Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.
You are given a string s consisting of lowercase English letters, and an integer k. Your task is to convert the string into an integer by a special process, and then transform it by summing its digits repeatedly k times. More specifically, perform the following steps:
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).k times in total.For example, if s = "zbax" and k = 2, then the resulting integer would be 8 by the following operations:
"zbax" ➝ "(26)(2)(1)(24)" ➝ "262124" ➝ 262124262124 ➝ 2 + 6 + 2 + 1 + 2 + 4 ➝ 1717 ➝ 1 + 7 ➝ 8Return 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 <= 1001 <= k <= 10s consists of lowercase English letters.Problem summary: You are given a string s consisting of lowercase English letters, and an integer k. Your task is to convert the string into an integer by a special process, and then transform it by summing its digits repeatedly k times. More specifically, perform the following steps: 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). Transform the integer by replacing it with the sum of its digits. Repeat the transform operation (step 2) 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: General problem-solving
"iiii" 1
"leetcode" 2
"zbax" 2
happy-number)add-digits)count-integers-with-even-digit-sum)minimum-element-after-replacement-with-digit-sum)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1945: Sum of Digits of String After Convert
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);
}
}
// Accepted solution for LeetCode #1945: Sum of Digits of String After Convert
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
}
# Accepted solution for LeetCode #1945: Sum of Digits of String After Convert
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)
// Accepted solution for LeetCode #1945: Sum of Digits of String After Convert
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()
}
}
// Accepted solution for LeetCode #1945: Sum of Digits of String After Convert
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);
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.