Binary Search¶

Question¶

Given an array of integers nums sorted in ascending order, and an integer target, return the index of target if it exists. Otherwise, return -1. The algorithm must run in \(O(\log n)\) time.

Solution¶

Pattern¶

Iterative Binary Search Maintain a search window defined by left and right pointers. Calculate the middle index, compare it to the target, and discard half of the search space at every step until the target is found or the window collapses.

How to Identify¶

The input is explicitly stated to be sorted (arrays, matrices).
The required time complexity is strictly \(O(\log n)\).
You are searching for a specific value, a boundary, or a minimum/maximum satisfying a condition.

Description¶

Step-by-step explanation:

Handle edge cases. If the array is empty or the target is strictly outside the bounds of the lowest and highest elements, return -1 immediately.
Initialize left to the 0th index and right to the last index (n - 1).
Loop as long as left is less than or equal to right. (The <= is critical to evaluate a search space of exactly 1 element).
Calculate mid = left + (right - left) / 2.
Check if the element at mid is the target. If yes, return mid.
If the target is strictly greater than the element at mid, the target must be in the right half. Shrink the window by moving left = mid + 1.
If the target is strictly less than the element at mid, the target must be in the left half. Shrink the window by moving right = mid - 1.
If the loop terminates without returning, the target does not exist. Return -1.

The Intuition¶

Think of finding a word in a physical dictionary. You don't read page 1, then page 2, then page 3 (Linear Search). Instead, you open the book exactly to the middle. If the word you are looking for comes alphabetically after the words on the current page, you know with 100% certainty that the word is in the right half of the book. You completely ignore the left half. You then open the right half to its middle, and repeat the process. Because you are throwing away exactly half of the remaining possibilities at every single step, you narrow down the answer exponentially fast.

Complexity¶

Label	Worst	Average
Time Complexity	\(O(\log n)\)	\(O(\log n)\)
Space Complexity	\(O(1)\)	\(O(1)\)

Time Complexity¶

In the worst case, the search space is divided by 2 repeatedly until only 1 element remains. The maximum number of divisions is \(\log_2(N)\). Therefore, time complexity is \(O(\log n)\).

Space Complexity¶

The iterative approach only requires three scalar integer pointers (left, right, mid). It modifies no arrays and uses no auxiliary data structures. Auxiliary space is strictly \(O(1)\). (Note: A recursive implementation would use \(O(\log n)\) space on the call stack).

Code¶

class Solution {
    public int search(int[] nums, int target) {
        if (nums == null || nums.length == 0 || target < nums[0] || target > nums[nums.length - 1]) {
            return -1;
        }

        int left = 0;
        int right = nums.length - 1;

        // `<=` is necessary to check the final element when left == right
        while (left <= right) {
            // Avoids integer overflow which occurs with (left + right) / 2
            int mid = left + (right - left) / 2;

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

        return -1;
    }
}

Caveats¶

Integer Overflow Bug: Using mid = (left + right) / 2 will crash in production systems if left + right exceeds 2,147,483,647 (the max 32-bit integer). It wraps around to a negative number, causing an ArrayIndexOutOfBoundsException. Always use left + (right - left) / 2 or the bitwise unsigned right shift (left + right) >>> 1.
Infinite Loops: If you forget to add/subtract 1 when updating pointers (e.g., writing left = mid instead of left = mid + 1), the window will never collapse when left and right are adjacent, resulting in an infinite loop (Time Limit Exceeded).

Concepts to Think About¶

Lower Bound (First Occurrence): If the array has duplicates and you need the first occurrence of the target, you do not return when nums[mid] == target. Instead, you record ans = mid and force the search to continue leftward by setting right = mid - 1.
Upper Bound (Last Occurrence): Conversely, to find the last occurrence, record ans = mid and force the search rightward by setting left = mid + 1.
Search Space on Answers: Binary Search isn't just for arrays. You can binary search an abstract mathematical range (like searching for the square root of \(X\) by binary searching between \(0\) and \(X\)).

Logical Follow-up¶

Question: How would you search for a target in a sorted array that has been rotated an unknown number of times? (LeetCode 33: Search in Rotated Sorted Array)

Solution: You can still achieve \(O(\log n)\) time. In any rotated sorted array, if you pick a mid point, at least one half of the array (either left to mid, or mid to right) is guaranteed to be perfectly sorted. You determine which half is sorted by comparing nums[left] to nums[mid]. Once you know which half is sorted, you check if the target falls within the bounds of that sorted half. If it does, you binary search that half. If it doesn't, the target must be in the other, unsorted half.

Question: What if the array has an unknown, infinite length (e.g., a data stream), and you cannot use nums.length?

Solution: You must first find bounds. Set left = 0 and right = 1. Check if target > nums[right]. If it is, exponentially expand your bounds: left = right and right = right * 2. Repeat this expansion until nums[right] >= target. Now you have a defined finite window [left, right] where standard Binary Search can be safely executed. Time complexity remains \(O(\log n)\) where \(n\) is the index of the target.