daily-2026-06-17-3614-process-string-with-special-operations-ii

problems/3614-process-string-with-special-operations-ii/analysis.md

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+# 3614. Process String with Special Operations II
+
+[LeetCode Link](https://leetcode.com/problems/process-string-with-special-operations-ii/)
+
+Difficulty: Hard
+Topics: String, Simulation
+
+Acceptance Rate: 26.9%
+
+## Hints
+
+### Hint 1
+
+Read the constraints before reaching for the obvious approach. The string `s` is short
+(at most `10^5`), but the final `result` can grow to `10^15` characters because of the
+duplication operator `'#'`. Actually building `result` is impossible — you would run out
+of memory long before you finished. Ask yourself: do you really need the *whole* string,
+or just *one* character of it?
+
+### Hint 2
+
+Since you only need the `k`-th character, think about working **backwards**. If you knew
+the length of `result` after every operation, could you trace where index `k` "came from"
+as you undo each operation one at a time? This is a classic trick for problems where the
+structure explodes in size but only a single position is queried.
+
+### Hint 3
+
+Do two passes. **Pass 1 (forward):** compute the length of `result` after each operation
+— this is cheap and never builds the string. **Pass 2 (backward):** start with the target
+index `pos = k` and walk the operations from last to first, rewriting `pos` to be the index
+it occupied in the *previous* `result`. Each operator has a simple inverse mapping on the
+index:
+
+- A letter appended at the end occupies index `prevLen`; if `pos == prevLen`, that letter
+  *is* your answer.
+- `'#'` (duplicate) doubled the string, so if `pos >= prevLen`, subtract `prevLen` to fold
+  it back into the first half.
+- `'%'` (reverse) maps `pos` to `prevLen - 1 - pos`.
+- `'*'` (remove) only chopped the tail, so a valid `pos` is unchanged.
+
+When `pos` lands exactly on an appended letter, you are done.
+
+## Approach
+
+The whole difficulty here is the size of `result`. Letters add one character, `'*'` removes
+one, `'%'` is length-preserving, and `'#'` *doubles* the length. After enough `'#'`
+operations the string is astronomically large, so simulation is out. But the question only
+asks for a single index `k`, which lets us avoid materializing the string entirely.
+
+**Pass 1 — record lengths.** Walk `s` left to right and maintain a running length `cur`:
+
+- lowercase letter: `cur++`
+- `'*'`: `cur--` if `cur > 0` (otherwise it stays 0)
+- `'#'`: `cur *= 2`
+- `'%'`: unchanged
+
+Store `lengths[i]` = length of `result` immediately after processing `s[i]`. The final
+length is `cur`. If `k >= cur`, index `k` is out of bounds and the answer is `'.'`.
+
+**Pass 2 — trace the index backwards.** Set `pos = k` (a valid index into the final
+`result`). Iterate `i` from `n-1` down to `0`. Let `prev` be the length *before* operation
+`i`, which is `lengths[i-1]` (or `0` when `i == 0`). Rewrite `pos` according to the inverse
+of `s[i]`:
+
+- **letter:** appending put that letter at index `prev`. If `pos == prev`, the letter
+  `s[i]` is exactly the character we are tracking — return it. Otherwise `pos < prev`, and
+  `pos` refers to a character already present before the append, so leave it unchanged.
+- **`'*'`:** the operation removed the last character, so every surviving index is identical
+  to its index in the previous string. Leave `pos` unchanged.
+- **`'#'`:** the previous string was concatenated with itself. Indices `[0, prev)` come from
+  the first copy; indices `[prev, 2*prev)` come from the second copy. If `pos >= prev`,
+  subtract `prev` to map it back onto the original copy.
+- **`'%'`:** the string was reversed, so index `pos` came from index `prev - 1 - pos`.
+
+Because every valid `pos` either resolves to an appended letter or is carried back to the
+start, by the time we finish the loop we will have returned a letter. (If we somehow fall
+through, returning `'.'` is a safe guard.)
+
+**Why it works:** at every step `pos` is maintained as "the index in the current `result`
+of the character we ultimately want." The inverse maps above are exact, so following them
+to the beginning lands us on the precise letter that produced that position. We never store
+more than the `lengths` array.
+
+**Walkthrough (Example 2):** `s = "cd%#*#"`, `k = 3`. Forward lengths are
+`[1, 2, 2, 4, 3, 6]`, final length `6`, so `k = 3` is in bounds. Tracing backward with
+`pos = 3`:
+
+- `i=5 '#'`, `prev=3`: `pos >= 3` → `pos = 0`
+- `i=4 '*'`, `prev=4`: unchanged → `pos = 0`
+- `i=3 '#'`, `prev=2`: `pos < 2` → `pos = 0`
+- `i=2 '%'`, `prev=2`: `pos = 2 - 1 - 0 = 1`
+- `i=1 'd'`, `prev=1`: `pos == 1` → return `'d'`. ✓
+
+## Complexity Analysis
+
+Time Complexity: O(n), where `n = len(s)` — one forward pass to build lengths and one
+backward pass to trace the index.
+Space Complexity: O(n) for the `lengths` array.
+
+## Edge Cases
+
+- **`k` out of bounds:** if `k >= finalLength` (including the case where `result` ends up
+  empty), return `'.'`. Example 3 (`"z*#"`, `k = 0`) ends with an empty string.
+- **`'*'` on an empty string:** removing from an empty `result` is a no-op; guard with
+  `if cur > 0` so the length never goes negative.
+- **`'#'` on an empty string:** duplicating an empty string stays empty (`0 * 2 = 0`), which
+  the length recurrence handles naturally.
+- **Large indices:** `k` can be up to `10^15` and the length up to `10^15`, so use 64-bit
+  integers (`int64`) for both the running length and `pos` to avoid overflow.
+- **`'%'` mapping near the boundary:** the reverse map `prev - 1 - pos` must use `prev`, the
+  length *before* the reverse (which equals the length after, since reverse preserves
+  length), to stay within bounds.

