# HackerRank: Repeated String

Tags:

## Repeated String

• python

• 先說我這解法會出包 XDD
• memory 爆炸!
``````# Complete the repeatedString function below.
def repeatedString(s, n):
# repeate the s 到最小長度 len(s) 剛剛 超過 n 就好
repeate_s = s * (int(n/len(s) + 1))
# trim 到剛好的長度
repeate_s = repeate_s[:n]
result = repeate_s.count('a')
return result

``````
• 多看、多學、多聽、多謙卑

• 如果沒記錯的話，之前掃 Angular 2019 年會活動照片
有看到一個很像 workshop brainstorm 的 海報 裡面有一句話我超級認同&共鳴!

• 多看別人的 coce

• 非本科系出生的兒~ 就是要多學、多看、多聽、多謙卑阿~~~

• 好扯多了~~ XDD

• 知識分享好棒棒

### 全新學習後的成果

• pyhton
``````# Complete the repeatedString function below.
def repeatedString(s, n):
intact_s = n // len(s)
remain_s = n % len(s)
count_a = s.count('a')
if remain_s != 0:
return intact_s * count_a + (s[:remain_s]).count('a')
else:
return intact_s * count_a
``````
• javascript

``````// Complete the repeatedString function below.
function repeatedString(s, n) {
let intact_s = Math.floor(n/(s.length))
let remain_s = n % (s.length)
let count_a = s.match(/a/g).length
if (remain_s !== 0) {
let left_s = s.substring(0, remain_s)
let left_a_count = (left_s.match(/a/g) || []).length
return (intact_s * count_a + left_a_count)
} else {
return intact_s * count_a
}
}
``````

Tags:

Updated: