Find The K-th Character In String Game Ii

Find the K‑th Character in String Game II
In the “String Game II” problem we are given an initial string s (consisting only of lowercase English letters) and an integer k. The game builds an infinite sequence of strings by repeatedly applying a transformation: each new string is formed by taking the previous string and incrementing every character by one (with ‘z’ wrapping to ‘a’). The final string after n rounds is the concatenation of all intermediate strings. The task is to return the k‑th character (1‑indexed) in this final concatenated string without actually constructing the huge string.

Below we walk through a complete, step‑by‑step solution that is both intuitive and efficient enough to pass the constraints ( |s| ≤ 100, k ≤ 10^9 ). The explanation includes the underlying mathematics, a recursive divide‑and‑conquer algorithm, pseudocode, complexity analysis, and ready‑to‑run implementations in Python, Java, and C++.

Table of Contents

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1. Problem Statement & Key Observations

Input

• A string s containing only lowercase letters ('a'‑'z').
• An integer k (1 ≤ k ≤ 10^9).

Game Rules

• Let S₀ = s.
• For each round i ≥ 1, define Sᵢ as the string obtained by incrementing every character of Sᵢ₋₁ by one (with wrap‑around: 'z' → 'a').
• The final string after n rounds is the concatenation T = S₀ + S₁ + S₂ + … + Sₙ.
• We are asked to return the character at position k in T.

Observations

1. The transformation is uniform: every character in Sᵢ₋₁ shifts by the same amount (+1). So, the relative order of characters inside each Sᵢ remains identical to s.
2. The length of each Sᵢ equals the length of s (let L = |s|). So naturally, each round contributes exactly L characters to the total length.
3. After n rounds, the total length is |T| = (n+1)·L.
4. Because k can be as large as 10^9 while L ≤ 100, we cannot materialise T. We need a way to jump directly to the round that contains the k‑th character and then compute the character after the appropriate number of shifts.

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2. Why a Direct Simulation Fails

A naïve approach would iteratively build S₀, S₁, … until the accumulated length exceeds k, then index into the last built string.
Even so, - In the worst case, k = 10^9 and L = 1 → we would need to generate ≈ 10^9 characters, which is impossible both time‑wise and memory‑wise. - Even with L = 100, we would need about 10⁷ rounds, still far too many for a typical 1‑second limit.

Thus we must avoid constructing the strings altogether and rely on mathematical properties.

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3. Core Idea: Lengths Grow Exponentially?

Actually, the lengths grow linearly with the number of rounds because each round adds a fixed‑size block (L). On the flip side, the character values evolve in a predictable cyclic manner: after 26 increments a character returns to its original form.

Key insight:

• The character at position p inside Sᵢ is simply the character s[p] shifted forward by i positions in the alphabet (mod 26).
• That's why, if we can determine which round (i) the k‑th character belongs to and the offset inside that round, we can compute the answer in O(1) time.

And yeah — that's actually more nuanced than it sounds Simple, but easy to overlook..

Finding the round reduces to a simple division: each round contributes L characters, so the round index is

zeroBasedRound = (k-1) // L          // integer division
offsetInRound  = (k-1) %  L          // 0‑based position inside the round

The answer is then

originalChar = s[offsetInRound]
shiftedChar  = (originalChar - 'a' + zeroBasedRound) % 26 + 'a'

That’s it! The problem is easier than it first appears because the transformation does not change the string’s length or internal ordering.

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4. Recursive Divide‑and‑Conquer Solution (Alternative View)

Although the direct formula solves the problem in O(1), many editorial solutions present a recursive divide‑and‑conquer approach that works for a more general variant where the transformation might change the string length (e.But , duplicating or reversing). g.Understanding that method helps solidify the reasoning.

After grasping this direct formula, one might ask whether a recursive divide‑and‑conquer strategy offers any advantage. Think about it: while such approaches become indispensable when the transformation alters the structure—like reversing the whole string or permuting its elements—the present problem remains tractable via the simple linear analysis outlined above. The elegance lies in the fact that each round merely appends a block of length L while uniformly shifting the underlying alphabet, leaving the global order untouched. As a result, locating the exact block containing the target character requires only integer division, and the required shift follows from the block index alone.

Simply put, the complete solution proceeds as follows: compute the zero‑based round number r = (k‑1) // L and the offset o = (k‑1) % L; retrieve the base character c = s[o]; finally map it through (c – ‘a’ + r) mod 26 + ‘a’ to obtain the answer. On top of that, this method runs in constant time and constant extra memory, comfortably handling the maximal constraints (k ≤ 10⁹, L ≤ 100). No intermediate strings are ever constructed, avoiding both time and space bottlenecks.

Conclusion: By exploiting the regularity of the per‑round transformation, we reduce the seemingly intractable task of simulating billions of iterations to a single arithmetic operation. The derived formula not only yields the correct result but also exemplifies how deep understanding of the problem’s

The derived formula not only yields the correct result but also exemplifies how deep understanding of the problem’s underlying periodicity can collapse an apparently exponential process into a single arithmetic operation. By recognizing that each round merely replicates the original string with a uniform Caesar shift, we avoid any need to simulate billions of steps. This pattern of “repeat‑and‑shift” appears in many other contexts—cryptographic rotations, periodic signals, and even certain data‑compression schemes—so the insight gained here is broadly applicable Surprisingly effective..

Implementing the solution is straightforward: compute r = (k-1) / L and o = (k-1) % L, fetch s[o], and apply the shift (s[o] - 'a' + r) % 26. All operations fit within standard 64‑bit integer ranges, and the algorithm uses constant extra memory, making it ideal for the given constraints (k ≤ 10⁹, L ≤ 100). Edge cases such as k landing exactly at a round boundary are handled naturally by the zero‑based indexing, and the modulo 26 arithmetic guarantees correct wrapping for the alphabet.

In practice, programmers should be mindful of integer overflow when performing the addition inside the modulo expression, especially if they port the solution to languages with fixed‑width integers. Using 64‑bit types or performing the modulo after the addition eliminates any risk.

Conclusion: The constant‑time formula transforms a seemingly intractable simulation into a simple, elegant calculation, showcasing the power of mathematical insight in algorithm design. Mastering this kind of pattern recognition equips developers to tackle massive input sizes efficiently, turning potential performance bottlenecks into trivial arithmetic steps Nothing fancy..