LeetCode #3614 — HARD

Process String with Special Operations II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode

The Problem

Problem Statement

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 k^th 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.

Step 01

Brute Force Baseline

Problem summary: 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 '.'.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"a#b%*"
1

Example 2

"cd%#*#"
3

Example 3

"z*#"
0

Step 02

Core Insight

What unlocks the optimal approach

Track the length of the string after each operation on <code>s</code>.
Walk backwards through <code>s</code>, undoing each # by using modulus on the tracked lengths, and undoing each % by mirroring across the midpoint, to pinpoint the <code>k</code>th character.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3614: Process String with Special Operations II
// Auto-generated Java example from go.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (go):
// // Accepted solution for LeetCode #3614: Process String with Special Operations II
// package main
// 
// // https://space.bilibili.com/206214
// func processStr1(s string, k int64) byte {
// 	n := len(s)
// 	size := make([]int64, n)
// 	sz := int64(0)
// 	for i, c := range s {
// 		if c == '*' {
// 			sz = max(sz-1, 0)
// 		} else if c == '#' {
// 			sz *= 2
// 		} else if c != '%' { // c 是字母
// 			sz++
// 		}
// 		size[i] = sz
// 	}
// 
// 	if k >= size[n-1] { // 下标越界
// 		return '.'
// 	}
// 
// 	// 迭代
// 	for i := n - 1; ; i-- {
// 		c := s[i]
// 		sz = size[i]
// 		if c == '#' {
// 			if k >= sz/2 { // k 在复制后的右半边
// 				k -= sz / 2
// 			}
// 		} else if c == '%' {
// 			k = sz - 1 - k // 反转前的下标为 sz-1-k 的字母就是答案
// 		} else if c != '*' && k == sz-1 { // 找到答案
// 			return c
// 		}
// 	}
// }
// 
// func processStr(s string, k int64) byte {
// 	sz := int64(0)
// 	for _, c := range s {
// 		if c == '*' {
// 			sz = max(sz-1, 0)
// 		} else if c == '#' {
// 			sz *= 2
// 		} else if c != '%' {
// 			sz++
// 		}
// 	}
// 
// 	if k >= sz {
// 		return '.'
// 	}
// 
// 	for i := len(s) - 1; ; i-- {
// 		c := s[i]
// 		if c == '*' {
// 			sz++
// 		} else if c == '#' {
// 			sz /= 2
// 			if k >= sz {
// 				k -= sz
// 			}
// 		} else if c == '%' {
// 			k = sz - 1 - k
// 		} else {
// 			sz--
// 			if k == sz {
// 				return c
// 			}
// 		}
// 	}
// }

// Accepted solution for LeetCode #3614: Process String with Special Operations II
package main

// https://space.bilibili.com/206214
func processStr1(s string, k int64) byte {
	n := len(s)
	size := make([]int64, n)
	sz := int64(0)
	for i, c := range s {
		if c == '*' {
			sz = max(sz-1, 0)
		} else if c == '#' {
			sz *= 2
		} else if c != '%' { // c 是字母
			sz++
		}
		size[i] = sz
	}

	if k >= size[n-1] { // 下标越界
		return '.'
	}

	// 迭代
	for i := n - 1; ; i-- {
		c := s[i]
		sz = size[i]
		if c == '#' {
			if k >= sz/2 { // k 在复制后的右半边
				k -= sz / 2
			}
		} else if c == '%' {
			k = sz - 1 - k // 反转前的下标为 sz-1-k 的字母就是答案
		} else if c != '*' && k == sz-1 { // 找到答案
			return c
		}
	}
}

func processStr(s string, k int64) byte {
	sz := int64(0)
	for _, c := range s {
		if c == '*' {
			sz = max(sz-1, 0)
		} else if c == '#' {
			sz *= 2
		} else if c != '%' {
			sz++
		}
	}

	if k >= sz {
		return '.'
	}

	for i := len(s) - 1; ; i-- {
		c := s[i]
		if c == '*' {
			sz++
		} else if c == '#' {
			sz /= 2
			if k >= sz {
				k -= sz
			}
		} else if c == '%' {
			k = sz - 1 - k
		} else {
			sz--
			if k == sz {
				return c
			}
		}
	}
}

