solution added to 984

dsa-solutions/lc-solutions/0900-0999/0984-string-without-aaa-or-bbb.md

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+---
+id: string-without-aaa-or-bbb
+title: String Without AAA or BBB
+sidebar_label: 984-String Without AAA or BBB
+tags:
+  - Greedy
+  - String
+  - LeetCode
+  - Java
+  - Python
+  - C++
+description: "This is a solution to the String Without AAA or BBB problem on LeetCode."
+sidebar_position: 3
+---
+
+## Problem Description
+
+Given two integers `a` and `b`, return any string `s` such that:
+
+- `s` has length `a + b` and contains exactly `a` 'a' letters and exactly `b` 'b' letters.
+- The substring 'aaa' does not occur in `s`.
+- The substring 'bbb' does not occur in `s`.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: a = 1, b = 2
+Output: "abb"
+Explanation: "abb", "bab", and "bba" are all correct answers.
+```
+
+**Example 2:**
+
+```
+Input: a = 4, b = 1
+Output: "aabaa"
+```
+
+### Constraints
+
+- `0 <= a, b <= 100`
+- It is guaranteed that such a string `s` exists for the given `a` and `b`.
+
+---
+
+## Solution for String Without AAA or BBB Problem
+
+To solve this problem, we need to construct a string of length `a + b` containing exactly `a` 'a' letters and `b` 'b' letters. The key is to ensure that no substring 'aaa' or 'bbb' is formed.
+
+### Approach
+
+1. **Greedy Construction:**
+   - Start by determining the majority character (the one with the higher count).
+   - Add two of the majority character if more than one remains, followed by one of the minority character.
+   - If both characters have equal counts, alternate between the two to avoid 'aaa' or 'bbb'.
+
+2. **Ensure No Consecutive Repetitions:**
+   - Keep track of the previous two characters to ensure that adding another of the same character won't create three consecutive identical characters.
+
+### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    string strWithout3a3b(int a, int b) {
+        string result;
+        while (a > 0 || b > 0) {
+            bool writeA = false;
+            int n = result.size();
+            if (n >= 2 && result[n - 1] == result[n - 2]) {
+                if (result[n - 1] == 'b') {
+                    writeA = true;
+                }
+            } else {
+                if (a >= b) {
+                    writeA = true;
+                }
+            }
+
+            if (writeA) {
+                result += 'a';
+                --a;
+            } else {
+                result += 'b';
+                --b;
+            }
+        }
+        return result;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public String strWithout3a3b(int a, int b) {
+        StringBuilder result = new StringBuilder();
+        while (a > 0 || b > 0) {
+            boolean writeA = false;
+            int n = result.length();
+            if (n >= 2 && result.charAt(n - 1) == result.charAt(n - 2)) {
+                if (result.charAt(n - 1) == 'b') {
+                    writeA = true;
+                }
+            } else {
+                if (a >= b) {
+                    writeA = true;
+                }
+            }
+
+            if (writeA) {
+                result.append('a');
+                --a;
+            } else {
+                result.append('b');
+                --b;
+            }
+        }
+        return result.toString();
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def strWithout3a3b(self, a: int, b: int) -> str:
+        result = []
+        while a > 0 or b > 0:
+            writeA = False
+            if len(result) >= 2 and result[-1] == result[-2]:
+                if result[-1] == 'b':
+                    writeA = True
+            else:
+                if a >= b:
+                    writeA = True
+
+            if writeA:
+                result.append('a')
+                a -= 1
+            else:
+                result.append('b')
+                b -= 1
+
+        return ''.join(result)
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- **Time Complexity**: $O(a + b)$, where `a` and `b` are the counts of 'a' and 'b' respectively. We iterate through each character once.
+- **Space Complexity**: $O(a + b)$, for storing the result string.
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['ImmidiSivani'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>
+```