Find position of an element in a sorted array of infinite numbers (Galloping Search)

Medium Binary Search Search on Answer Range Original on GeeksforGeeks

Find position of an element in a sorted array of infinite numbers (Galloping Search) is a medium binary search problem solved with the search on answer range pattern. The best approach, optimal (galloping / exponential search), runs in O(log p) time and O(1) space. Below are 2 approaches in Java, from linear scan up.

Problem

A sorted array is so large that its length is unknown (treat it as infinite). You can only read values by index. Find the position of target, or return -1.

Examples

Example 1

Input: arr = [3, 5, 7, 9, 10, 90, 100, 130, 140, 160, ...], target = 10
Output: 4

Example 2

Input: same array, target = 8
Output: -1

Constraints

The array is sorted and too large to know its length.
You may only read arr.get(i).

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

Approach	Time	Space
Linear scan	`O(p)`	`O(1)`
Optimal (galloping / exponential search)	`O(log p)`	`O(1)`

1Linear scan

TimeO(p)p is the position of the target.

SpaceO(1)

Walk from index 0 until you reach a value >= target.

i = 0; while arr[i] < target: i++.
Return i if arr[i] == target, else -1.

Java

class Solution {
    interface ArrayReader { int get(int i); }

    int search(ArrayReader arr, int target) {
        int i = 0;
        while (arr.get(i) < target) i++;
        return arr.get(i) == target ? i : -1;
    }
}

2Optimal (galloping / exponential search)

TimeO(log p)log p doublings, then a binary search over a range of size p.

SpaceO(1)

Find bounds by doubling hi until arr.get(hi) >= target; the previous hi is a safe lo. Then run an ordinary binary search on [lo, hi].

lo = 0, hi = 1; while arr.get(hi) < target: lo = hi; hi *= 2.
Binary search on [lo, hi].

Java

class Solution {
    interface ArrayReader { int get(int i); }

    int search(ArrayReader arr, int target) {
        int lo = 0, hi = 1;
        while (arr.get(hi) < target) {
            lo = hi;
            hi *= 2;
        }
        while (lo <= hi) {
            int mid = lo + (hi - lo) / 2;
            int v = arr.get(mid);
            if (v == target) return mid;
            if (v < target) lo = mid + 1; else hi = mid - 1;
        }
        return -1;
    }
}

Edge cases to test

Target at index 0
Target smaller than arr[0]

Hints

Hint 1

Double hi (1, 2, 4, 8, ...) until arr[hi] >= target. Then binary search inside [hi / 2, hi].

FAQ

What is the best time complexity for Find position of an element in a sorted array of infinite numbers (Galloping Search)?

Optimal (galloping / exponential search) runs in O(log p) time and O(1) extra space. log p doublings, then a binary search over a range of size p.

Which pattern does Find position of an element in a sorted array of infinite numbers (Galloping Search) use?

It is a binary search problem that uses the search on answer range pattern. Other problems with the same pattern: Pow(x, n) (Binary Exponentiation), Find Nth root of M, Sqrt(x).

Is there a brute force solution for Find position of an element in a sorted array of infinite numbers (Galloping Search)?

Yes. Linear scan takes O(p) time and O(1) space. Walk from index 0 until you reach a value = target.

Which edge cases should I test for Find position of an element in a sorted array of infinite numbers (Galloping Search)?

Target at index 0; Target smaller than arr[0].