Given a sorted array and a value x, the ceiling of x is the smallest element in array greater than or equal to x, and the floor is the greatest element smaller than or equal to x. Assume than the array is sorted in non-decreasing order. Write efficient functions to find floor and ceiling of x.
Examples :
For example, let the input array be {1, 2, 8, 10, 10, 12, 19}
For x = 0: floor doesn't exist in array, ceil = 1
For x = 1: floor = 1, ceil = 1
For x = 5: floor = 2, ceil = 8
For x = 20: floor = 19, ceil doesn't exist in array
In below methods, we have implemented only ceiling search functions. Floor search can be implemented in the same way.
Method 1 (Linear Search)
Algorithm to search ceiling of x:
1) If x is smaller than or equal to the first element in array then return 0(index of first element)
2) Else Linearly search for an index i such that x lies between arr[i] and arr[i+1].
3) If we do not find an index i in step 2, then return -1
Java
class Main{ /* Function to get index of ceiling of x in arr[low..high] */ static int ceilSearch(int arr[], int low, int high, int x) { int i; /* If x is smaller than or equal to first element,then return the first element */ if(x <= arr[low]) return low; /* Otherwise, linearly search for ceil value */ for(i = low; i < high; i++) { if(arr[i] == x) return i; /* if x lies between arr[i] and arr[i+1] including arr[i+1], then return arr[i+1] */ if(arr[i] < x && arr[i+1] >= x) return i+1; } /* If we reach here then x is greater than the last element of the array, return -1 in this case */ return -1; } /* Driver program to check above functions */ public static void main (String[] args) { int arr[] = {1, 2, 8, 10, 10, 12, 19}; int n = arr.length; int x = 3; int index = ceilSearch(arr, 0, n-1, x); if(index == -1) System.out.println("Ceiling of "+x+" doesn't exist in array"); else System.out.println("ceiling of "+x+" is "+arr[index]); } } |
Output :
ceiling of 3 is 8
Time Complexity : O(n)
Auxiliary Space: O(1)
As constant extra space is used.
Method 2 (Binary Search)
Instead of using linear search, binary search is used here to find out the index. Binary search reduces time complexity to O(Logn).
Java
class Main{ /* Function to get index of ceiling of x in arr[low..high]*/ static int ceilSearch(int arr[], int low, int high, int x) { int mid; /* If x is smaller than or equal to the first element, then return the first element */ if(x <= arr[low]) return low; /* If x is greater than the last element, then return -1 */ if(x > arr[high]) return -1; /* get the index of middle element of arr[low..high]*/ mid = (low + high)/2; /* low + (high - low)/2 */ /* If x is same as middle element, then return mid */ if(arr[mid] == x) return mid; /* If x is greater than arr[mid], then either arr[mid + 1] is ceiling of x or ceiling lies in arr[mid+1...high] */ else if(arr[mid] < x) { if(mid + 1 <= high && x <= arr[mid+1]) return mid + 1; else return ceilSearch(arr, mid+1, high, x); } /* If x is smaller than arr[mid], then either arr[mid] is ceiling of x or ceiling lies in arr[low...mid-1] */ else { if(mid - 1 >= low && x > arr[mid-1]) return mid; else return ceilSearch(arr, low, mid - 1, x); } } /* Driver program to check above functions */ public static void main (String[] args) { int arr[] = {1, 2, 8, 10, 10, 12, 19}; int n = arr.length; int x = 8; int index = ceilSearch(arr, 0, n-1, x); if(index == -1) System.out.println("Ceiling of "+x+" doesn't exist in array"); else System.out.println("ceiling of "+x+" is "+arr[index]); } } |
Output :
Ceiling of 20 doesn't exist in array
Time Complexity: O(Logn)
Auxiliary Space: O(Logn)
The extra space is used in recursive call stack.
Related Articles:
Floor in a Sorted Array
Find floor and ceil in an unsorted array
Please write comments if you find any of the above codes/algorithms incorrect, or find better ways to solve the same problem, or want to share code for floor implementation.
Please refer complete article on Ceiling in a sorted array for more details!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!