We always use nested loops for printing the patterns.
- For the outer loop, we count the number of lines/rows and loop for them.
- Next, for the inner loop, we focus on the number of columns and somehow connect them to the rows by forming a logic such that for each row we get the required number of columns to be printed.
- We print the character inside the inner loop.
- Observe symmetry in the pattern or check if a pattern is a combination of two or more similar patterns.
To analyze the time complexity of this program, let's break it down step by step:
- Outer Loop (i): The outer loop runs N times, where N is the input parameter.
- First Inner Loop (j): This loop runs from
1 to N-i, where i is the current value of the outer loop variable. - Second Inner Loop (j): This loop runs from
1 to 2*i, where i is the current value of the outer loop variable. - Third Inner Loop (j): Similar to the first inner loop, it runs from
1 to N-iiterations.
Now, let's calculate the total number of operations across all iterations of the outer loop:
- The first inner loop contributes a total of
N-1 + 3 + 2...0iterations, which is equivalent to$N*(N-1)/2$ . - The second inner loop contributes a total of
1 + 3 + 5 + ... + 2*N-1iterations, which is equivalent to$N^2$ . - The third inner loop contributes a total of
N-1 + 3 + 2...0iterations, which is equivalent to$N*(N-1)/2$ .
The total operations can be represented as
Therefore, the overall time complexity of the code, using the frequency count method, is
where N is the number of rows/lines (horizontally).
- The additional space used by the code is mainly for temporary variables and the input values.
- The memory used by the nested loops is constant and does not depend on the input size N.
- So, the space complexity of the code is
$O(1)$ , constant space complexity.