Most Frequent Digit Sum
Description
A step(x)
operation works like this: it changes a number x
into x - s(x)
, where s(x)
is the sum of x
’s digits. You like applying functions to numbers, so given the number n, you decide to build a decreasing sequence of numbers: n
, step(n)
, step(step(n))
, etc., with 0
as the last element.
Building a single sequence isn’t enough for you, so you replace all elements of the sequence with the sums of their digits (s(x)
). Now you’re curious as to which number appears in the new sequence most often. If there are several answers, return the maximal one.
Example
-
For
n = 88
, the output should bemostFrequentDigitSum(n) = 9
.- Here is the first sequence you built:
88, 72, 63, 54, 45, 36, 27, 18, 9, 0
; - And here is
s(x)
for each of its elements:16, 9, 9, 9, 9, 9, 9, 9, 9, 0
.
As you can see, the most frequent number in the second sequence is
9
. - Here is the first sequence you built:
-
For
n = 8
, the output should bemostFrequentDigitSum(n) = 8
.- At first you built the following sequence:
8, 0
s(x)
for each of its elements is:8, 0
- At first you built the following sequence:
As you can see, the answer is 8
(it appears as often as 0
, but is greater than it).
Input/Output
-
[execution time limit] 4 seconds (js)
-
[input] integer n
Guaranteed constraints:
1 ≤ n ≤ 105
. -
[output] integer
- The most frequent number in the sequence
s(n)
,s(step(n))
,s(step(step(n)))
, etc.
- The most frequent number in the sequence
[JavaScript (ES6)] Syntax Tips
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Solution
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function mostFrequentDigitSum(n) {
var count = [1];
var sx;
while (n) {
sx = String(n)
.split("")
.map(Number)
.reduce((a, b) => a + b, 0);
count[sx] = -~count[sx];
n -= sx;
}
var max = 1;
var maxn = 0;
for (var i in count) {
if (count[i] >= max) {
max = count[i];
maxn = i;
}
}
return Number(maxn);
}