Positions of Large Groups, by LeetCode

-
Hi folks, long time no see! Problem is here: https://leetcode.com/problems/positions-of-large-groups/description/:

In a string S of lowercase letters, these letters form consecutive groups of the same character.
For example, a string like S = "abbxxxxzyy" has the groups "a", "bb", "xxxx", "z" and "yy".
Call a group large if it has 3 or more characters.  We would like the starting and ending positions of every large group.
The final answer should be in lexicographic order.
The solution is straightforward: index manipulation, understanding some of the data structures of your favorite language (in my case C#), linear pass keeping track of the beginning of the current block (in the code below expressed by "previousIndex"), and that's pretty much it! Have fun, code is below, many hugs!!! Boris

public class Solution
{
public IList<IList<int>> LargeGroupPositions(string S)
{
List<IList<int>> retVal = new List<IList<int>>();

 int previousIndex = 0;
for (int i = 1; i < S.Length; i++)
{
if (S[i] != S[previousIndex])
{
if (i - previousIndex >= 3)
{
List<int> list = new List<int>();
list.Add(previousIndex);
list.Add(i - 1);
retVal.Add(list);
}
previousIndex = i;
}
else if (i == S.Length - 1 && S[i] == S[previousIndex] && i - previousIndex + 1 >= 3)
{
List<int> list = new List<int>();
list.Add(previousIndex);
list.Add(i);
retVal.Add(list);
}
}

 return retVal;
}
}

Comments

  1. TarasSeptember 3, 2019 at 7:24 PM

    3 liner :)

    class Solution {
    public:
    vector> largeGroupPositions(const string& S) {
    vector> groups;
    for (int start = 0, end = 1; end < S.size(); start = end, ++end) {
    while (S[end] == S[end-1]) end += 1;
    if (end - start >= 3) groups.push_back({start, end-1});
    }
    return groups;
    }
    };

    ReplyDelete

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