Diamond with Plus Pattern

Print a diamond pattern formed by asterisks and plus pattern, surrounded by hyphens as empty spaces

Understand the Problem

Problem Statement

The program must accept an integer N as the input. The program must print (2*N)-1 lines of pattern as shown in the Example Input/Output section. The asterisks in the pattern indicate the diamond and plus symbols. The hyphens in the pattern indicate the empty spaces.

Constraints

2 ≤ N ≤ 50
Pattern consists of exactly (2*N)-1 lines
Each line contains only asterisks (*) and hyphens (-)

Examples

Example 1

Input

Output

--------*
------*-*-*
----*---*---*
--*-----*-----*
*-*-*-*-*-*-*-*-*
--*-----*-----*
----*---*---*
------*-*-*
--------*

Explanation

For N=5, the pattern has 9 lines. The middle line (5th) contains 17 characters with alternating '*' and '-' starting with '*'. The upper and lower halves form a diamond shape with asterisks at the edges and center column.

Example 2

Input

Output

----*
--*-*-*
*-*-*-*-*
--*-*-*
----*

Explanation

For N=3, the pattern has 5 lines. The middle line has 9 characters with alternating '*' and '-' pattern. The diamond shape is smaller but follows the same logic.

Example 3

Input

Output

----------*
--------*-*-*
------*---*---*
----*-----*-----*
--*-------*-------*
*-*-*-*-*-*-*-*-*-*-*
--*-------*-------*
----*-----*-----*
------*---*---*
--------*-*-*
----------*

Explanation

For N=6, the pattern has 11 lines. The middle line has 21 characters with the full horizontal bar pattern. The diamond extends further with more spacing between asterisks in the upper and lower halves.

Solution

#include <stdio.h>

int main() {
    int N;
    scanf("%d", &N);
    
    // Upper half including middle line
    for (int i = 0; i < N; i++) {
        // Print leading hyphens
        for (int j = 0; j < 2 * (N - 1 - i); j++) {
            printf("-");
        }
        
        if (i == N - 1) {
            // Middle line: horizontal bar of plus
            for (int k = 0; k < 2 * N - 1; k++) {
                printf("*");
                if (k < 2 * N - 2) printf("-");
            }
        } else {
            // Diamond pattern with vertical bar
            printf("*");
            // Left inner part
            for (int j = 0; j < 2 * i - 1; j++) {
                printf("-");
            }
            if (i > 0) printf("*");
            // Right inner part
            for (int j = 0; j < 2 * i - 1; j++) {
                printf("-");
            }
            if (i > 0) printf("*");
        }
        printf("\n");
    }
    
    // Lower half
    for (int i = N - 2; i >= 0; i--) {
        // Print leading hyphens
        for (int j = 0; j < 2 * (N - 1 - i); j++) {
            printf("-");
        }
        
        if (i == 0) {
            printf("*");
        } else {
            printf("*");
            // Left inner part
            for (int j = 0; j < 2 * i - 1; j++) {
                printf("-");
            }
            printf("*");
            // Right inner part
            for (int j = 0; j < 2 * i - 1; j++) {
                printf("-");
            }
            printf("*");
        }
        printf("\n");
    }
    
    return 0;
}

Time:O(N²) - We have nested loops where the outer loop runs (2*N)-1 times and inner loops run up to O(N) times each

Space:O(1) - Only using a constant amount of extra space for variables

Approach:

The C solution follows the same logic as described in the approach. It uses nested loops to:

Print leading hyphens for each line
Handle the middle line specially with the horizontal bar pattern
Print the diamond pattern for other lines with vertical bar
Mirror the upper half for the lower half

The key is managing the spacing correctly with the formula 2*(N-1-i) for leading hyphens and 2*i-1 for inner hyphens.

Visual Explanation

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