Ceil in a Sorted Array
Ceil in a Sorted Array is a easy binary search problem solved with the bisect pattern.
The best approach, optimal (bisect_left), 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 ceil of x: the smallest element that is greater 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
2- Why
- The smallest value >= 5 is 8, at index 2.
Example 2
- Input
arr = [1, 2, 8], x = 20- Output
-1
Constraints
1 <= arr.length <= 10^5, sorted non-decreasing.- With duplicates, return the first index of the ceil 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_left) | O(log n) | O(1) |
1Linear scan
O(n)O(1)Return the first index whose value is at least x.
- Walk until arr[i] >= x.
class Solution {
public int findCeil(int[] arr, int x) {
for (int i = 0; i < arr.length; i++) if (arr[i] >= x) return i;
return -1;
}
}2Optimal (bisect_left)
O(log n)O(1)The first index with a value >= x is the ceil. If it equals n, nothing qualifies.
- lo = bisect_left(arr, x).
- Return lo < n ? lo : -1.
class Solution {
public int findCeil(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 < arr.length ? lo : -1;
}
}Edge cases to test
- x larger than every element (-1)
- x equal to an element
Hints
Hint 1
The ceil is at bisect_left(x), if that index is inside the array.
FAQ
What is the best time complexity for Ceil in a Sorted Array?
Optimal (bisect_left) runs in O(log n) time and O(1) extra space.
Which pattern does Ceil 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, Floor in a Sorted Array.
Is there a brute force solution for Ceil in a Sorted Array?
Yes. Linear scan takes O(n) time and O(1) space. Return the first index whose value is at least x.
Which edge cases should I test for Ceil in a Sorted Array?
x larger than every element (-1); x equal to an element.