✅ C++ || 4 SOLUTIONS || T.C. : O(logn), S.C.: O(1)
\u2705 APPROACH 1:\n\n1.  TIME COMPLEXITY : O(logn)\n2.  SPACE COMPLEXITY : O(1)\n\n# BINARY SEARCH SOLUTION\n\n\nint findMin(vector<int>& nums) {\n        int 
* \u2705 ***APPROACH 1:***\n\n**1.  TIME COMPLEXITY : O(logn)\n2.  SPACE COMPLEXITY : O(1)**\n\n# **BINARY SEARCH SOLUTION**\n\n```\nint findMin(vector<int>& nums) {\n        int left=0,right=nums.size()-1;\n        \n        while(left<right){\n            int mid=left+(right-left)/2;\n\n            if(nums[mid]>nums[right])left=mid+1;              // in this case we assure the minimum is in the right side\n            else right=mid;                                  // in this case we assure the minimum is in the left side\n        }\n        return nums[left];\n    }\n```\n\n* ***\u2705 APPROACH 2 :***\n\n**1.  TIME COMPLEXITY : O(n)\n2.  SPACE COMPLEXITY : O(1)**\n\n# **SIMPLE TRAVERSAL**\n\n```\nint findMin(vector<int>& nums) {\n        int ans=INT_MAX;\n        for(int i=0;i<nums.size();i++){\n            ans=min(ans,nums[i]);\n        }\n        return ans;\n    }\n```\n\n* ***\u2705APPROACH 3 :***\n\n**1.  TIME COMPLEXITY : O(n)\n2.  SPACE COMPLEXITY : O(1)**\n\n# **TRAVERSE FROM END**\n\n```\nint findMin(vector<int>& nums) {\n        \n        for(int i=nums.size()-1;i>0;i--){\n            if(nums[i]<nums[i-1])return nums[i];\n        }\n        return nums[0];\n    }\n```\n\n* ***\u2705APPROACH 4 :***\n\n**1.  TIME COMPLEXITY : O(nlogn)\n2.  SPACE COMPLEXITY : O(1)**\n\n# **SORTING**\n\n```\nint findMin(vector<int>& nums) {\n        sort(nums.begin(),nums.end());\n        return nums[0];\n    }\n```\n\n![image](https://assets.leetcode.com/users/images/d0bc009b-5bb4-4333-a525-1e366d34ba75_1654317187.9661255.png)\n\n\n\uD83D\uDE4C\uD83D\uDE4C HAPPY CODING !!
['Binary Search', 'C', 'Binary Tree', 'C++']