problems/3614-process-string-with-special-operations-ii/problem.md

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+---
+number: "3614"
+frontend_id: "3614"
+title: "Process String with Special Operations II"
+slug: "process-string-with-special-operations-ii"
+difficulty: "Hard"
+topics:
+  - "String"
+  - "Simulation"
+acceptance_rate: 2686.3
+is_premium: false
+created_at: "2026-06-17T05:23:09.270497+00:00"
+fetched_at: "2026-06-17T05:23:09.270497+00:00"
+link: "https://leetcode.com/problems/process-string-with-special-operations-ii/"
+date: "2026-06-17"
+---
+
+# 3614. Process String with Special Operations II
+
+You are given a string `s` consisting of lowercase English letters and the special characters: `'*'`, `'#'`, and `'%'`.
+
+You are also given an integer `k`.
+
+Build a new string `result` by processing `s` according to the following rules from left to right:
+
+  * If the letter is a **lowercase** English letter append it to `result`.
+  * A `'*'` **removes** the last character from `result`, if it exists.
+  * A `'#'` **duplicates** the current `result` and **appends** it to itself.
+  * A `'%'` **reverses** the current `result`.
+
+
+
+Return the `kth` character of the final string `result`. If `k` is out of the bounds of `result`, return `'.'`.
+
+
+
+**Example 1:**
+
+**Input:** s = "a#b%*", k = 1
+
+**Output:** "a"
+
+**Explanation:**
+
+`i` | `s[i]` | Operation | Current `result`  
+---|---|---|---  
+0 | `'a'` | Append `'a'` | `"a"`  
+1 | `'#'` | Duplicate `result` | `"aa"`  
+2 | `'b'` | Append `'b'` | `"aab"`  
+3 | `'%'` | Reverse `result` | `"baa"`  
+4 | `'*'` | Remove the last character | `"ba"`  
+
+The final `result` is `"ba"`. The character at index `k = 1` is `'a'`.
+
+**Example 2:**
+
+**Input:** s = "cd%#*#", k = 3
+
+**Output:** "d"
+
+**Explanation:**
+
+`i` | `s[i]` | Operation | Current `result`  
+---|---|---|---  
+0 | `'c'` | Append `'c'` | `"c"`  
+1 | `'d'` | Append `'d'` | `"cd"`  
+2 | `'%'` | Reverse `result` | `"dc"`  
+3 | `'#'` | Duplicate `result` | `"dcdc"`  
+4 | `'*'` | Remove the last character | `"dcd"`  
+5 | `'#'` | Duplicate `result` | `"dcddcd"`  
+
+The final `result` is `"dcddcd"`. The character at index `k = 3` is `'d'`.
+
+**Example 3:**
+
+**Input:** s = "z*#", k = 0
+
+**Output:** "."
+
+**Explanation:**
+
+`i` | `s[i]` | Operation | Current `result`  
+---|---|---|---  
+0 | `'z'` | Append `'z'` | `"z"`  
+1 | `'*'` | Remove the last character | `""`  
+2 | `'#'` | Duplicate the string | `""`  
+
+The final `result` is `""`. Since index `k = 0` is out of bounds, the output is `'.'`.
+
+
+
+**Constraints:**
+
+  * `1 <= s.length <= 105`
+  * `s` consists of only lowercase English letters and special characters `'*'`, `'#'`, and `'%'`.
+  * `0 <= k <= 1015`
+  * The length of `result` after processing `s` will not exceed `1015`.

