Binary TreesBinary Trees36

  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Postorder Traversal of a Binary Tree Using Recursion
  4. 4Postorder Traversal of a Binary Tree using Iteration
  5. 5Level Order Traversal of a Binary Tree using Recursion
  6. 6Level Order Traversal of a Binary Tree using Iteration
  7. 7Reverse Level Order Traversal of a Binary Tree using Iteration
  8. 8Reverse Level Order Traversal of a Binary Tree using Recursion
  9. 9Find Height of a Binary Tree
  10. 10Find Diameter of a Binary Tree
  11. 11Find Mirror of a Binary Tree - Todo
  12. 12Inorder Traversal of a Binary Tree using Recursion
  13. 13Inorder Traversal of a Binary Tree using Iteration
  14. 14Left View of a Binary Tree
  15. 15Right View of a Binary Tree
  16. 16Top View of a Binary Tree
  17. 17Bottom View of a Binary Tree
  18. 18Zigzag Traversal of a Binary Tree
  19. 19Check if a Binary Tree is Balanced
  20. 20Diagonal Traversal of a Binary Tree
  21. 21Boundary Traversal of a Binary Tree
  22. 22Construct a Binary Tree from a String with Bracket Representation
  23. 23Convert a Binary Tree into a Doubly Linked List
  24. 24Convert a Binary Tree into a Sum Tree
  25. 25Find Minimum Swaps Required to Convert a Binary Tree into a BST
  26. 26Check if a Binary Tree is a Sum Tree
  27. 27Check if All Leaf Nodes are at the Same Level in a Binary Tree
  28. 28Lowest Common Ancestor (LCA) in a Binary Tree
  29. 29Solve the Tree Isomorphism Problem
  30. 30Check if a Binary Tree Contains Duplicate Subtrees of Size 2 or More
  31. 31Check if Two Binary Trees are Mirror Images
  32. 32Calculate the Sum of Nodes on the Longest Path from Root to Leaf in a Binary Tree
  33. 33Print All Paths in a Binary Tree with a Given Sum
  34. 34Find the Distance Between Two Nodes in a Binary Tree
  35. 35Find the kth Ancestor of a Node in a Binary Tree
  36. 36Find All Duplicate Subtrees in a Binary Tree

Lower Bound in Sorted Array
Binary Search Approach



Problem Statement

Given a sorted array of integers and a target number x, your task is to find the lower bound of x.

Examples

Input ArrayKey (x)Lower Bound IndexDescription
[10, 20, 30, 40, 50]302Exact match at index 2
[10, 20, 30, 40, 50]353First value ≥ 35 is 40 at index 3
[10, 20, 30, 40, 50]50All values ≥ 5, so index 0 is answer
[10, 20, 30, 40, 50]605No values ≥ 60, return array length 5
[10]100Single element equals key
[10]50Key smaller than only element
[10]151Key larger than only element
[]100Empty array, default lower bound is 0

Solution

To find the lower bound of a number x in a sorted array, we want to locate the first element that is greater than or equal to x. Binary search is the ideal approach for this because it efficiently narrows the range to find this position without scanning every element.

Understanding Lower Bound

Let’s understand it through a few common scenarios:

  • If x exists in the array, the lower bound is the index of its first occurrence.
  • If x is smaller than all elements, the lower bound is 0.
  • If x is greater than all elements, there's no valid element ≥ x, so the lower bound is the length of the array.
  • If the array is empty, the result is naturally 0, since there's nowhere to insert the element.

How Binary Search Helps

Binary search helps us find the lower bound efficiently by:

  • Maintaining two pointers: low and high, covering the current search range.
  • If we find arr[mid] ≥ x, it could be a valid answer, but we still check the left half to find a better one.
  • If arr[mid] < x, we discard the left half and move to the right.

We continue until all positions are checked. The moment we find an element ≥ x, we store its index as a possible lower bound. Eventually, we return the smallest index that satisfies the condition.

Why This is Useful

Lower bound is useful in range queries, insert positions, and understanding element distributions. Using binary search keeps the time complexity at O(log n), which is essential for large datasets.

Visualization

Algorithm Steps

  1. Given a sorted array arr and a target integer x.
  2. Initialize two pointers: low = 0 and high = arr.length - 1.
  3. Initialize ans = arr.length (in case x is greater than all elements).
  4. Repeat while low ≤ high:
  5. → Calculate mid = Math.floor((low + high) / 2).
  6. → If arr[mid] ≥ x, set ans = mid and move high = mid - 1.
  7. → Else, move low = mid + 1.
  8. Return ans as the index of the lower bound.

Code

Python
JavaScript
Java
C++
C
def lower_bound(arr, x):
    low, high = 0, len(arr) - 1
    ans = len(arr)
    while low <= high:
        mid = (low + high) // 2
        if arr[mid] >= x:
            ans = mid
            high = mid - 1
        else:
            low = mid + 1
    return ans

# Sample Input
arr = [1, 2, 4, 4, 5, 6]
x = 4
print("Lower Bound Index:", lower_bound(arr, x))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)The lower bound is found at the very first mid index comparison in the binary search.
Average CaseO(log n)With each iteration, the binary search reduces the search space by half to locate the correct lower bound.
Average CaseO(log n)The binary search may run until the search space is reduced to a single element when the target is not present or is larger than all elements.

Space Complexity

O(1)

Explanation: The algorithm operates in-place and uses a fixed number of variables (like low, high, mid, and ans), resulting in constant extra space usage.



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