Is there a more efficient method than storing the numbers in an array and using .sort() to arrange them?
Is there a more efficient method than storing the numbers in an array and using .sort() to arrange them?
If you want to optimize performance and avoid unnecessary string conversions, consider implementing counting sort for sorting integers based on their digits. This approach takes advantage of the fact that there are only 10 possible digits (0-9) in a number.
function countingSortDigits(n) {
const count = Array(10).fill(0);
for (; n; n = (n - n % 10) / 10) {
count[n % 10]++;
}
for (let digit = 0; digit < 10; digit++) {
for (let j = count[digit]; j; j--) {
n = n*10 + digit;
}
}
return n;
}
A comparison of execution times is provided below:
function toStringSort(n) { // Initial implementation
return parseInt(n.toString().split("").sort().join(""));
}
function countingSort(n) {
const count = Array(10).fill(0);
for (; n; n = (n - n % 10) / 10) {
count[n % 10]++;
}
for (let digit = 0; digit < 10; digit++) {
for (let i = count[digit]; i; i--) {
n = n*10 + digit;
}
}
return n;
}
function timeIt(f, repeat) {
const start = performance.now();
while (repeat--) f();
return performance.now() - start;
}
const timing = { toStringSort: 0, countingSort: 0 };
// Generate random integers and measure execution time
for (let i = 0; i < 1000; i++) {
let n = Math.floor(Math.random() * 1e15);
timing.toStringSort += timeIt(toStringSort.bind(null, n), 1000);
timing.countingSort += timeIt(countingSort.bind(null, n), 1000);
}
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