|
| 1 | +--- |
| 2 | +id: reverse-only-letters |
| 3 | +title: Reverse Only Letters |
| 4 | +sidebar_label: 0917 - Reverse Only Letters |
| 5 | +tags: |
| 6 | + - Two Pointers |
| 7 | + - String |
| 8 | + - Stack |
| 9 | +description: "This is a solution to the Reverse Only Letters problem on LeetCode." |
| 10 | +--- |
| 11 | + |
| 12 | +## Problem Description |
| 13 | + |
| 14 | +Given a string `s`, reverse the string according to the following rules: |
| 15 | + |
| 16 | +- All the characters that are not English letters remain in the same position. |
| 17 | +- All the English letters (lowercase or uppercase) should be reversed. |
| 18 | + |
| 19 | +Return `s` after reversing it. |
| 20 | + |
| 21 | +### Examples |
| 22 | + |
| 23 | +**Example 1:** |
| 24 | + |
| 25 | +``` |
| 26 | +Input: s = "ab-cd" |
| 27 | +Output: "dc-ba" |
| 28 | +``` |
| 29 | +**Example 2:** |
| 30 | + |
| 31 | +``` |
| 32 | +Input: s = "a-bC-dEf-ghIj" |
| 33 | +Output: "j-Ih-gfE-dCba" |
| 34 | +``` |
| 35 | + |
| 36 | +### Constraints |
| 37 | + |
| 38 | +- $1 \leq s.length \leq 100$ |
| 39 | +- `s` consists of characters with ASCII values in the range `[33, 122]`. |
| 40 | +- `s` does not contain `'\"'` or `'\\'`. |
| 41 | + |
| 42 | +## Solution for Reverse Only Letters |
| 43 | + |
| 44 | +## Approach: Stack of Letters |
| 45 | +### Intuition and Algorithm |
| 46 | + |
| 47 | +Collect the letters of `S` separately into a stack, so that popping the stack reverses the letters. (Alternatively, we could have collected the letters into an array and reversed the array.) |
| 48 | + |
| 49 | +Then, when writing the characters of `S`, any time we need a letter, we use the one we have prepared instead. |
| 50 | + |
| 51 | +### Code in Different Languages |
| 52 | + |
| 53 | +<Tabs> |
| 54 | +<TabItem value="cpp" label="C++"> |
| 55 | + <SolutionAuthor name="@Shreyash3087"/> |
| 56 | + |
| 57 | +```cpp |
| 58 | +class Solution { |
| 59 | +public: |
| 60 | + string reverseOnlyLetters(string S) { |
| 61 | + stack<char> letters; |
| 62 | + for (char c : S) |
| 63 | + if (isalpha(c)) |
| 64 | + letters.push(c); |
| 65 | + |
| 66 | + string ans; |
| 67 | + for (char c : S) { |
| 68 | + if (isalpha(c)) |
| 69 | + ans += letters.top(), letters.pop(); |
| 70 | + else |
| 71 | + ans += c; |
| 72 | + } |
| 73 | + |
| 74 | + return ans; |
| 75 | + } |
| 76 | +}; |
| 77 | + |
| 78 | +``` |
| 79 | +</TabItem> |
| 80 | +<TabItem value="java" label="Java"> |
| 81 | + <SolutionAuthor name="@Shreyash3087"/> |
| 82 | + |
| 83 | +```java |
| 84 | +class Solution { |
| 85 | + public String reverseOnlyLetters(String S) { |
| 86 | + Stack<Character> letters = new Stack(); |
| 87 | + for (char c: S.toCharArray()) |
| 88 | + if (Character.isLetter(c)) |
| 89 | + letters.push(c); |
| 90 | + |
| 91 | + StringBuilder ans = new StringBuilder(); |
| 92 | + for (char c: S.toCharArray()) { |
| 93 | + if (Character.isLetter(c)) |
| 94 | + ans.append(letters.pop()); |
| 95 | + else |
| 96 | + ans.append(c); |
| 97 | + } |
| 98 | + |
| 99 | + return ans.toString(); |
| 100 | + } |
| 101 | +} |
| 102 | +``` |
| 103 | + |
| 104 | +</TabItem> |
| 105 | +<TabItem value="python" label="Python"> |
| 106 | + <SolutionAuthor name="@Shreyash3087"/> |
| 107 | + |
| 108 | +```python |
| 109 | +class Solution(object): |
| 110 | + def reverseOnlyLetters(self, S): |
| 111 | + letters = [c for c in S if c.isalpha()] |
| 112 | + ans = [] |
| 113 | + for c in S: |
| 114 | + if c.isalpha(): |
| 115 | + ans.append(letters.pop()) |
| 116 | + else: |
| 117 | + ans.append(c) |
| 118 | + return "".join(ans) |
| 119 | +``` |
| 120 | +</TabItem> |
| 121 | +</Tabs> |
| 122 | + |
