Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/702.html

702. Search in a Sorted Array of Unknown Size (Medium)

Given an integer array sorted in ascending order, write a function to search target in nums.  If target exists, then return its index, otherwise return -1. However, the array size is unknown to you. You may only access the array using an ArrayReader interface, where ArrayReader.get(k) returns the element of the array at index k (0-indexed).

You may assume all integers in the array are less than 10000, and if you access the array out of bounds, ArrayReader.get will return 2147483647.

 

Example 1:

Input: array = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: array = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

 

Note:

1. You may assume that all elements in the array are unique.
2. The value of each element in the array will be in the range [-9999, 9999].

Companies:
Google

Related Topics:
Binary Search

Similar Questions:

• Binary Search (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/
// Time: O(1)
// Space: O(1)
class Solution {
public:
    int search(const ArrayReader& reader, int target) {
        int minVal = reader.get(0);
        int L = 0, R = 9999 - minVal;
        while (L <= R) {
            int M = (L + R) / 2;
            int val = reader.get(M);
            if (val == target) return M;
            if (val > target) R = M - 1;
            else L = M + 1;
        }
        return -1;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/
// Time: O(1)
// Space: O(1)
// Ref: https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/discuss/151685/Shortest-and-cleanest-Java-solution-so-far
class Solution {
public:
    int search(const ArrayReader& reader, int target) {
        int L, R = 1;
        while (reader.get(R) < target) R *= 2;
        L = R / 2;
        while (L <= R) {
            int M = (L + R) / 2;
            int val = reader.get(M);
            if (val == target) return M;
            if (val > target) R = M - 1;
            else L = M + 1;
        }
        return -1;
    }
};

• Java
• C++
• Python
• Go

• class Solution {
      public int search(ArrayReader reader, int target) {
          if (reader.get(0) == target)
              return 0;
          int low = 0, high = 1;
          while (reader.get(high) < target) {
              low = high + 1;
              high <<= 1;
          }
          while (low <= high) {
              int mid = ((high - low) >> 1) + low;
              int num = reader.get(mid);
              if (num == target)
                  return mid;
              else if (num > target)
                  high--;
              else
                  low++;
          }
          return -1;
      }
  }
  
  ############
  
  /**
   * // This is ArrayReader's API interface.
   * // You should not implement it, or speculate about its implementation
   * interface ArrayReader {
   *     public int get(int index) {}
   * }
   */
  
  class Solution {
      public int search(ArrayReader reader, int target) {
          int left = 0, right = 20000;
          while (left < right) {
              int mid = left + right >> 1;
              if (reader.get(mid) >= target) {
                  right = mid;
              } else {
                  left = mid + 1;
              }
          }
          return reader.get(left) == target ? left : -1;
      }
  }
  

• // OJ: https://leetcode.com/problems/search-in-a-sorted-array-of-unknown-size/
  // Time: O(1)
  // Space: O(1)
  class Solution {
  public:
      int search(const ArrayReader& reader, int target) {
          int minVal = reader.get(0);
          int L = 0, R = 9999 - minVal;
          while (L <= R) {
              int M = (L + R) / 2;
              int val = reader.get(M);
              if (val == target) return M;
              if (val > target) R = M - 1;
              else L = M + 1;
          }
          return -1;
      }
  };
  

• # """
  # This is ArrayReader's API interface.
  # You should not implement it, or speculate about its implementation
  # """
  # class ArrayReader:
  #    def get(self, index: int) -> int:
  
  
  class Solution:
      def search(self, reader, target):
          """
          :type reader: ArrayReader
          :type target: int
          :rtype: int
          """
          left, right = 0, 20000
          while left < right:
              mid = (left + right) >> 1
              if reader.get(mid) >= target:
                  right = mid
              else:
                  left = mid + 1
          return left if reader.get(left) == target else -1
  
  
  

• /**
   * // This is the ArrayReader's API interface.
   * // You should not implement it, or speculate about its implementation
   * type ArrayReader struct {
   * }
   *
   * func (this *ArrayReader) get(index int) int {}
   */
  
  func search(reader ArrayReader, target int) int {
  	left, right := 0, 20000
  	for left < right {
  		mid := (left + right) >> 1
  		if reader.get(mid) >= target {
  			right = mid
  		} else {
  			left = mid + 1
  		}
  	}
  	if reader.get(left) == target {
  		return left
  	}
  	return -1
  }
  

All Problems

All Solutions