Find Target in Sorted Array (Binary Search)

🧩 Problem Statement

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.

🔍 Approaches

1. Binary Search (Iterative)

Since the array is sorted, we can use Binary Search.
We define a search space [left, right]. We check the mid element.

• If nums[mid] == target, we found it.
• If nums[mid] < target, the target must be in the right half, so we move left = mid + 1.
• If nums[mid] > target, the target must be in the left half, so we move right = mid - 1.

✨ Intuition

Binary search works by repeatedly dividing the search interval in half.
Imagine looking for a word in a dictionary. You open the middle.

• If the word is alphabetically after, you discard the first half.
• If before, you discard the second half.
• Repeat until found.

🔥 Algorithm Steps

1. Initialize left = 0, right = n - 1.
2. While left <= right:

• Calculate mid = left + (right - left) / 2. (Avoids overflow compared to (left + right) / 2)
• If nums[mid] == target, return mid.
• If nums[mid] < target, left = mid + 1.
• If nums[mid] > target, right = mid - 1.

3. If loop finishes, return -1.

⏳ Time & Space Complexity

• Time Complexity: $O(\log n)$, halving search space each step.
• Space Complexity: $O(1)$, iterative.

🚀 Code Implementations

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 - left) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] < target) left = mid + 1;
else right = mid - 1;
}
return -1;
}
};

Python

class Solution:
def search(self, nums: List[int], target: int) -> int:
left, right = 0, len(nums) - 1
while left <= right:
mid = (left + right) // 2
if nums[mid] == target:
return mid
elif nums[mid] < target:
left = mid + 1
else:
right = mid - 1
return -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 - left) / 2;
if (nums[mid] == target) return mid;
if (nums[mid] < target) left = mid + 1;
else right = mid - 1;
}
return -1;
}
}

🌍 Real-World Analogy

Guess the Number Game:

Someone thinks of a number between 1 and 100.

• You guess 50.
• They say "Higher".
• You know it's between 51-100. You guess 75.
• They say "Lower".
• You know it's between 51-74.
• You keep halving the range efficiently.

🎯 Summary

✅ O(log n): Most efficient search for sorted data.
✅ Overflow Safety: Use left + (right - left) / 2.