# Accepted solution for LeetCode #3614: Process String with Special Operations II
# Time:  O(n)
# Space: O(1)

# backward simulation
class Solution(object):
    def processStr(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: str
        """
        l = 0
        for x in s:
            if x == '*':
                l = max(l-1, 0)
            elif x == '#':
                l <<= 1
            elif x == '%':
                continue
            else:
                l += 1
        if k >= l:
            return '.'
        for x in reversed(s):
            if x == '*':
                l += 1
            elif x == '#':
                l >>= 1
                if k >= l:
                    k -= l
            elif x == '%':
                k = (l-1)-k
            else:
                l -= 1
                if l == k:
                    return x

// Accepted solution for LeetCode #3614: Process String with Special Operations II
// Rust example auto-generated from go reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (go):
// // Accepted solution for LeetCode #3614: Process String with Special Operations II
// package main
// 
// // https://space.bilibili.com/206214
// func processStr1(s string, k int64) byte {
// 	n := len(s)
// 	size := make([]int64, n)
// 	sz := int64(0)
// 	for i, c := range s {
// 		if c == '*' {
// 			sz = max(sz-1, 0)
// 		} else if c == '#' {
// 			sz *= 2
// 		} else if c != '%' { // c 是字母
// 			sz++
// 		}
// 		size[i] = sz
// 	}
// 
// 	if k >= size[n-1] { // 下标越界
// 		return '.'
// 	}
// 
// 	// 迭代
// 	for i := n - 1; ; i-- {
// 		c := s[i]
// 		sz = size[i]
// 		if c == '#' {
// 			if k >= sz/2 { // k 在复制后的右半边
// 				k -= sz / 2
// 			}
// 		} else if c == '%' {
// 			k = sz - 1 - k // 反转前的下标为 sz-1-k 的字母就是答案
// 		} else if c != '*' && k == sz-1 { // 找到答案
// 			return c
// 		}
// 	}
// }
// 
// func processStr(s string, k int64) byte {
// 	sz := int64(0)
// 	for _, c := range s {
// 		if c == '*' {
// 			sz = max(sz-1, 0)
// 		} else if c == '#' {
// 			sz *= 2
// 		} else if c != '%' {
// 			sz++
// 		}
// 	}
// 
// 	if k >= sz {
// 		return '.'
// 	}
// 
// 	for i := len(s) - 1; ; i-- {
// 		c := s[i]
// 		if c == '*' {
// 			sz++
// 		} else if c == '#' {
// 			sz /= 2
// 			if k >= sz {
// 				k -= sz
// 			}
// 		} else if c == '%' {
// 			k = sz - 1 - k
// 		} else {
// 			sz--
// 			if k == sz {
// 				return c
// 			}
// 		}
// 	}
// }

// Accepted solution for LeetCode #3614: Process String with Special Operations II
// Auto-generated TypeScript example from go.
function exampleSolution(): void {
}
// Reference (go):
// // Accepted solution for LeetCode #3614: Process String with Special Operations II
// package main
// 
// // https://space.bilibili.com/206214
// func processStr1(s string, k int64) byte {
// 	n := len(s)
// 	size := make([]int64, n)
// 	sz := int64(0)
// 	for i, c := range s {
// 		if c == '*' {
// 			sz = max(sz-1, 0)
// 		} else if c == '#' {
// 			sz *= 2
// 		} else if c != '%' { // c 是字母
// 			sz++
// 		}
// 		size[i] = sz
// 	}
// 
// 	if k >= size[n-1] { // 下标越界
// 		return '.'
// 	}
// 
// 	// 迭代
// 	for i := n - 1; ; i-- {
// 		c := s[i]
// 		sz = size[i]
// 		if c == '#' {
// 			if k >= sz/2 { // k 在复制后的右半边
// 				k -= sz / 2
// 			}
// 		} else if c == '%' {
// 			k = sz - 1 - k // 反转前的下标为 sz-1-k 的字母就是答案
// 		} else if c != '*' && k == sz-1 { // 找到答案
// 			return c
// 		}
// 	}
// }
// 
// func processStr(s string, k int64) byte {
// 	sz := int64(0)
// 	for _, c := range s {
// 		if c == '*' {
// 			sz = max(sz-1, 0)
// 		} else if c == '#' {
// 			sz *= 2
// 		} else if c != '%' {
// 			sz++
// 		}
// 	}
// 
// 	if k >= sz {
// 		return '.'
// 	}
// 
// 	for i := len(s) - 1; ; i-- {
// 		c := s[i]
// 		if c == '*' {
// 			sz++
// 		} else if c == '#' {
// 			sz /= 2
// 			if k >= sz {
// 				k -= sz
// 			}
// 		} else if c == '%' {
// 			k = sz - 1 - k
// 		} else {
// 			sz--
// 			if k == sz {
// 				return c
// 			}
// 		}
// 	}
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.