Search in Rotated Sorted Array II in C++

C++Server Side ProgrammingProgramming

Consider we have an array sorted in ascending order. That is rotated at some pivot unknown to us beforehand. For example, if the array is like [0,0,1,2,2,5,6], this might become [2,5,6,0,0,1,2]. We have a target value to search. If that is found in the array, then return true, otherwise return false. So if the array is like [2,5,6,0,0,1,2], and target is 0, then the output will be 0

Let us see the steps −

  • low := 0 and high := size of array

  • while low < high

    • mid := low + (high - low)/2

    • if nums[mid] = target, then return true

    • if nums[low] = nums[mid] and nums[high - 1] = nums[mid], then

      • increase low by 1 and decrease high by 1, and continue for the next iteration

    • if nums[low] <= nums[mid], then

      • if target >= nums[low] and target < nums[mid], then high := mid, otherwise low := mid + 1

    • Otherwise

      • if target <= nums[high - 1] and target > nums[mid], then low := mid + 1, otherwise high := mid

    • return false

Example

Let us see the following implementation to get better understanding −

class Solution(object):
   def search(self, nums, target):
      """
      :type nums: List[int]
      :type target: int
      :rtype: int
      """
      low = 0
      high = len(nums)
      while low<high:
         mid = low + (high-low)//2
         print(mid)
      if nums[mid] == target:
         return True
      if nums[low] == nums[mid] and nums[high-1] == nums[mid]:
         low +=1
         high -=1
         continue
      if nums[low]<=nums[mid]:
         if target >=nums[low] and target <nums[mid]:
            high = mid
         else:
            low = mid+1
      else:
         if target<=nums[high-1] and target>nums[mid]:
            low = mid+1
      else:
         high = mid
   return False

Input

[2,5,6,0,0,1,2]
0

Output

true
raja
Published on 03-Feb-2020 14:47:34
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