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Search in rotated sorted arrays leetcode solution.

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 Search in rotated sorted arrays leetcode solution.

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
 

Constraints:

1 <= nums.length <= 5000
-104 <= nums[i] <= 104
All values of nums are unique.
nums is an ascending array that is possibly rotated.
-104 <= target <= 104



Solution in java :- 

class Solution {
public int search(int[] nums, int target) {
int pivot = pivotElement(nums);
if(nums[pivot] == target){
return pivot;
}
int lpivot = bs(nums,0,pivot-1,target);
int rpivot = bs(nums,pivot+1,nums.length-1,target);

      if(lpivot == -1 && rpivot == -1){
          return -1;
      }
      if(lpivot > -1){
          return lpivot;
      }
      else{
          return rpivot;
      }
    
}

//start of Binary Search
public int bs(int[] nums,int start,int end,int target)
{
    while(start <= end){
        
        int mid = start + (end - start)/2;
        if(nums[mid] == target){
            
            return mid;
            
        }
        else if(nums[mid] > target){
            
            end = mid - 1;
        }
        else{
            
            start = mid+1;
        }
        
        
    }
    
    return -1;
    
}
//end of Binary Search



//start of pivot function
public int pivotElement(int[] nums){
    int start = 0;
    int end = nums.length-1;
     int min = -1;
    while(start <= end)
     {
           int mid = start +(end-start)/2;
           int prev = (mid+nums.length-1)% nums.length;
           int next = (mid+1)%nums.length;
   
           if(nums[mid] <= nums[prev] && nums[mid] <= nums[next]){
                  min = mid;
                  return min;
        
            }
           else if(nums[start] <= nums[mid]){
                 if(nums[start] <= nums[end]){
                         end = mid-1;
                 }
                else{
                start = mid + 1;    
                } 
           }
          else{
             end = mid-1;
          }
        
    }       

   return min;
    
}
//End of pivot function
}


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