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Binary Search

Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

 

Example 1:

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

Example 2:

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

 

Constraints:

  • 1 <= nums.length <= 104
  • -104 < nums[i], target < 104
  • All the integers in nums are unique.
  • nums is sorted in ascending order.

This problem requires finding the index of a target integer within a sorted array. The most efficient approach is to use binary search, which has a time complexity of O(log n).

Algorithm:

  1. Initialization: Set two pointers, left (l) and right (r), to the start and end of the array, respectively.

  2. Iteration: While left is less than right:

    • Calculate the middle index mid as (left + right) / 2 (or using bitwise shift (left + right) >> 1 for slightly better performance).
    • Compare the element at nums[mid] with the target:
      • If nums[mid] is greater than or equal to target, the target (if it exists) must be in the left half or at mid. Therefore, update right to mid.
      • Otherwise, the target (if it exists) must be in the right half. Update left to mid + 1.

  3. Result: After the loop terminates (left >= right), there are two possibilities:

    • If nums[left] equals target, return left (the index of the target).
    • Otherwise, the target is not present in the array, so return -1.

Time Complexity: O(log n) because in each iteration, we are halving the search space.

Space Complexity: O(1) because we are using a constant amount of extra space (only the left, right, and mid variables).

Code Examples in Multiple Languages:

Python:

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left + right) >> 1  # Bitwise right shift for efficiency
            if nums[mid] >= target:
                right = mid
            else:
                left = mid + 1
        return left if nums[left] == target else -1

Java:

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

C++:

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

JavaScript:

var search = function(nums, target) {
    let left = 0, right = nums.length -1;
    while(left < right){
        let mid = Math.floor((left + right) / 2);
        if(nums[mid] >= target){
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return nums[left] === target ? left : -1;
};

The other languages (Go, TypeScript, Rust, C#) follow a very similar structure, using the same core binary search algorithm with minor syntax adjustments specific to each language. The efficiency remains the same across all implementations.