Floor in a Sorted Array

Easy Binary Search Bisect Original on GeeksforGeeks

Floor in a Sorted Array is a easy binary search problem solved with the bisect pattern. The best approach, optimal (bisect_right - 1), runs in O(log n) time and O(1) space. Below are 2 approaches in Java, from linear scan up.

Problem

Given a sorted array and a value x, return the index of the floor of x: the largest element that is less than or equal to x. If none exists, return -1.

Examples

Example 1

Input: arr = [1, 2, 8, 10, 10, 12, 19], x = 5
Output: 1
Why: The largest value <= 5 is 2, at index 1.

Example 2

Input: arr = [1, 2, 8], x = 0
Output: -1

Constraints

1 <= arr.length <= 10^6, sorted non-decreasing.
With duplicates, return the last index of the floor value.

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(n)`	`O(1)`
Optimal (bisect_right - 1)	`O(log n)`	`O(1)`

1Linear scan

TimeO(n)

SpaceO(1)

Keep the last index whose value is at most x.

Walk while arr[i] <= x; answer i - 1.

Java

class Solution {
    public int findFloor(int[] arr, int x) {
        int i = 0;
        while (i < arr.length && arr[i] <= x) i++;
        return i - 1;
    }
}

2Optimal (bisect_right - 1)

TimeO(log n)

SpaceO(1)

bisect_right(x) is the first index with a value greater than x, so the index just before it holds the largest value <= x, or -1 if there is none.

Find the first index with arr[i] > x.
Return that index - 1.

Java

class Solution {
    public int findFloor(int[] arr, int x) {
        int lo = 0, hi = arr.length;
        while (lo < hi) {
            int mid = (lo + hi) >>> 1;
            if (arr[mid] > x) hi = mid; else lo = mid + 1;
        }
        return lo - 1;
    }
}

Edge cases to test

x smaller than every element (-1)
x equal to a repeated value (last occurrence)

Hints

Hint 1

The floor is at bisect_right(x) - 1.

FAQ

What is the best time complexity for Floor in a Sorted Array?

Optimal (bisect_right - 1) runs in O(log n) time and O(1) extra space.

Which pattern does Floor in a Sorted Array use?

It is a binary search problem that uses the bisect pattern. Other problems with the same pattern: bisect_left, bisect_right, Search Insert Position, Ceil in a Sorted Array.

Is there a brute force solution for Floor in a Sorted Array?

Yes. Linear scan takes O(n) time and O(1) space. Keep the last index whose value is at most x.

Which edge cases should I test for Floor in a Sorted Array?

x smaller than every element (-1); x equal to a repeated value (last occurrence).