Ceil in a Sorted Array

Author: Kondareddy1209

Leetcode/Ceil_in_a_Sorted_Array.py

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+"""
+Problem: Ceil in a Sorted Array
+
+Given a sorted array nums[] and an integer x,
+find the index (0-based) of the smallest element
+that is greater than or equal to x.
+
+This element is called the ceil of x.
+
+If such an element does not exist, return -1.
+
+Examples:
+Input:
+nums = [1, 2, 8, 10, 11, 12, 19]
+x = 5
+
+Output:
+2
+
+Explanation:
+The smallest element >= 5 is 8,
+which is present at index 2.
+
+--------------------------------------------------
+Approach: Binary Search
+--------------------------------------------------
+
+Since the array is sorted, Binary Search can be used.
+
+Observation:
+
+1. If nums[mid] >= x:
+   - nums[mid] can be a possible ceil.
+   - Store mid in ans.
+   - Search on the left side to find a smaller value
+     that is still >= x.
+
+2. If nums[mid] < x:
+   - Current element cannot be the ceil.
+   - Search on the right side.
+
+The last valid stored index will be the answer.
+
+--------------------------------------------------
+Dry Run
+--------------------------------------------------
+
+nums = [1, 2, 8, 10, 11, 12, 19]
+x = 5
+
+Initial:
+l = 0
+r = 6
+ans = -1
+
+Iteration 1:
+mid = 3
+nums[mid] = 10
+
+10 >= 5
+
+Possible ceil found
+ans = 3
+
+Search left side
+
+r = 2
+
+----------------------------------
+
+Iteration 2:
+l = 0
+r = 2
+
+mid = 1
+nums[mid] = 2
+
+2 < 5
+
+Move right
+
+l = 2
+
+----------------------------------
+
+Iteration 3:
+l = 2
+r = 2
+
+mid = 2
+nums[mid] = 8
+
+8 >= 5
+
+Possible ceil found
+ans = 2
+
+Search left side
+
+r = 1
+
+Loop Ends
+
+Answer = 2
+
+--------------------------------------------------
+Time Complexity
+--------------------------------------------------
+
+Binary Search reduces the search space by half
+in every iteration.
+
+TC = O(log n)
+
+--------------------------------------------------
+Space Complexity
+--------------------------------------------------
+
+Only a few variables are used.
+
+SC = O(1)
+
+--------------------------------------------------
+"""
+
+
+class Solution:
+    def findCeil(self, nums, x):
+        """
+        Returns the index of the smallest element
+        greater than or equal to x.
+
+        Parameters:
+        nums (List[int]): Sorted array
+        x (int): Target value
+
+        Returns:
+        int: Index of ceil element, or -1 if not found
+        """
+
+        ans = -1
+
+        l = 0
+        r = len(nums) - 1
+
+        while l <= r:
+            mid = l + (r - l) // 2
+
+            if nums[mid] >= x:
+                # Possible ceil found
+                ans = mid
+
+                # Search for a smaller valid ceil
+                r = mid - 1
+
+            else:
+                # Current element is too small
+                l = mid + 1
+
+        return ans
+
+
+# --------------------------------------------------
+# Example Usage
+# --------------------------------------------------
+
+if __name__ == "__main__":
+
+    nums = [1, 2, 8, 10, 11, 12, 19]
+    x = 5
+
+    sol = Solution()
+
+    result = sol.findCeil(nums, x)
+
+    print("Ceil Index:", result)
+
+    if result != -1:
+        print("Ceil Value:", nums[result])
+    else:
+        print("No ceil exists.")