Majority Element II

Author: rishabhv12

10-October/05-Majority Element II.md

@@ -1,4 +1,4 @@
-## 04. Majority Element II
+## 06. Majority Element II
 
 
 The problem can be found at the following link: [Question Link](https://leetcode.com/problems/majority-element-ii/description/)

10-October/06-Integer Break.md

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+## 06. Integer Break 
+
+
+The problem can be found at the following link: [Question Link](https://leetcode.com/problems/majority-element-ii/description/)
+
+
+### My Approach
+
+
+1. Initialize variables:
+   - `n` to store the size of the input vector `nums`.
+   - Create an unordered map `cnt` to store the count of occurrences of each element.
+   - Initialize an empty vector `ans` to store the majority elements.
+
+2. Iterate through the `nums` vector using a for loop:
+   - Use the `cnt` map to count the occurrences of each element in the `nums` vector.
+
+3. Iterate through the elements in the `cnt` map:
+   - Check if the count (`j->second`) of an element is greater than one-third of the total elements (`n/3`).
+   - If the count meets the condition, add the element (`j->first`) to the `ans` vector.
+
+4. After both loops have finished, the `ans` vector will contain the majority elements (elements that appear more than one-third of the time) in the input vector `nums`.
+
+5. Return the `ans` vector as the result.
+
+
+
+### Time and Auxiliary Space Complexity
+
+- **Time Complexity**: `O(n)` 
+- **Auxiliary Space Complexity**: `O(n)`
+
+
+
+### Code (C++)
+
+```cpp
+
+class Solution {
+public:
+    vector<int> majorityElement(vector<int>& nums) {  
+        int n=nums.size();
+        unordered_map<int, int> cnt;
+        vector<int> ans;
+        for(int i=0;i<n;i++){
+            cnt[nums[i]]++;
+        }
+        for(auto j=cnt.begin();j!=cnt.end();j++){
+            if(j->second>n/3)ans.push_back(j->first);
+        }
+        return ans;
+    }
+};
+
+```
+
+### Contribution and Support
+
+For discussions, questions, or doubts related to this solution, please visit our [discussion section](https://leetcode.com/discuss/general-discussion). We welcome your input and aim to foster a collaborative learning environment.
+
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