🧒 Kids With the Greatest Number of Candies

---

🔹 Problem Summary:

You’re given:

• A vector candies, where candies[i] is the number of candies the i-th kid has.

• An integer extraCandies.

Task:

• Return a boolean array result of the same length as candies.

• For each kid, add extraCandies to their current candies.

• If the new total is greater than or equal to the maximum among all kids, set result[i] = true; else false.

• Note: Multiple kids can have the greatest number of candies.

---

📌 Example:

Input: candies = [1, 2, 3], extraCandies = 1

• Maximum current candies = 3

Output: [false, true, true]

---

🔍 Approach:

1. Find the maximum number of candies any kid currently has.

2. Loop through each kid’s candies and:

   • Add the extra candies.

   • Compare with maxVal.

   • Store true or false in result vector.

3. Return the result vector.

---

✅ Final Code (LeetCode-style only):

class Solution {
public:
    vector<bool> kidsWithCandies(vector<int>& candies, int extraCandies) {
        int maxVal = *max_element(candies.begin(), candies.end());
        vector<bool> res;
        for (int candy : candies) {
            res.push_back(candy + extraCandies >= maxVal);
        }
        return res;
    }
};

---

🧠 Quick Note on max_element:

To find the maximum candies:

int maxVal = *max_element(candies.begin(), candies.end());

• max_element (from <algorithm>) returns an iterator to the largest element.

• We use * to dereference and get the value.

• Time complexity: O(n)

---

keep coding keep optimising 🚀