| 123 | +### Complexity Analysis |
| 124 | + |
| 125 | +#### Time Complexity: $O(N)$ |
| 126 | + |
| 127 | +> **Reason**: where `N` is the length of `S`. |
| 128 | +
|
| 129 | +#### Space Complexity: $O(N)$ |
| 130 | + |
| 131 | +## Approach: Reverse Pointer |
| 132 | +### Intuition |
| 133 | + |
| 134 | +Write the characters of `S` one by one. When we encounter a letter, we want to write the next letter that occurs if we iterated through the string backwards. |
| 135 | + |
| 136 | +So we do just that: keep track of a pointer `j` that iterates through the string backwards. When we need to write a letter, we use it. |
| 137 | + |
| 138 | +### Code in Different Languages |
| 139 | + |
| 140 | +<Tabs> |
| 141 | +<TabItem value="cpp" label="C++"> |
| 142 | + <SolutionAuthor name="@Shreyash3087"/> |
| 143 | + |
| 144 | +```cpp |
| 145 | +#include <string> |
| 146 | +#include <cctype> |
| 147 | +#include <sstream> |
| 148 | + |
| 149 | +class Solution { |
| 150 | +public: |
| 151 | + std::string reverseOnlyLetters(std::string S) { |
| 152 | + std::stringstream ans; |
| 153 | + int j = S.length() - 1; |
| 154 | + for (int i = 0; i < S.length(); ++i) { |
| 155 | + if (std::isalpha(S[i])) { |
| 156 | + while (!std::isalpha(S[j])) |
| 157 | + j--; |
| 158 | + ans << S[j--]; |
| 159 | + } else { |
| 160 | + ans << S[i]; |
| 161 | + } |
| 162 | + } |
| 163 | + |
| 164 | + return ans.str(); |
| 165 | + } |
| 166 | +}; |
| 167 | + |
| 168 | +``` |
| 169 | +</TabItem> |
| 170 | +<TabItem value="java" label="Java"> |
| 171 | + <SolutionAuthor name="@Shreyash3087"/> |
| 172 | + |
| 173 | +```java |
| 174 | +class Solution { |
| 175 | + public String reverseOnlyLetters(String S) { |
| 176 | + StringBuilder ans = new StringBuilder(); |
| 177 | + int j = S.length() - 1; |
| 178 | + for (int i = 0; i < S.length(); ++i) { |
| 179 | + if (Character.isLetter(S.charAt(i))) { |
| 180 | + while (!Character.isLetter(S.charAt(j))) |
| 181 | + j--; |
| 182 | + ans.append(S.charAt(j--)); |
| 183 | + } else { |
| 184 | + ans.append(S.charAt(i)); |
| 185 | + } |
| 186 | + } |
| 187 | + |
| 188 | + return ans.toString(); |
| 189 | + } |
| 190 | +} |
| 191 | +``` |
| 192 | + |
| 193 | +</TabItem> |
| 194 | +<TabItem value="python" label="Python"> |
| 195 | + <SolutionAuthor name="@Shreyash3087"/> |
| 196 | + |
| 197 | +```python |
| 198 | +class Solution(object): |
| 199 | + def reverseOnlyLetters(self, S): |
| 200 | + ans = [] |
| 201 | + j = len(ans) - 1 |
| 202 | + for i, x in enumerate(S): |
| 203 | + if x.isalpha(): |
| 204 | + while not S[j].isalpha(): |
| 205 | + j -= 1 |
| 206 | + ans.append(S[j]) |
| 207 | + j -= 1 |
| 208 | + else: |
| 209 | + ans.append(x) |
| 210 | + |
| 211 | + return "".join(ans) |
| 212 | +``` |
| 213 | +</TabItem> |
| 214 | +</Tabs> |
| 215 | + |
| 216 | +### Complexity Analysis |
| 217 | + |
| 218 | +#### Time Complexity: $O(N)$ |
| 219 | + |
| 220 | +> **Reason**: where `N` is the length of `S`. |
| 221 | +
|
| 222 | +#### Space Complexity: $O(N)$ |
| 223 | + |
| 224 | +## Video Solution |
| 225 | + |
| 226 | +<LiteYouTubeEmbed |
| 227 | + id="M2TwbZCpJpw" |
| 228 | + params="autoplay=1&autohide=1&showinfo=0&rel=0" |
| 229 | + title="LeetCode Reverse Only Letters Solution Explained - Java" |
| 230 | + poster="hqdefault" |
| 231 | + webp /> |
| 232 | + |
| 233 | +## References |
| 234 | + |
| 235 | +- **LeetCode Problem**: [Reverse Only Letters](https://leetcode.com/problems/reverse-only-letters/description/) |
| 236 | + |
| 237 | +- **Solution Link**: [Reverse Only Letters](https://leetcode.com/problems/reverse-only-letters/solutions/) |
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