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@GreggSetzer
Last active August 3, 2018 16:42
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Javascript Interview Question: Pyramid
/*
This function builds a pyramid using ES6 repeat function available on the String prototype.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/repeat
colSize: is the width of the grid in columns. Also think of it as the width of the last row in the pyramid.
level: This represents the width of the pyramid level at each iteration.
numSpaces: This represents the number of spaces to use on both sides of the pyramid.
spaces: This variable is used to cache the result of String#repeat(), no need to do the same work twice.
*/
function pyramid(n) {
const colSize = 2 * n - 1;
for (let i = 1; i <= n; i++) {
const level = 2 * i - 1;
const numSpaces = (colSize - level) / 2;
const spaces = ' '.repeat(numSpaces);
console.log(`'${spaces + '#'.repeat(level) + spaces}'`);
}
}
//Yet another possible solution. This approach seems strange to me, requiring the console.log to print
//the output, and only afterwards, would we update the numSteps counter. It seems more natural for me to
//to see the console.log as the last step in the for loop.
function pyramid(n) {
let numSpaces = n - 1;
let numSteps = 1;
for (let i = 0; i < n; i++) {
const spaces = ' '.repeat(numSpaces);
const steps = '#'.repeat(numSteps)
numSpaces -= 1;
console.log(`'${spaces + "#".repeat(numSteps) + spaces}'`);
numSteps += 2;
}
}
pyramid(5);
/*
Draw this pyramid.
' # '
' ### '
' ##### '
' ####### '
'#########'
Notes:
Step 1: Calculate the number of characters to print
for each row in the pyramid. Use the formula 2n - 1
to caculate this value. The examples below show the
pattern.
const colSize = (2 * n) - 1; //where n is provided as an arg.
Formula: 2n - 1
(2 * 1) - 1 = 1
'#'
(2 * 2) - 1 = 3
' # '
'###'
(2 * 3) - 1 = 5
' # '
' ### '
'#####'
(2 * 4) - 1 = 7
' # '
' ### '
' ##### '
'#######'
(2 * 5) - 1 = 9
' # '
' ### '
' ##### '
' ####### '
'#########'
Step 2: Calculate the mid point, or the center of
the pyramid. Use the formula, (2n - 1) / 2 to
find the mid point. We use Math.floor() to round
down to whole integer for results that contain a
fraction.
const colSize = (2 * n) - 1;
const midPoint = Math.floor(colSize / 2);
Step 3: Iterate both rows and cols, check if we are
within the bounds of the colSize +/- the offset.
If so, print a '#' character. Otherwise, print a
space character.
*/
function pyramid(n) {
const colSize = 2 * n - 1;
const midPoint = Math.floor(colSize / 2);
for (let row = 0; row < n; row++) {
let step = '';
const leftSide = midPoint - row;
const rightSide = midPoint + row;
for (let col = 0; col < colSize; col++) {
step += (col >= leftSide && col <= rightSide) ? '#' : ' ';
}
console.log(step);
}
}
//Try it
pyramid(5);
//Example 2: Create a pyramid using recursion.
function pyramid(n, row = 0, step = '') {
//Base case
if (n === row) {
return;
}
//Calculate the column size and mid point.
const colSize = 2 * n - 1;
const midPoint = Math.floor(colSize / 2);
//Are we at the end of the row?
if (step.length === colSize) {
console.log(step);
return pyramid(n, row + 1);
}
//Calculate left and right boundaries.
const leftSide = midPoint - row;
const rightSide = midPoint + row;
//Append character.
step += (step.length >= leftSide && step.length <= rightSide) ? '#' : ' ';
pyramid(n, row, step);
}
pyramid(5);
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