Added Search in rotated array (O(logn) complexity)

Medium/SearchInRotatedArray.cpp

@@ -0,0 +1,105 @@
+/*
+Problem Description :
+
+There is an integer array nums sorted in ascending order (with distinct values).
+Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].
+Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
+You must write an algorithm with O(log n) runtime complexity.
+
+Examples:
+
+Example 1:
+Input: nums = [4,5,6,7,0,1,2], target = 0
+Output: 4
+
+Example 2:
+Input: nums = [4,5,6,7,0,1,2], target = 3
+Output: -1
+
+Example 3:
+Input: nums = [1], target = 0
+Output: -1
+*/
+
+class Solution {
+public:
+    int search(vector<int>& nums, int target) {
+        //find break point
+        int break_p=0, min=0, max=nums.size()-1, flag=0;
+
+        if(nums[0]>nums[nums.size()-1]){
+            while(min<max){
+                if(nums[(min+max)/2]>nums[((min+max)/2)+1]){
+                    break_p=(min+max)/2;
+                    flag=1;
+                    break;
+                }
+                else if(nums[(min+max)/2]>nums[0])
+                    min=(min+max)/2;
+                else
+                    max=(min+max)/2;
+            }
+        }
+        if(flag==1){
+
+            //search in first half
+            min=0, max=break_p;
+            if(target>=nums[min] && target<=nums[max]){
+                while(min<max){
+                    if(nums[(min+max)/2]==target){
+                        return (min+max)/2;
+                    }
+                    else if(nums[(min+max)/2]>target)
+                        max=(min+max)/2;
+                    else{
+                        min=(min+max+1)/2;
+                        cout<<min<<endl;
+                    }
+                } 
+                if(nums[(min+max)/2]==target){
+                        return (min+max)/2;
+                    }
+            }
+
+          //search in second half
+            min=break_p+1, max=nums.size()-1;
+            if(target>=nums[min] && target<=nums[max]){
+                while(min<max){
+                    if(nums[(min+max)/2]==target){
+                        return (min+max)/2;
+                    }
+                    else if(nums[(min+max)/2]>target)
+                        max=(min+max)/2;
+                    else{
+                        min=(min+max+1)/2;
+                        cout<<min<<endl;
+                    }
+                } 
+                if(nums[(min+max)/2]==target){
+                        return (min+max)/2;
+                    }
+            }
+        }
+
+        else{
+            min=0, max=nums.size()-1;
+            if(target>=nums[min] && target<=nums[max]){
+                while(min<max){
+                    if(nums[(min+max)/2]==target){
+                        return (min+max)/2;
+                    }
+                    else if(nums[(min+max)/2]>target)
+                        max=(min+max)/2;
+                    else{
+                        min=(min+max+1)/2;
+                        cout<<min<<endl;
+                    }
+                } 
+                if(nums[(min+max)/2]==target){
+                        return (min+max)/2;
+                    }
+            }
+        }
+        return -1;
+    }
+};