From 68f0bfda10d73c59ba0159cb8f1738d45a0378ff Mon Sep 17 00:00:00 2001
From: ImmidiSivani <147423543+ImmidiSivani@users.noreply.github.com>
Date: Sun, 28 Jul 2024 15:47:20 +0530
Subject: [PATCH] solution added to 984
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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
+
+
+
+
+
+```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;
+ }
+};
+```
+
+
+
+
+
+```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();
+ }
+}
+```
+
+
+
+
+
+```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)
+```
+
+
+
+
+#### 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.
+
+---
+
+