🧠 #Day11 of #90DaysOfDSA — Problem Solving & String Manipulation 🚀

Today’s DSA grind was all about string manipulation and logical arrays. I tackled four interesting problems that tested my attention to detail and logical reasoning.

---

🔸 Problem 1: Can Place Flowers

🪻 Concept: Greedy checking of adjacent spots
📌 Key Takeaway: A simple yet elegant greedy check to ensure flowers aren’t placed next to each other.

class Solution:
    def canPlaceFlowers(self, flowerbed: List[int], n: int) -> bool:
        if n == 0:
            return True  

        length = len(flowerbed)

        for i in range(length):
            if flowerbed[i] == 0:
                empty_left = (i == 0 or flowerbed[i - 1] == 0)
                empty_right = (i == length - 1 or flowerbed[i + 1] == 0)

                if empty_left and empty_right:
                    flowerbed[i] = 1
                    n -= 1
                    if n == 0:
                        return True

        return n == 0

---

🔸 Problem 2: Kids With the Greatest Number of Candies

🍬 Concept: Comparing each element to the max
📌 Key Takeaway: A simple one-liner with list comprehension helped make this very Pythonic.

class Solution:
    def kidsWithCandies(self, candies: List[int], extraCandies: int) -> List[bool]:
        maximum = max(candies)
        result = []
        for i in candies:
            if (i + extraCandies) >= maximum:
                result.append(True)
            else:
                result.append(False)
        return result

---

🔸 Problem 3: Reverse Vowels of a String

🔄 Concept: Two-pointer technique
📌 Key Takeaway: Efficient vowel reversal using two pointers for better space optimization.

class Solution:
    def reverseVowels(self, s: str) -> str:
        vowels = set("aeiouAEIOU")  
        s = list(s)
        left, right = 0, len(s) - 1

        while left < right:
            if s[left] not in vowels:
                left += 1
            elif s[right] not in vowels:
                right -= 1
            else:
                s[left], s[right] = s[right], s[left]
                left += 1
                right -= 1

        return "".join(s)

🔸 Problem 4: Reverse Words in a String

📝 Concept: Trim & reverse technique
📌 Key Takeaway: Strip and reverse split words while handling multiple spaces.

class Solution:
    def reverseWords(self, s: str) -> str:
        words = s.strip().split()

        left, right = 0, len(words) - 1
        while left < right:
            words[left], words[right] = words[right], words[left]
            left += 1
            right -= 1

        return " ".join(words)

---

💬 Reflection:
Day 11 was about sharpening basic operations in strings and arrays. I'm gaining more confidence with in-place modifications and optimizing my logic.

---

📌 Stay tuned for Day 12 where I’ll dive deeper into array manipulations and more intermediate problems.

🔗 GitHub Repository - All Code