Process String with Special Operations II

Link: https://leetcode.com/problems/process-string-with-special-operations-ii/

📖 Interpreting the Problem

We’re given a string s composed of lowercase letters and three special characters: *, #, and %. Each character modifies an initially empty string result as we iterate from left to right.

• Lowercase letter: Append it to result.
• *: Delete the last character from result, if any.
• #: Duplicate the current result and append it to itself.
• %: Reverse the current result.

After processing the entire string, we’re to return the k-th character of result (0-based). If k is out of bounds, return .(dot).

🧠 First Attempts and Their Pitfalls

A natural first approach is to simulate this directly. Start with an empty result string and apply each character’s rule step by step. It works flawlessly for small input sizes:

string result = "";
for (char c : s) {
    if (islower(c)) result += c;
    else if (c == '*') if (!result.empty()) result.pop_back();
    else if (c == '#') result += result;
    else if (c == '%') reverse(result.begin(), result.end());
}

This is conceptually simple but fails catastrophically as result grows, especially with multiple # or % operators. Duplication doubles the size exponentially: after 20 # operations, result could reach over a million characters. That’s unsustainable for time and memory.

Even worse, k can be as large as 10<sup>18</sup>. We can never afford to build such a string fully. Constructing the entire string to extract one character would be inefficient and likely infeasible.

This leads to a deeper realization: we don’t care about the whole string. We care about only one specific character. So why build the entire house just to look through one window?

💡 The Breakthrough Insight

The key realization is that we don’t need the whole string to find the k-th character. Instead, we need a way to reverse-engineer the character at index k based on the operations that generated it. This leads to a clever simulation in reverse.

First, we simulate the operations forward to compute the final length of result. We don’t store result, only track its length:

• # doubles the length (if non-zero)
• % keeps the length the same
• * decreases it by 1 (if non-zero)
• lowercase letters increase it by 1

Then we process the string in reverse (right to left). At each step, we simulate the effect of undoing that operation. Our goal: track where the index k came from.

• For #, if result was doubled, then k came from the first half if k < length/2; otherwise, subtract length/2.
• For %, reverse means k becomes length - 1 - k.
• For *, it means the previous string had one more character at the end we now ignore.
• For lowercase letters, if k refers to the last added character, return it. Otherwise, reduce the length and continue.

This process rewinds how the final string was constructed until we pinpoint the character at k.

📘 About Reverse Simulation
By treating the operations as reversible transformations, we can avoid building massive intermediate states. This technique works especially well for problems where the output is too large to construct but only partial information is needed.

🔍 Walking Through the Algorithm (with Full-Length Example)

Let’s take s = "a#b%" and k = 2.

Forward simulation (length only):

• a → length 1
• # → length 2 (a duplicated)
• b → length 3
• % → length 3 (unchanged)

Final length = 3. Now reverse-process:

• %: reverse index → k = 3 - 1 - 2 = 0
• b: is k == length-1? No (k=0, length=3) → length– (now 2)
• #: was length doubled, now half → length=1, k=0 (unchanged)
• a: k == length-1? Yes → return 'a'

Answer is 'a'.

🖼️ Suggested Diagram
Show a timeline of result length evolution and how k is transformed at each reverse step. Use arrows to indicate k shifting due to #, %, or shrinking from * or letter removal.

🧾 Pseudocode (Code Only)

finalLength = 0
for c in s:
    update finalLength based on rules
if finalLength <= k: return '.'

reverse s
currentK = k
currentLength = finalLength
for c in s:
    reverse-transform currentK and currentLength
    if letter and currentK == currentLength - 1:
        return c
return '.'

💻 Source Code

class Solution {
public:
    char processStr(string s, long long k) {
        long long finalLength = 0LL;

        // First pass: simulate the operations to calculate final string length
        for (char c : s) {
            if (c == '#') {
                if (finalLength > 0) finalLength *= 2;  // duplicate the current result
            } else if (c == '*') {
                if (finalLength > 0) finalLength--;     // delete last character if exists
            } else {
                finalLength++;                          // add new character
            }
        }

        // If k is outside the bounds of the final result
        if (finalLength <= k) return '.';

        // Reverse simulation: trace back to original character that became the k-th
        reverse(s.begin(), s.end());
        long long currentK = k, currentLength = finalLength;

        for (char c : s) {
            if (c == '#') {
                long long half = currentLength / 2;
                if (currentK >= half) currentK -= half;  // map k to first half of duplicated result
                currentLength = half;
            } else if (c == '%') {
                currentK = currentLength - 1 - currentK; // reverse effect
            } else if (c == '*') {
                currentLength++;                         // undo deletion: assume a char was removed
            } else {
                if (currentK == currentLength - 1) return c; // found the character
                currentLength--;                         // move to previous state
            }
        }

        return '.';  // shouldn't reach here
    }
};

📌 Summary and Key Takeaways

What makes this problem interesting is how deceptively simple operations can lead to exponential blowups. The trick was realizing that we don’t need the whole string — just the character at one index. By simulating operations in reverse and tracking how k evolves, we sidestep the explosion in size.

This solution was guided by the insight that the output string’s full content is irrelevant, and only its transformation history matters for retrieving a single character. Recognizing this allowed us to treat the string construction as a mathematical transformation rather than a physical structure.