The Second Most Popular Dynamic Programming Problem

-
How many Edit Distance problems are there? I guess a very high number.. This is the second most popular Dynamic Programming problem out there (guess the #1? Yeah, Fibonacci. By a mile).

Here is another version of the Edit Distance problem: https://leetcode.com/problems/one-edit-distance/

161. One Edit Distance
Medium
Given two strings s and t, determine if they are both one edit distance apart.
Note: 
There are 3 possiblities to satisify one edit distance apart:
  1. Insert a character into s to get t
  2. Delete a character from s to get t
  3. Replace a character of s 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 = "cab", t = "ad"
Output: false
Explanation: We cannot get t from s by only one step.
Example 3:
Input: s = "1203", t = "1213"
Output: true
Explanation: We can replace '0' with '1' to get t.
Accepted
96,242
Submissions
299,225

This is 100% Edit Distance: build the DP table, and at the very end, just make sure that the last cell in your DP table is equal to... to... yes, 1 (one). Code is below, cheers, ACC.


public class Solution
{
    public bool IsOneEditDistance(string s, string t)
    {
        if (String.IsNullOrEmpty(s) && String.IsNullOrEmpty(t)) return false;
        if (String.IsNullOrEmpty(s)) return t.Length == 1;
        if (String.IsNullOrEmpty(t)) return s.Length == 1;

         int[][] dp = new int[s.Length + 1][];
        for (int i = 0; i < dp.Length; i++) dp[i] = new int[t.Length + 1];

         dp[0][0] = 0;
        for (int i = 1; i < dp.Length; i++) dp[i][0] = i;
        for (int j = 1; j < dp[0].Length; j++) dp[0][j] = j;

         for (int i = 1; i < dp.Length; i++)
        {
            for (int j = 1; j < dp[0].Length; j++)
            {
                if (s[i - 1] == t[j - 1])
                {
                    dp[i][j] = dp[i - 1][j - 1];
                }
                else
                {
                    dp[i][j] = Math.Min(Math.Min(dp[i - 1][j] + 1, dp[i][j - 1] + 1), dp[i - 1][j - 1] + 1);
                }
            }
        }

         return dp[dp.Length - 1][dp[0].Length - 1] == 1;
    }
}

Comments

Post a Comment

Popular posts from this blog

Count Binary Substrings

-
Image
Problem by Leetcode:  https://leetcode.com/problems/count-binary-substrings/description/ Give a string  s , count the number of non-empty (contiguous) substrings that have the same number of 0's and 1's, and all the 0's and all the 1's in these substrings are grouped consecutively. Substrings that occur multiple times are counted the number of times they occur. Example 1: Input: "00110011" Output: 6 Explanation: There are 6 substrings that have equal number of consecutive 1's and 0's: "0011", "01", "1100", "10", "0011", and "01". Notice that some of these substrings repeat and are counted the number of times they occur. Also, "00110011" is not a valid substring because all the 0's (and 1's) are not grouped together. Example 2: Input: "10101" Output: 4 Explanation: There are 4 substrings: "10", "01", "10", "01&
Read more

Count Vowels Permutation: Standard DP

-
Image
Standard DP problem:  https://leetcode.com/problems/count-vowels-permutation/ Given an integer  n , your task is to count how many strings of length  n  can be formed under the following rules: Each character is a lower case vowel ( 'a' ,  'e' ,  'i' ,  'o' ,  'u' ) Each vowel  'a'  may only be followed by an  'e' . Each vowel  'e'  may only be followed by an  'a'  or an  'i' . Each vowel  'i'   may not  be followed by another  'i' . Each vowel  'o'  may only be followed by an  'i'  or a  'u' . Each vowel  'u'  may only be followed by an  'a'. Since the answer may be too large, return it modulo  10^9 + 7. Example 1: Input: n = 1 Output: 5 Explanation: All possible strings are: "a", "e", "i" , "o" and "u". Example 2: Input: n = 2 Output: 10 Explanation: All possible strings are
Read more

Maximum Number of Balloons

-
Image
#431 in my solved list:  https://leetcode.com/problems/maximum-number-of-balloons/ Given a string  text , you want to use the characters of  text  to form as many instances of the word  "balloon"  as possible. You can use each character in  text   at most once . Return the maximum number of instances that can be formed.   Example 1: Input: text = "nlaebolko" Output: 1 Example 2: Input: text = "loonbalxballpoon" Output: 2 Example 3: Input: text = "leetcode" Output: 0 Constraints: 1 <= text.length <= 10^4 text  consists of lower case English letters only. Solution is as follows:  - Count the number of occurrences of the "balloon" letters in the input string  - For "l" and "o", divide the count by 2  - If the balloon string is not fully covered, return 0  - Return the min number across all occurrences Code is below, cheers, ACC. public class Solution {     public int M
Read more