search ortated arr

Author: sachuverma

DSA Crack Sheet/README.md

@@ -95,6 +95,7 @@
 
 - [First and last occurrences of X](https://practice.geeksforgeeks.org/problems/first-and-last-occurrences-of-x/0# "view question") - [Cpp Solution](./solutions/First%20and%20last%20occurrences%20of%20X.cpp)
 - [Value equal to index value](https://practice.geeksforgeeks.org/problems/value-equal-to-index-value1330/1# "view question") - [Cpp Solution](./solutions/Value%20equal%20to%20index%20value.cpp)
+- [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/ "view question") - [Cpp Solution](./solutions/Search%20in%20Rotated%20Sorted%20Array.cpp)
 - []( "view question") - [Cpp Solution](./solutions/.cpp)
 
 ### Linked-List

DSA Crack Sheet/solutions/Search in Rotated Sorted Array.cpp

@@ -0,0 +1,62 @@
+/*
+Search in Rotated Sorted Array
+==============================
+
+There is an integer array nums sorted in ascending order (with distinct values).
+
+Prior to being passed to your function, nums is rotated at an unknown pivot index k (0 <= 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 rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.
+
+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
+
+Constraints:
+1 <= nums.length <= 5000
+-104 <= nums[i] <= 104
+All values of nums are unique.
+nums is guaranteed to be rotated at some pivot.
+-104 <= target <= 104
+
+Follow up: Can you achieve this in O(log n) time complexity?
+*/
+
+class Solution
+{
+public:
+  int search(vector<int> &nums, int target)
+  {
+    int i = 0, j = nums.size() - 1, mid = -1;
+    while (i <= j)
+    {
+      mid = (i + j) / 2;
+
+      if (target == nums[mid])
+        return mid;
+      if (nums[i] <= nums[mid])
+      {
+        if (target <= nums[mid] && target >= nums[i])
+          j = mid - 1;
+        else
+          i = mid + 1;
+      }
+      else
+      {
+        if (target >= nums[mid] && target <= nums[j])
+          i = mid + 1;
+        else
+          j = mid - 1;
+      }
+    }
+    return -1;
+  }
+};