Binary TreesBinary Trees36
  1. 1Preorder Traversal of a Binary Tree using Recursion
  2. 2Preorder Traversal of a Binary Tree using Iteration
  3. 3Inorder Traversal of a Binary Tree using Recursion
  4. 4Inorder Traversal of a Binary Tree using Iteration
  5. 5Postorder Traversal of a Binary Tree Using Recursion
  6. 6Postorder Traversal of a Binary Tree using Iteration
  7. 7Level Order Traversal of a Binary Tree using Recursion
  8. 8Level Order Traversal of a Binary Tree using Iteration
  9. 9Reverse Level Order Traversal of a Binary Tree using Iteration
  10. 10Reverse Level Order Traversal of a Binary Tree using Recursion
  11. 11Find Height of a Binary Tree
  12. 12Find Diameter of a Binary Tree
  13. 13Find Mirror of a Binary Tree
  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
GraphsGraphs46
  1. 1Breadth-First Search in Graphs
  2. 2Depth-First Search in Graphs
  3. 3Number of Provinces in an Undirected Graph
  4. 4Connected Components in a Matrix
  5. 5Rotten Oranges Problem - BFS in Matrix
  6. 6Flood Fill Algorithm - Graph Based
  7. 7Detect Cycle in an Undirected Graph using DFS
  8. 8Detect Cycle in an Undirected Graph using BFS
  9. 9Distance of Nearest Cell Having 1 - Grid BFS
  10. 10Surrounded Regions in Matrix using Graph Traversal
  11. 11Number of Enclaves in Grid
  12. 12Word Ladder - Shortest Transformation using Graph
  13. 13Word Ladder II - All Shortest Transformation Sequences
  14. 14Number of Distinct Islands using DFS
  15. 15Check if a Graph is Bipartite using DFS
  16. 16Topological Sort Using DFS
  17. 17Topological Sort using Kahn's Algorithm
  18. 18Cycle Detection in Directed Graph using BFS
  19. 19Course Schedule - Task Ordering with Prerequisites
  20. 20Course Schedule 2 - Task Ordering Using Topological Sort
  21. 21Find Eventual Safe States in a Directed Graph
  22. 22Alien Dictionary Character Order
  23. 23Shortest Path in Undirected Graph with Unit Distance
  24. 24Shortest Path in DAG using Topological Sort
  25. 25Dijkstra's Algorithm Using Set - Shortest Path in Graph
  26. 26Dijkstra’s Algorithm Using Priority Queue
  27. 27Shortest Distance in a Binary Maze using BFS
  28. 28Path With Minimum Effort in Grid using Graphs
  29. 29Cheapest Flights Within K Stops - Graph Problem
  30. 30Number of Ways to Reach Destination in Shortest Time - Graph Problem
  31. 31Minimum Multiplications to Reach End - Graph BFS
  32. 32Bellman-Ford Algorithm for Shortest Paths
  33. 33Floyd Warshall Algorithm for All-Pairs Shortest Path
  34. 34Find the City With the Fewest Reachable Neighbours
  35. 35Minimum Spanning Tree in Graphs
  36. 36Prim's Algorithm for Minimum Spanning Tree
  37. 37Disjoint Set (Union-Find) with Union by Rank and Path Compression
  38. 38Kruskal's Algorithm - Minimum Spanning Tree
  39. 39Minimum Operations to Make Network Connected
  40. 40Most Stones Removed with Same Row or Column
  41. 41Accounts Merge Problem using Disjoint Set Union
  42. 42Number of Islands II - Online Queries using DSU
  43. 43Making a Large Island Using DSU
  44. 44Bridges in Graph using Tarjan's Algorithm
  45. 45Articulation Points in Graphs
  46. 46Strongly Connected Components using Kosaraju's Algorithm

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

Visualization Player

Solution

Understanding the Problem

We are given a rotated sorted array — an array that was originally sorted in increasing order but has been rotated at some pivot point. Our task is to find the minimum element in this array efficiently.

For example, a sorted array [1, 2, 3, 4, 5] might become [4, 5, 1, 2, 3] after rotation. The goal is to identify the smallest number — in this case, 1 — without checking each element one by one.

Step-by-Step Solution with Example

Step 1: Observe if the array is already sorted

If the array is not rotated at all, the first element will be less than the last. In that case, the first element is the smallest. For example, [1, 2, 3, 4, 5] — here, 1 < 5, so the minimum is 1.

Step 2: Initialize Binary Search

We'll use binary search because the array has a sorted structure. Initialize two pointers: low = 0 and high = array.length - 1.

Step 3: Loop until the search space is narrowed

While low < high, do the following:

  • Calculate mid = Math.floor((low + high) / 2).
  • If array[mid] > array[high], it means the minimum is to the right of mid. So, set low = mid + 1.
  • Otherwise, the minimum is at mid or to the left. Set high = mid.

Step 4: Return the minimum element

After the loop ends, low == high, and we have found the index of the minimum element. Return array[low].

Step 5: Example Walkthrough

Let’s take [4, 5, 6, 1, 2, 3] as an example:

  • Initial: low = 0, high = 5
  • mid = 2 → array[2] = 6, array[5] = 3 → 6 > 3 → minimum must be on the right → low = 3
  • mid = 4 → array[4] = 2, array[5] = 3 → 2 ≤ 3 → high = 4
  • mid = 3 → array[3] = 1, array[4] = 2 → 1 ≤ 2 → high = 3
  • Now low = high = 3 → minimum = array[3] = 1

Edge Cases

  • Array not rotated: [1, 2, 3, 4, 5] → min = 1
  • Single element: [7] → min = 7
  • Two elements: [10, 5] → min = 5
  • Empty array: Return -1 or handle as an invalid input

Finally

This approach is optimal because it uses binary search, which has a time complexity of O(log n). By recognizing whether to move left or right based on the middle and end values, we efficiently narrow the search space. This method is both intuitive and scalable for large datasets.

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.
Worst 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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