Move from brute-force thinking to an efficient approach using two pointers strategy.
A magical string s consists of only '1' and '2' and obeys the following rule:
'1' and '2' generates the string s itself.The first few elements of s is s = "1221121221221121122……". If we group the consecutive 1's and 2's in s, it will be "1 22 11 2 1 22 1 22 11 2 11 22 ......" and counting the occurrences of 1's or 2's in each group yields the sequence "1 2 2 1 1 2 1 2 2 1 2 2 ......".
You can see that concatenating the occurrence sequence gives us s itself.
Given an integer n, return the number of 1's in the first n number in the magical string s.
Example 1:
Input: n = 6 Output: 3 Explanation: The first 6 elements of magical string s is "122112" and it contains three 1's, so return 3.
Example 2:
Input: n = 1 Output: 1
Constraints:
1 <= n <= 105Problem summary: A magical string s consists of only '1' and '2' and obeys the following rule: Concatenating the sequence of lengths of its consecutive groups of identical characters '1' and '2' generates the string s itself. The first few elements of s is s = "1221121221221121122……". If we group the consecutive 1's and 2's in s, it will be "1 22 11 2 1 22 1 22 11 2 11 22 ......" and counting the occurrences of 1's or 2's in each group yields the sequence "1 2 2 1 1 2 1 2 2 1 2 2 ......". You can see that concatenating the occurrence sequence gives us s itself. Given an integer n, return the number of 1's in the first n number in the magical string s.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers
6
1
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #481: Magical String
class Solution {
public int magicalString(int n) {
List<Integer> s = new ArrayList<>(List.of(1, 2, 2));
for (int i = 2; s.size() < n; ++i) {
int pre = s.get(s.size() - 1);
int cur = 3 - pre;
for (int j = 0; j < s.get(i); ++j) {
s.add(cur);
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
if (s.get(i) == 1) {
++ans;
}
}
return ans;
}
}
// Accepted solution for LeetCode #481: Magical String
func magicalString(n int) (ans int) {
s := []int{1, 2, 2}
for i := 2; len(s) < n; i++ {
pre := s[len(s)-1]
cur := 3 - pre
for j := 0; j < s[i]; j++ {
s = append(s, cur)
}
}
for _, c := range s[:n] {
if c == 1 {
ans++
}
}
return
}
# Accepted solution for LeetCode #481: Magical String
class Solution:
def magicalString(self, n: int) -> int:
s = [1, 2, 2]
i = 2
while len(s) < n:
pre = s[-1]
cur = 3 - pre
s += [cur] * s[i]
i += 1
return s[:n].count(1)
// Accepted solution for LeetCode #481: Magical String
impl Solution {
pub fn magical_string(n: i32) -> i32 {
let mut s = vec![1, 2, 2];
let mut i = 2;
while s.len() < n as usize {
let pre = s[s.len() - 1];
let cur = 3 - pre;
for _ in 0..s[i] {
s.push(cur);
}
i += 1;
}
s.iter().take(n as usize).filter(|&&x| x == 1).count() as i32
}
}
// Accepted solution for LeetCode #481: Magical String
function magicalString(n: number): number {
const s: number[] = [1, 2, 2];
for (let i = 2; s.length < n; ++i) {
let pre = s[s.length - 1];
let cur = 3 - pre;
for (let j = 0; j < s[i]; ++j) {
s.push(cur);
}
}
return s.slice(0, n).filter(x => x === 1).length;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.