48 49 44 31 10 21 38 47 46 45 32 11 20 37 36 35 34 33 12 19 18 17 16 15 14 13
To solve this, we will use a 2D matrix of size n x n, In this example, we have taken n = 7. Then fill the matrix from the top-left corner in a spiral fashion. And at the end print the entire matrix. Here, we are printing from 1 to 7 on the first row, then the flow is changing its direction, moving towards the bottom up to 13, then again towards left up till 19, and finally going upward up till 24, then right again, and so on. Let us see the algorithm for a better understanding.
Algorithm
- Take input s as the number of lines
- create one s x s matrix and initialize them with 0
- num := 1
- initialize i, j, m with 0
- initialize n := s - 1, p := 0 and q := s - 1
- while num is not exceeding s * s, do
- for j ranging from p to q, do
- mat[ m, j ] := num
- num := num + 1
- end for
- m := m + 1
- for i ranging from m to n, do
- mat[ i, q ] := num
- num := num + 1
- end for
- q := q - 1
- for j ranging from q to p, decreasing j by 1, do
- mat[ n, j ] := num
- num := num + 1
- end for
- n := n - 1
- for i ranging from n to m, decrease i by 1, do
- mat[ i, p ] := num
- num := num + 1
- end for
- p := p + 1
- for j ranging from p to q, do
- end while
- for i ranging from 0 to s - 1, do
- for j ranging from 0 to s - 1, do
- display mat[ i, j ]
- end for
- move the cursor to the next line
- for j ranging from 0 to s - 1, do
- end for
Example
#include <iostream>
using namespace std;
void solve( int s ){
int mat[ s ][ s ] = {0};
int i, j, m, n, p, q, num;
num = 1; // start count from 1
i = 0;
j = 0;
m = 0; // row index lower limit
n = s - 1; // row index upper limit
p = 0; // column index lower limit
q = s - 1; // column index upper limit
while ( num <= s * s ) {
// place numbers horizontally left to right
for ( j = p; j <= q; j++ ) {
mat[ m ][ j ] = num;
num = num + 1;
}
m = m + 1;
// fill vertically from top to bottom
for ( i = m; i <= n; i++ ) {
mat[ i ][ q ] = num;
num = num + 1;
}
q = q - 1;
// fill horizontally from right to left
for ( j = q; j >= p; j-- ) {
mat[ n ][ j ] = num;
num = num + 1;
}
n = n - 1;
// fill vertically from bottom to top
for ( i = n; i >= m; i-- ) {
mat[ i ][ p ] = num;
num++;
}
p = p + 1;
}
// display the mat
for ( i = 0; i < s; i++ ) {
for ( j = 0; j < s; j++ ) {
printf("%d\t", mat[i][j]);
}
printf("\n");
}
}
int main(){
int n = 5;
cout << "Spiral numbers for " << n << " lines." << endl;
solve( n );
}
Output
Spiral numbers for 5 lines. 1 2 3 4 5 16 17 18 19 6 15 24 25 20 7 14 23 22 21 8 13 12 11 10 9
Output (With n = 12)
Spiral numbers for 12 lines. 1 2 3 4 5 6 7 8 9 10 11 12 44 45 46 47 48 49 50 51 52 53 54 13 43 80 81 82 83 84 85 86 87 88 55 14 42 79 108 109 110 111 112 113 114 89 56 15 41 78 107 128 129 130 131 132 115 90 57 16 40 77 106 127 140 141 142 133 116 91 58 17 39 76 105 126 139 144 143 134 117 92 59 18 38 75 104 125 138 137 136 135 118 93 60 19 37 74 103 124 123 122 121 120 119 94 61 20 36 73 102 101 100 99 98 97 96 95 62 21 35 72 71 70 69 68 67 66 65 64 63 22 34 33 32 31 30 29 28 27 26 25 24 23
Conclusion
Displaying number patterns is a fairly common problem while learning programming languages. In this article, we have seen how to display numbers in a square where elements are printed in the spiral form in C++. Starting from the top left corner we move towards right then at the end of n column, we go down, then at the end of n row, go left, and then after reaching the first row go up till 2nd row, then do the same again and again until completing the entire square. Unlike other number pattern problems, it requires a 2D array to solve this problem in an efficient manner.
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