Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
Alice and Bob are playing a game. Initially, Alice has a string word = "a".
You are given a positive integer k. You are also given an integer array operations, where operations[i] represents the type of the ith operation.
Now Bob will ask Alice to perform all operations in sequence:
operations[i] == 0, append a copy of word to itself.operations[i] == 1, generate a new string by changing each character in word to its next character in the English alphabet, and append it to the original word. For example, performing the operation on "c" generates "cd" and performing the operation on "zb" generates "zbac".Return the value of the kth character in word after performing all the operations.
Note that the character 'z' can be changed to 'a' in the second type of operation.
Example 1:
Input: k = 5, operations = [0,0,0]
Output: "a"
Explanation:
Initially, word == "a". Alice performs the three operations as follows:
"a" to "a", word becomes "aa"."aa" to "aa", word becomes "aaaa"."aaaa" to "aaaa", word becomes "aaaaaaaa".Example 2:
Input: k = 10, operations = [0,1,0,1]
Output: "b"
Explanation:
Initially, word == "a". Alice performs the four operations as follows:
"a" to "a", word becomes "aa"."bb" to "aa", word becomes "aabb"."aabb" to "aabb", word becomes "aabbaabb"."bbccbbcc" to "aabbaabb", word becomes "aabbaabbbbccbbcc".Constraints:
1 <= k <= 10141 <= operations.length <= 100operations[i] is either 0 or 1.word has at least k characters after all operations.Problem summary: Alice and Bob are playing a game. Initially, Alice has a string word = "a". You are given a positive integer k. You are also given an integer array operations, where operations[i] represents the type of the ith operation. Now Bob will ask Alice to perform all operations in sequence: If operations[i] == 0, append a copy of word to itself. If operations[i] == 1, generate a new string by changing each character in word to its next character in the English alphabet, and append it to the original word. For example, performing the operation on "c" generates "cd" and performing the operation on "zb" generates "zbac". Return the value of the kth character in word after performing all the operations. Note that the character 'z' can be changed to 'a' in the second type of operation.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Bit Manipulation
5 [0,0,0]
10 [0,1,0,1]
shifting-letters)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3307: Find the K-th Character in String Game II
class Solution {
public char kthCharacter(long k, int[] operations) {
long n = 1;
int i = 0;
while (n < k) {
n *= 2;
++i;
}
int d = 0;
while (n > 1) {
if (k > n / 2) {
k -= n / 2;
d += operations[i - 1];
}
n /= 2;
--i;
}
return (char) ('a' + (d % 26));
}
}
// Accepted solution for LeetCode #3307: Find the K-th Character in String Game II
func kthCharacter(k int64, operations []int) byte {
n := int64(1)
i := 0
for n < k {
n *= 2
i++
}
d := 0
for n > 1 {
if k > n/2 {
k -= n / 2
d += operations[i-1]
}
n /= 2
i--
}
return byte('a' + (d % 26))
}
# Accepted solution for LeetCode #3307: Find the K-th Character in String Game II
class Solution:
def kthCharacter(self, k: int, operations: List[int]) -> str:
n, i = 1, 0
while n < k:
n *= 2
i += 1
d = 0
while n > 1:
if k > n // 2:
k -= n // 2
d += operations[i - 1]
n //= 2
i -= 1
return chr(d % 26 + ord("a"))
// Accepted solution for LeetCode #3307: Find the K-th Character in String Game II
impl Solution {
pub fn kth_character(mut k: i64, operations: Vec<i32>) -> char {
let mut n = 1i64;
let mut i = 0;
while n < k {
n *= 2;
i += 1;
}
let mut d = 0;
while n > 1 {
if k > n / 2 {
k -= n / 2;
d += operations[i - 1] as i64;
}
n /= 2;
i -= 1;
}
((b'a' + (d % 26) as u8) as char)
}
}
// Accepted solution for LeetCode #3307: Find the K-th Character in String Game II
function kthCharacter(k: number, operations: number[]): string {
let n = 1;
let i = 0;
while (n < k) {
n *= 2;
i++;
}
let d = 0;
while (n > 1) {
if (k > n / 2) {
k -= n / 2;
d += operations[i - 1];
}
n /= 2;
i--;
}
return String.fromCharCode('a'.charCodeAt(0) + (d % 26));
}
Use this to step through a reusable interview workflow for this problem.
Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.
Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.