Floor in a Sorted Array

Author: Kondareddy1209

Leetcode/Floor_in_a_Sorted_Array.py

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