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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the  One Edit Distance in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
  • Arrays
  • Strings
  • Hash Table
  • Dynamic Programming
  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the Problem One Edit Distance– LeetCode Problem

 One Edit Distance– LeetCode Problem

Problem:

Given two strings s and t, return true if they are both one edit distance apart, otherwise return false.

A string s is said to be one distance apart from a string t if you can:

  • Insert exactly one character into s to get t.
  • Delete exactly one character from s to get t.
  • Replace exactly one character of s with a different character to get t.

Example 1:

Input: s = "ab", t = "acb"
Output: true
Explanation: We can insert 'c' into s to get t.

Example 2:

Input: s = "", t = ""
Output: false
Explanation: We cannot get t from s by only one step.

Example 3:

Input: s = "a", t = ""
Output: true

Example 4:

Input: s = "", t = "A"
Output: true

Constraints:

  • 0 <= s.length <= 104
  • 0 <= t.length <= 104
  • s and t consist of lower-case letters, upper-case letters and/or digits.
 One Edit Distance– LeetCode Solutions
 One Edit Distance Solution in C++:
class Solution {
 public:
  bool isOneEditDistance(string s, string t) {
    const int m = s.length();
    const int n = t.length();
    if (m > n)  // make sure len(s) <= len(t)
      return isOneEditDistance(t, s);

    for (int i = 0; i < m; ++i)
      if (s[i] != t[i]) {
        if (m == n)
          return s.substr(i + 1) == t.substr(i + 1);  // replace s[i] with t[i]
        return s.substr(i) == t.substr(i + 1);        // delete t[i]
      }

    return m + 1 == n;  // delete t[-1]
  }
};
 One Edit Distance Solution in Java:
class Solution {
  public boolean isOneEditDistance(String s, String t) {
    final int m = s.length();
    final int n = t.length();
    if (m > n) // make sure len(s) <= len(t)
      return isOneEditDistance(t, s);

    for (int i = 0; i < m; ++i)
      if (s.charAt(i) != t.charAt(i)) {
        if (m == n)
          return s.substring(i + 1).equals(t.substring(i + 1)); // replace s[i] with t[i]
        return s.substring(i).equals(t.substring(i + 1));       // delete t[i]
      }

    return m + 1 == n; // delete t[-1]
  }
}
 One Edit Distance Solution in Python:
class Solution:
  def isOneEditDistance(self, s: str, t: str) -> bool:
    m = len(s)
    n = len(t)
    if m > n:  # make sure len(s) <= len(t)
      return self.isOneEditDistance(t, s)

    for i in range(m):
      if s[i] != t[i]:
        if m == n:
          return s[i + 1:] == t[i + 1:]  # replace s[i] with t[i]
        return s[i:] == t[i + 1:]  # delete t[i]

    return m + 1 == n  # delete t[-1]

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