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

Search Insert Position in Sorted Array
Lower Bound Binary Search



Problem Statement

Given a sorted array of distinct integers and a target number x, your task is to find the insert position of x.

This means you must return the index where x is located in the array. If it is not present, return the index where it should be inserted in order to keep the array sorted.

This problem is also known as finding the lower bound — the first index where an element is greater than or equal to x.

Examples

Input ArrayTarget (x)Insert PositionDescription
[10, 20, 30, 40, 50]35335 should be inserted before 40
[10, 20, 30, 40, 50]201Exact match at index 1
[10, 20, 30, 40, 50]50Smaller than all elements, inserted at beginning
[10, 20, 30, 40, 50]555Larger than all elements, inserted at end
[5, 15, 25, 35]252Exact match at index 2
[5]20Inserted before the only element
[5]50Exact match at index 0
[5]81Inserted after the only element

Solution

To find the position where a target value x should be inserted in a sorted array, we use a binary search approach known as the lower bound.

Understanding the Goal

The goal is to find the first index where the value in the array is greater than or equal to x. If such a number exists, we return its index. If not, we return the position after the last element.

Step-by-Step Explanation

We start by setting up two pointers, low and high, at the start and end of the array respectively. We also initialize an answer variable ans as the length of the array. This variable keeps track of the best candidate index where x could be inserted.

In each step of the binary search:

  • We calculate the mid index between low and high.
  • If arr[mid] is greater than or equal to x, it means this position might be a valid place to insert x, so we update ans = mid and search the left half.
  • If arr[mid] is less than x, then we need to move right to find a higher or equal value, so we set low = mid + 1.

The loop continues until low > high. At this point, the ans holds the index where the target should be inserted to keep the array sorted.

Why Binary Search?

Since the array is already sorted, binary search helps us find the insert position in O(log n) time, which is much more efficient than scanning every element linearly.

Visualization

Algorithm Steps

  1. Given a sorted array arr of distinct integers and a target value x.
  2. Initialize: low = 0, high = arr.length - 1, ans = arr.length.
  3. While low ≤ high:
  4. → Compute mid = Math.floor((low + high) / 2).
  5. → If arr[mid] ≥ x, update ans = mid and move left: high = mid - 1.
  6. → Else, move right: low = mid + 1.
  7. Return ans as the correct insert position.

Code

Python
JavaScript
Java
C++
C
def search_insert_position(arr, x):
    n = len(arr)
    low, high = 0, n - 1
    ans = n
    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, 3, 5, 6]
target = 2
print("Insert Position:", search_insert_position(arr, target))

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)The target is found at the middle index on the first comparison, requiring only one iteration.
Average CaseO(log n)Binary search halves the search space in each iteration, so on average, it takes log n steps to find the insert position.
Average CaseO(log n)In the worst case, the entire array must be searched logarithmically to determine where the target should be inserted.

Space Complexity

O(1)

Explanation: The algorithm operates in-place using only a constant amount of memory — just a few integer variables for low, high, mid, and ans.



Welcome to ProgramGuru

Sign up to start your journey with us

Support ProgramGuru.org

Mention your name, and programguru.org in the message. Your name shall be displayed in the sponsers list.

PayPal

UPI

PhonePe QR

MALLIKARJUNA M