problems/3614-process-string-with-special-operations-ii/solution_daily_20260617.go

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+package main
+
+// Process String with Special Operations II (LeetCode 3614)
+//
+// The final `result` can grow to 10^15 characters because of the '#'
+// duplication operator, so we never build it. Instead we:
+//
+//  1. Forward pass: record the length of result after each operation.
+//  2. Backward pass: starting from index k, undo each operation, rewriting
+//     the index to the position it occupied in the previous string. When the
+//     index lands on an appended letter, that letter is the answer.
+//
+// Runs in O(n) time and O(n) space, where n = len(s).
+func processStr(s string, k int64) byte {
+	n := len(s)
+
+	// Pass 1: length of result after processing each character.
+	lengths := make([]int64, n)
+	var cur int64
+	for i := 0; i < n; i++ {
+		switch s[i] {
+		case '*':
+			if cur > 0 {
+				cur--
+			}
+		case '#':
+			cur *= 2
+		case '%':
+			// reverse: length unchanged
+		default:
+			cur++ // lowercase letter
+		}
+		lengths[i] = cur
+	}
+
+	// Out of bounds (covers the empty-result case as well).
+	if k >= cur {
+		return '.'
+	}
+
+	// Pass 2: trace index k back to the letter that produced it.
+	pos := k
+	for i := n - 1; i >= 0; i-- {
+		var prev int64
+		if i > 0 {
+			prev = lengths[i-1]
+		}
+		switch s[i] {
+		case '*':
+			// Removed the tail; surviving indices are unchanged.
+		case '#':
+			// Duplicated; fold the second copy back onto the first.
+			if pos >= prev {
+				pos -= prev
+			}
+		case '%':
+			// Reversed; mirror the index.
+			pos = prev - 1 - pos
+		default:
+			// Letter appended at index prev.
+			if pos == prev {
+				return s[i]
+			}
+		}
+	}
+
+	return '.'
+}

problems/3614-process-string-with-special-operations-ii/solution_daily_20260617_test.go

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+package main
+
+import "testing"
+
+func TestSolution(t *testing.T) {
+	tests := []struct {
+		name     string
+		s        string
+		k        int64
+		expected byte
+	}{
+		// Examples from the problem statement.
+		{"example 1: remove after reverse, result \"ba\"", "a#b%*", 1, 'a'},
+		{"example 2: duplicate/remove/duplicate, result \"dcddcd\"", "cd%#*#", 3, 'd'},
+		{"example 3: result empty, k out of bounds", "z*#", 0, '.'},
+
+		// Additional edge cases.
+		{"edge case: single letter, k in bounds", "x", 0, 'x'},
+		{"edge case: single letter, k out of bounds", "x", 1, '.'},
+		{"edge case: remove on empty result stays empty", "*", 0, '.'},
+		{"edge case: duplicate then index into second copy", "ab#", 2, 'a'},
+		{"edge case: reverse a simple string", "abc%", 0, 'c'},
+		{"edge case: large k via duplication", "a##", 3, 'a'},
+		{"edge case: every char removed", "abc***", 0, '.'},
+	}
+
+	for _, tt := range tests {
+		t.Run(tt.name, func(t *testing.T) {
+			result := processStr(tt.s, tt.k)
+			if result != tt.expected {
+				t.Errorf("processStr(%q, %d) = %q, want %q",
+					tt.s, tt.k, result, tt.expected)
+			}
+		})
+	}
+}