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Binary Search Implementation

Certification: Basic Level Accuracy: 0% Submissions: 0 Points: 8

Write a JavaScript program to implement binary search algorithm to find the target value in a sorted array. Return the index of the target if found, otherwise return -1.

Example 1
    
  
  • Input: nums = [-1,0,3,5,9,12], target = 9
    • 
  
  • Output: 4
    • 
  
  • Explanation: 
    • We search for target 9 in the sorted array. 
    • Using binary search, we check the middle element and narrow down the search space. 
    • We find that 9 exists at index 4 in the array. 
    • Therefore, we return index 4.
    •
Example 2
    
  
  • Input: nums = [-1,0,3,5,9,12], target = 2
    • 
  
  • Output: -1
    • 
  
  • Explanation: 
    • We search for target 2 in the sorted array. 
    • Using binary search, we systematically eliminate half the array in each iteration. 
    • After checking all possible positions, 2 is not found in the array. 
    • Therefore, we return -1 to indicate the target is not present.
    •
Constraints
    
  
  • 1 ≤ nums.length ≤ 10^4
    • 
  
  • -10^4 < nums[i], target < 10^4
    • 
  
  • All integers in nums are unique
    • 
  
  • nums is sorted in ascending order
    • 
  
  • Time Complexity: O(log n)
    • 
  
  • Space Complexity: O(1)
    •
ArraysAlgorithmsHCL TechnologiesSamsung
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Hint
Solution

Solution Hints

    
  
  • Use two pointers: left starting at 0 and right starting at array length - 1
    • 
  
  • Calculate the middle index as (left + right) / 2 in each iteration
    • 
  
  • Compare the middle element with the target value
    • 
  
  • If middle element equals target, return the middle index
    • 
  
  • If middle element is less than target, search in the right half
    • 
  
  • If middle element is greater than target, search in the left half
    • 
  
  • Continue until left pointer exceeds right pointer
    •

Steps to solve by this approach:

 Step 1: Initialize left pointer to 0 and right pointer to array length - 1.

 Step 2: Enter a loop that continues while left is less than or equal to right.

 Step 3: Calculate the middle index using (left + right) / 2.

 Step 4: Compare the element at middle index with the target value.

 Step 5: If they match, return the middle index as the result.

 Step 6: If middle element is smaller than target, update left = mid + 1.

 Step 7: If middle element is larger than target, update right = mid - 1.

 Step 8: If loop ends without finding target, return -1.

Submitted Code :