add codeforces solutions folder

leetcode/lc33.cpp

@@ -0,0 +1,45 @@
+class Solution{
+    public:
+        int search(vector<int> nums, int target){
+            int start = 0, end = nums.size() - 1;
+            while(left<=right){
+                int mid = start + (end - start)/2;
+                if(nums[mid]==target) return mid;
+
+                if(nums[start]<=nums[mid]){
+                    if(target>=nums[start] && target< nums[mid])
+                        end = mid-1
+                    else
+                        start = mid+1;
+                }
+                else{
+                      if(target>nums[mid] && target<= nums[end])
+                        start = mid+1;
+                    else
+                        end = mid-1;
+
+                }
+            return -1;
+        }
+};
+
+//TC : o(logn), SC: o(1)
+
+/*
+Approach:
+
+This is a modified binary search for a rotated sorted array.
+The array was originally sorted in ascending order and then rotated once,
+which means there is exactly one pivot point. Because of this, at any
+given mid index, at least one half (left or right) must still be sorted.
+
+At each step:
+1) Find the middle index.
+2) Determine which half is sorted.
+3) Check if the target lies within the sorted half.
+   - If yes, move the search to that half.
+   - If no, move to the other half.
+
+By always eliminating half of the array, we maintain
+O(log n) time complexity without explicitly finding the pivot.
+*/