day9

Author: TharunDharmaraj

Search in rotated sorted array/modifiedBinary.java

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+class Solution {
+    public int search(int[] nums, int target) {
+        int left = 0, right = nums.length - 1 ;
+        while(left <= right){
+            int mid = (left + right) / 2 ;
+            if(nums[mid] == target){
+                return mid;
+            }
+            if (nums[left] <= nums[mid]){
+                if(target >= nums[left] && target < nums[mid]) {
+                    right = mid - 1;
+                }else{
+                    left = mid + 1;
+                }
+            }else{
+                if(target > nums[mid] && target <= nums[right]){
+                    left = mid + 1;
+                }else{
+                    right = mid - 1;
+                }
+            }
+        }
+        return -1;
+    }
+}

Search in rotated sorted array/ques.java

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+Link: https://leetcode.com/problems/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 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.
+
+ 
+
+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