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

Longest Common Prefix in Array of Strings
Optimal Vertical Scanning



Problem Statement

Given an array of strings, your task is to find the longest common prefix shared among all the strings.

This problem is commonly asked in coding interviews to test your understanding of string manipulation and iteration logic.

Examples

Input StringsLongest Common PrefixDescription
["flower", "flow", "flight"]"fl""fl" is the longest starting part common in all three strings
["dog", "racecar", "car"]""No common prefix exists
["interview", "internet", "internal"]"inte"All strings share "inte" as a common starting sequence
["a"]"a"Only one string — the prefix is the entire string
[""]""Single empty string, no prefix
[]""Empty input array — return empty string
["apple", "apple", "apple"]"apple"All strings are exactly the same
["test", "testing", "tester", "testify"]"test""test" is the longest starting match across all strings
["abc", ""]""One string is empty, so no common prefix

Solution

The goal is to find the longest portion at the beginning of all strings that is identical. This is called the longest common prefix.

Imagine writing all the words in a vertical column and comparing them letter by letter, from top to bottom. We go through the characters of the first string one at a time and check whether the same character appears in the same position in all other strings.

Let’s explore the different cases you might encounter:

  • Normal case: If the strings have a common start — like ["flower", "flow", "flight"] — we start comparing character by character. At position 0, all start with 'f'. At position 1, all have 'l'. At position 2, 'flower' and 'flow' have 'o', but 'flight' has 'i'. So the common prefix ends just before the mismatch: "fl".
  • No match at all: If the first characters don’t even match — like in ["dog", "racecar", "car"] — there is no common prefix, and we return "".
  • Single string: If there’s only one string, like ["a"], that string itself is the longest common prefix.
  • Some string is empty: If even one of the strings is empty — like ["abc", ""] — there can’t be a common prefix. Return "".
  • Empty array: If the input array has no strings at all — return "" right away.
  • All strings are identical: In cases like ["apple", "apple", "apple"], all characters will match and the whole string is the common prefix.

This method is known as vertical scanning because it checks each character vertically across all strings at a particular index. It stops as soon as it finds a mismatch, making it efficient and easy to understand.

Edge cases like empty arrays or strings are simple to handle with early checks. This way, we avoid unnecessary comparisons and ensure correctness in all scenarios.

Visualization

Algorithm Steps

  1. Check if the input array is empty. If so, return "".
  2. Loop through each character index i of the first string.
  3. For each string in the array, check if the character at position i matches that in the first string.
  4. If a mismatch is found or if a string is shorter than i, return the prefix found up to that point.
  5. If no mismatch is found, the entire first string is the common prefix.

Code

Java
Python
JavaScript
C
C++
C#
Kotlin
Swift
Go
Php
public class LongestCommonPrefix {
  public static String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";

    for (int i = 0; i < strs[0].length(); i++) {
      char ch = strs[0].charAt(i); // Take character from first string
      for (int j = 1; j < strs.length; j++) {
        // If i is out of bounds or character doesn't match, return prefix
        if (i >= strs[j].length() || strs[j].charAt(i) != ch) {
          return strs[0].substring(0, i);
        }
      }
    }
    return strs[0];
  }

  public static void main(String[] args) {
    String[] input = {"flower", "flow", "flight"};
    System.out.println("Longest Common Prefix: " + longestCommonPrefix(input));
  }
}

Time Complexity

CaseTime ComplexityExplanation
Best CaseO(n)When all strings are identical, only the length of one string is scanned.
Average CaseO(n × m)Where n is the number of strings and m is the length of the shortest string.
Average CaseO(n × m)In the worst case, all strings are compared fully until a mismatch is found at the end.

Space Complexity

O(1)

Explanation: Only a few variables are used for iteration and prefix tracking. No extra memory required.



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