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 Square Root Using Binary Search

Problem Statement

Given a non-negative integer x, your task is to find its integer square root. The integer square root of x is the greatest integer r such that r * r ≤ x.

You must solve this problem without using any built-in square root functions. If x is negative, return -1 (square root of negative numbers is not defined in real numbers).

Examples

Input (x) Output Description
0 0 Square root of 0 is 0
1 1 Square root of 1 is 1
4 2 Perfect square: 2 * 2 = 4
10 3 3 * 3 = 9, 4 * 4 = 16 > 10. So answer is 3
100 10 Perfect square: 10 * 10 = 100
99 9 9 * 9 = 81, 10 * 10 = 100 > 99. So answer is 9
1000000 1000 Large number, 1000 * 1000 = 1000000
-25 -1 Negative input: square root not defined
-1 Empty input or undefined: treated as invalid

Visualization Player

Solution

Understanding the Problem

We are given a number x, and we need to compute its integer square root. That means we want the largest integer r such that:

  • r * r ≤ x
  • and (r + 1) * (r + 1) > x

For example, the square root of 10 is approximately 3.16, but since we want only the integer part, we return 3.

This problem becomes interesting when x is very large, because brute force checking every number up to x would be too slow. That’s where binary search helps — it allows us to quickly narrow down our search.

Step-by-Step Solution with Example

Step 1: Identify base cases

First, we handle easy cases directly. If x is 0 or 1, then the square root is just x itself.

Step 2: Initialize binary search range

We set low = 0 and high = x. These define the search range for our answer. We'll repeatedly guess a number mid between low and high.

Step 3: Apply binary search

At each step, we calculate mid = (low + high) / 2, and check mid * mid:

  • If mid * mid == x, then we’ve found the exact square root.
  • If mid * mid < x, it means mid might be the answer, but we try to find a larger one: low = mid + 1.
  • If mid * mid > x, then mid is too large: high = mid - 1.

We keep track of the last value where mid * mid <= x as our best answer so far.

Step 4: Example – x = 10

Let’s see how this works for x = 10:

  1. low = 0, high = 10
  2. mid = 5 → 5 * 5 = 25 → too big → high = 4
  3. mid = 2 → 2 * 2 = 4 → too small → low = 3, answer = 2
  4. mid = 3 → 3 * 3 = 9 → still small → low = 4, answer = 3
  5. mid = 4 → 4 * 4 = 16 → too big → high = 3

Loop ends, and the last valid answer is 3.

Edge Cases

  • Perfect squares: For inputs like 25 or 100, the binary search will land exactly on the square root.
  • Non-perfect squares: For 10 or 99, the result will be the floor of the real square root.
  • Large numbers: Binary search handles even huge numbers like 1,000,000 efficiently in ~20 steps.
  • 0 or 1: Directly return the same number.
  • Negative numbers: There’s no real square root for negative inputs. We return -1 to indicate invalid input.
  • Invalid or empty input: If the input is null, undefined, or not a number, return -1.

Finally

This is a classic problem where binary search really shines. Instead of looping through all values up to x, we reduce the problem space by half in each iteration.

By understanding how square roots behave, and tracking midpoints smartly, we get a clean and efficient solution that works even for very large numbers — all in O(log x) time.

Algorithm Steps

  1. Set low = 0 and high = x.
  2. Initialize ans = -1 to keep track of closest answer.
  3. While low ≤ high, do:
  4. → Find mid = (low + high) / 2.
  5. → If mid * mid == x, return mid.
  6. → If mid * mid < x, update ans = mid, and move low = mid + 1.
  7. → Else, move high = mid - 1.
  8. Return ans.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class SqrtBinarySearch {
  public static int sqrt(int x) {
    if (x == 0 || x == 1) return x;

    int start = 0, end = x, ans = -1;
    while (start <= end) {
      int mid = start + (end - start) / 2;
      long square = (long) mid * mid;

      if (square == x)
        return mid;
      else if (square < x) {
        ans = mid;          // Store potential answer
        start = mid + 1;    // Look right for a closer higher sqrt
      } else {
        end = mid - 1;      // Move left to reduce square
      }
    }
    return ans;
  }

  public static void main(String[] args) {
    int x = 15;
    System.out.println("Square root of " + x + " is: " + sqrt(x));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(1)If the square root is found in the first mid calculation.
Average CaseO(log x)Binary search narrows down the possible square root range in logarithmic time.
Worst CaseO(log x)Search continues halving the range until low > high.

Space Complexity

O(1)

Explanation: No additional space is used; only constant variables are needed.


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