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

Find Minimum in Rotated Sorted Array
Optimal Binary Search Approach



Problem Statement

Given a rotated sorted array of unique integers, your task is to find the minimum element in the array.

A rotated sorted array is an array that was originally sorted in increasing order, but then some leading elements were moved to the end. For example, [1, 2, 3, 4, 5] could become [3, 4, 5, 1, 2].

The array contains no duplicate elements. If the array is empty, return -1.

Examples

Input ArrayOutputDescription
[4, 5, 6, 7, 0, 1, 2]0Array is rotated, minimum is at the pivot (index 4)
[3, 4, 5, 1, 2]1Minimum element is in the middle of the array
[1, 2, 3, 4, 5]1Array is not rotated, minimum is at index 0
[2, 1]1Two elements, rotated once
[1]1Single element array
[]-1Empty array, no elements to check

Solution

To find the minimum element in a rotated sorted array, we can take advantage of the array's sorted structure using a smart approach that avoids checking every element.

Understanding the Problem

A rotated sorted array is one that has been shifted from its original sorted position. For example, the sorted array [1, 2, 3, 4, 5] might become [4, 5, 1, 2, 3]. Our goal is to find the smallest element in this modified structure efficiently.

What Are the Possible Cases?

1. The array is not rotated at all: If the first element is less than the last element, then the array is still sorted in order. The minimum is simply the first element.

2. The array is rotated: Somewhere in the array, there is a drop from a large number to a smaller one. That drop is where the rotation happened, and the smaller number is the minimum.

3. Small arrays: For arrays with one or two elements, we can directly check them. If there’s one element, it’s the answer. If there are two, the smaller of the two is the answer.

4. Empty array: If the array has no elements, there's no minimum to return. In such cases, we return -1 as a signal that the input is invalid or empty.

How Binary Search Helps

Since the array is mostly sorted (but rotated), we can use binary search to narrow down the search range:

  • If the middle element is greater than the last element, it means the minimum must be on the right side.
  • If the middle element is less than or equal to the last element, then the minimum could be at mid or to the left.

We continue this process until our search space is narrowed down to one element — and that’s the minimum!

Why This Approach is Efficient

Binary search cuts the search range in half at every step. So instead of looking at all n elements, we find the answer in O(log n) time — making this method ideal for large arrays.

This strategy gives us a clear and optimal way to find the minimum element in any rotated sorted array.

Visualization

Algorithm Steps

  1. Initialize two pointers: start = 0 and end = n - 1.
  2. While start < end:
  3. → Compute mid = start + (end - start) / 2.
  4. → If arr[mid] > arr[end], it means the minimum is in the right half, so set start = mid + 1.
  5. → Else, the minimum is in the left half including mid, so set end = mid.
  6. When the loop ends, arr[start] is the minimum element.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class RotatedArrayMinimum {
  public static int findMin(int[] nums) {
    int start = 0;
    int end = nums.length - 1;

    while (start < end) {
      int mid = start + (end - start) / 2;

      if (nums[mid] > nums[end]) {
        // Minimum must be in the right half
        start = mid + 1;
      } else {
        // Minimum could be at mid or to the left
        end = mid;
      }
    }

    return nums[start]; // or nums[end], both are same here
  }

  public static void main(String[] args) {
    int[] nums = {4, 5, 6, 7, 0, 1, 2};
    System.out.println("Minimum Element: " + findMin(nums));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)When the array is not rotated and the first element is the minimum.
Average CaseO(log n)Each iteration of binary search reduces the array size by half, giving logarithmic performance.
Average CaseO(log n)In the worst case, the binary search continues until only one element remains.

Space Complexity

O(1)

Explanation: Only a few pointers are used, and no extra space is required.



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