String
Simulation
Description
You are given a string s
consisting of lowercase English letters, and an integer k
.
First, convert s
into an integer by replacing each letter with its position in the alphabet (i.e., replace 'a'
with 1
, 'b'
with 2
, ..., 'z'
with 26
). Then, transform the integer by replacing it with the sum of its digits . Repeat the transform operation k
times in total.
For example, if s = "zbax"
and k = 2
, then the resulting integer would be 8
by the following operations:
Convert : "zbax" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124
Transform #1 : 262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 17
Transform #2 : 17 ➝ 1 + 7 ➝ 8
Return the resulting integer after performing the operations described above .
Example 1:
Input: s = "iiii", k = 1
Output: 36
Explanation: The operations are as follows:
- Convert: "iiii" ➝ "(9)(9)(9)(9)" ➝ "9999" ➝ 9999
- Transform #1: 9999 ➝ 9 + 9 + 9 + 9 ➝ 36
Thus the resulting integer is 36.
Example 2:
Input: s = "leetcode", k = 2
Output: 6
Explanation: The operations are as follows:
- Convert: "leetcode" ➝ "(12)(5)(5)(20)(3)(15)(4)(5)" ➝ "12552031545" ➝ 12552031545
- Transform #1: 12552031545 ➝ 1 + 2 + 5 + 5 + 2 + 0 + 3 + 1 + 5 + 4 + 5 ➝ 33
- Transform #2: 33 ➝ 3 + 3 ➝ 6
Thus the resulting integer is 6.
Example 3:
Input: s = "zbax", k = 2
Output: 8
Constraints:
1 <= s.length <= 100
1 <= k <= 10
s
consists of lowercase English letters.
Solutions
Solution 1
Python3 Java C++ Go TypeScript Rust PHP
class Solution :
def getLucky ( self , s : str , k : int ) -> int :
s = '' . join ( str ( ord ( c ) - ord ( 'a' ) + 1 ) for c in s )
for _ in range ( k ):
t = sum ( int ( c ) for c in s )
s = str ( t )
return int ( s )
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17 class Solution {
public int getLucky ( String s , int k ) {
StringBuilder sb = new StringBuilder ();
for ( char c : s . toCharArray ()) {
sb . append ( c - 'a' + 1 );
}
s = sb . toString ();
while ( k -- > 0 ) {
int t = 0 ;
for ( char c : s . toCharArray ()) {
t += c - '0' ;
}
s = String . valueOf ( t );
}
return Integer . parseInt ( s );
}
}
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14 class Solution {
public :
int getLucky ( string s , int k ) {
string t ;
for ( char c : s ) t += to_string ( c - 'a' + 1 );
s = t ;
while ( k -- ) {
int t = 0 ;
for ( char c : s ) t += c - '0' ;
s = to_string ( t );
}
return stoi ( s );
}
};
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17 func getLucky ( s string , k int ) int {
var t strings . Builder
for _ , c := range s {
t . WriteString ( strconv . Itoa ( int ( c - 'a' + 1 )))
}
s = t . String ()
for k > 0 {
k --
t := 0
for _ , c := range s {
t += int ( c - '0' )
}
s = strconv . Itoa ( t )
}
ans , _ := strconv . Atoi ( s )
return ans
}
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14 function getLucky ( s : string , k : number ) : number {
let ans = '' ;
for ( const c of s ) {
ans += c . charCodeAt ( 0 ) - 'a' . charCodeAt ( 0 ) + 1 ;
}
for ( let i = 0 ; i < k ; i ++ ) {
let t = 0 ;
for ( const v of ans ) {
t += Number ( v );
}
ans = ` ${ t } ` ;
}
return Number ( ans );
}
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16 impl Solution {
pub fn get_lucky ( s : String , k : i32 ) -> i32 {
let mut ans = String :: new ();
for c in s . as_bytes () {
ans . push_str ( & ( c - b'a' + 1 ). to_string ());
}
for _ in 0 .. k {
let mut t = 0 ;
for c in ans . as_bytes () {
t += ( c - b'0' ) as i32 ;
}
ans = t . to_string ();
}
ans . parse (). unwrap ()
}
}
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23 class Solution {
/**
* @param String $s
* @param Integer $k
* @return Integer
*/
function getLucky($s, $k) {
$rs = '';
for ($i = 0; $i < strlen($s); $i++) {
$num = ord($s[$i]) - 96;
$rs = $rs . strval($num);
}
while ($k != 0) {
$sum = 0;
for ($j = 0; $j < strlen($rs); $j++) {
$sum += intval($rs[$j]);
}
$rs = strval($sum);
$k--;
}
return intval($rs);
}
}