Binary Search First Occurrence

Find the first copy of a duplicated target by recording matches and continuing to search the left half.

Algorithm

Basic Implementation

Basic.java

public class Basic {
    public static void main(String[] args) {
        int[] arr = {1, 2, 4, 4, 4, 7, 9};
        int target = 4;
        int lo = 0;
        int hi = arr.length - 1;
        int result = -1;
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            if (arr[mid] == target) {
                result = mid;
                hi = mid - 1;
            } else if (arr[mid] < target) {
                lo = mid + 1;
            } else {
                hi = mid - 1;
            }
        }
        System.out.println(result);
    }
}

Complexity

Time: O(log n)
Space: O(1)

Implementation notes

Java stores the sorted input in a primitive int[], so arr[mid] reads copy primitive values from checked array slots while lo, hi, mid, and result remain local int variables.
The midpoint uses the overflow-safe form lo + (hi - lo) / 2 inside the while (lo <= hi) bounds guard. Window updates keep future indexed reads inside 0 through arr.length - 1.
On a match, the code assigns result = mid and then moves hi = mid - 1, continuing left to find the first occurrence among duplicate 4 values. Lower values move lo = mid + 1; higher values move hi = mid - 1.
The replay-visible states show result changing from -1 to 3, then to 2, before lo > hi ends the loop. Aside from the initial array, the search uses primitive locals and creates no ongoing JVM GC pressure.

execution replay The checked-in replay follows the language-neutral state table for `search-binary-first`.

cross-language comparison This Java DSA version keeps the same data and final output as every other DSA book in this wave.