How to Print Star Pattern in JavaScript in a Very Simple Manner

Build a pyramid of stars using map function in javascript

You can use map, yes, but I think string repeat based on the index is the bigger challenge. As you map, keep track of the index and pad your stars on the left and right accordingly:

function pyramid(h) {  return Array(h).fill('*')    .map((s, i) =>      ' '.repeat(h - i - 1) +      s.repeat(i + 1).split('').join(' ') +      ' '.repeat(h - i - 1))    .join('\n');}
console.log(pyramid(3));console.log(pyramid(4));console.log(pyramid(5));

Printing Simple Diamond Pattern in Python

Since the middle and largest row of stars has 9 stars, you should make n equal to 9. You were able to print out half of the diamond, but now you have to try to make a function that prints a specific number of spaces, then a specific number of stars. So try to develop a pattern with the number of spaces and stars in each row,

Row1: 4 spaces, 1 star, 4 spaces
Row2: 3 spaces, 3 stars, 3 spaces
Row3: 2 spaces, 5 stars, 2 spaces
Row4: 1 space, 7 stars, 1 space
Row5: 0 spaces, 9 stars, 0 spaces
Row6: 1 space, 7 stars, 1 space
Row7: 2 spaces, 5 stars, 2 spaces
Row8: 3 spaces, 3 stars, 3 spaces
Row9: 4 spaces, 1 star, 4 spaces

So what can you deduce? From row 1 to (n+1)/2, the number of spaces decreases as the number of stars increase. So from 1 to 5, the # of stars = (row number * 2) - 1, while # of spaces before stars = 5 - row number.

Now from row (n+1)/2 + 1 to row 9, the number of spaces increase while the number of stars decrease. So from 6 to n, the # of stars = ((n+1 - row number) * 2) - 1, while # of spaces before stars = row number - 5.

From this information, you should be able to make a program that looks like this,

n = 9
print("Pattern 1")
for a1 in range(1, (n+1)//2 + 1): #from row 1 to 5
for a2 in range((n+1)//2 - a1):
print(" ", end = "")
for a3 in range((a1*2)-1):
print("*", end = "")
print()

for a1 in range((n+1)//2 + 1, n + 1): #from row 6 to 9
for a2 in range(a1 - (n+1)//2):
print(" ", end = "")
for a3 in range((n+1 - a1)*2 - 1):
print("*", end = "")
print()

Note that you can replace n with any odd number to create a perfect diamond of that many lines.

How to check if a String contains another String in a case insensitive manner in Java?

Yes, contains is case sensitive. You can use java.util.regex.Pattern with the CASE_INSENSITIVE flag for case insensitive matching:

Pattern.compile(Pattern.quote(wantedStr), Pattern.CASE_INSENSITIVE).matcher(source).find();

EDIT: If s2 contains regex special characters (of which there are many) it's important to quote it first. I've corrected my answer since it is the first one people will see, but vote up Matt Quail's since he pointed this out.



Related Topics



Leave a reply



Submit