Jump Search Algorithm In Java

// SOLVING THIS WITH AN AI ASSISTANT (2026)

If you are working through this problem with an AI coding assistant — Claude, ChatGPT, Cursor chat, Gemini, GitHub Copilot, Aider, or any agent — the goal isn’t to ask for the answer. It is to use the tool to understand the pattern. The prompt sequence I’d run:

Spec it back to me first. “In your own words, what is this problem actually testing? What’s the smallest example that fails the naive approach?”
Brute-force first, optimize after. “Write the simplest correct solution, even if it’s O(n²). Don’t optimize. Just make it correct, with comments explaining each step.”
Ask for the upgrade. “Now show me the optimal solution. What insight makes it possible? What pattern is this an instance of?”
Stress-test it. “Generate 10 edge cases — empty input, single element, duplicates, max size, sorted, reverse-sorted. Run my solution against each.”

The pattern matters more than the answer. If the agent just hands you optimized code, you’ve trained yourself to lose interviews.

Jump search algorithm is a pretty new algorithm to search for an element in a sorted array.

The idea of jump search is to skip the number of comparisons by jumping the indices by length m while performing the searching thus getting a better time complexity than linear search.

For jump search m is determined by the square root of the length of the array m= √n

Let’s take the below sorted array A[], the target to find is 16, in a linear search algorithm it would take O(n) time complexity to find the number 16, in jump search we would jump the indices of size m= √n from 0->4->8->12->16 and would be able to find in O(√n)

Sorted Array

Algorithm:

Find jump block size m=√n.
If A[i] == target return i (i= index to compare).
If A[i] < target i= m and increment m + √n.
Repeat the above step till m<n-1.
If A[i] >target perform linear search from A[i- √n] to A[i].

Heres the video explanation

Now we have seen the algorithm let’s see how the code looks

public static int jumpSearch(int[] a, int target) {
    int m = (int) Math.floor(Math.sqrt(a.length));

    int currentLastIndex = m-1;
    
    // Jump to next block as long as target element is > currentLastIndex
    while (currentLastIndex < a.length && target > a[currentLastIndex]) {
        currentLastIndex += m;
    }

    // Find the accurate position of target number using Linear Search
    for (int currentSearchIndex = currentLastIndex - m + 1;
         currentSearchIndex <= currentLastIndex && currentSearchIndex < a.length; currentSearchIndex++) {
        if (target == a[currentSearchIndex]) {
            return currentSearchIndex;
        }
    }
    // If target number not found, return negative integer as element position.
    return -1;
}

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rajendra