The problem is: Print out how much odd numbers there are in a given number.
function oddCount(n) {
var odd = [];
for (i = 0; i < n; i++) {
if (i % 2 == 1) {
odd.push([i]);
}
}
return odd.length;
}
console.log(oddCount(8));As we can see, it works properly, however, on codewars, it wants me to optimize it to run faster. Can someone show me how so I can learn it quickly please.
Neither "ceil" or "floor" is a correct answer as a one liner. "ceil" will make the division Base 1, "floor" will make the division Base 0. So both could be used in an implementation, but the "polarity" of n matters.
It's necessary to check whether the input number is odd or even.
function oddCount(n) {
// odd / even check
if (n % 2 == 0) {
// its even, we can divide by 2
return n / 2
}
else {
// n is odd, so we must include n as a count+1 itself
return ((n - 1) / 2) + 1
}
}
// Disclaimer: Are negative numbers odd or even? In this code they
// apparently aren't handled. So the set of numbers are integers from
// 0 to +Infinity
// Test cases:
console.log( oddCount(8) ); // 4
console.log( oddCount(9) ); // 5But this code "breaks" if n itself is 0 or less. So we need to fix it:
Right after we say function oddCount(n) {, put:
if (n < 1) return 0;
All worries solved. But still debate on whether 0 is odd or even, and whether -1 is odd and -2 is even.
function oddCount(n) {
var odd = [];
for (i = 0; i < n; i++) {
if (i & 0x1 == 1) {
odd.push([i]);
}
}
return odd.length;
}
console.log(oddCount(8));
or
function oddCount(n) {
return (n - (n & 0x01)) / 2;
}
console.log(oddCount(8));
If the number is an even number, there's n/2 odd numbers
Eg if n is 6
*1*,2,*3*,4,*5*,6
If it is an odd number, there's n/2+1 odd numbers. Because n-1 would be even
Eg if n is 5
*1*,2,*3*,4, + *5*
So basically
if (n%2==0) return n/2
else return (n-1)/2+1
The for loops aren't needed Also like the others pointed out, ceiling is a more concise way to do it
return Math.ceil(n/2)