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Formatted question description: https://leetcode.ca/all/830.html

# 830. Positions of Large Groups (Easy)

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.

Example 1:

Input: "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the single large group with starting  3 and ending positions 6.


Example 2:

Input: "abc"
Output: []
Explanation: We have "a","b" and "c" but no large group.


Example 3:

Input: "abcdddeeeeaabbbcd"
Output: [[3,5],[6,9],[12,14]]

Note:  1 <= S.length <= 1000

Companies:

Related Topics:
Array

## Solution 1.

• class Solution {
public List<List<Integer>> largeGroupPositions(String S) {
List<List<Integer>> largeGroupPositions = new ArrayList<List<Integer>>();
if (S.length() < 3)
return largeGroupPositions;
int begin = 0;
char prev = S.charAt(0);
int length = S.length();
for (int i = 1; i < length; i++) {
char c = S.charAt(i);
if (c != prev) {
if (i - begin >= 3) {
int end = i - 1;
List<Integer> largeGroupPosition = new ArrayList<Integer>();
}
begin = i;
}
prev = c;
}
if (length - begin >= 3) {
int end = length - 1;
List<Integer> largeGroupPosition = new ArrayList<Integer>();
}
return largeGroupPositions;
}
}

############

class Solution {
public List<List<Integer>> largeGroupPositions(String s) {
int n = s.length();
int i = 0;
List<List<Integer>> ans = new ArrayList<>();
while (i < n) {
int j = i;
while (j < n && s.charAt(j) == s.charAt(i)) {
++j;
}
if (j - i >= 3) {
ans.add(Arrays.asList(i, j - 1));
}
i = j;
}
return ans;
}
}

• // OJ: https://leetcode.com/problems/positions-of-large-groups/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<vector<int>> largeGroupPositions(string S) {
vector<vector<int>> ans;
for (int N = S.size(), i = 0; i < N;) {
int j = i + 1;
while (j < N && S[j] == S[i]) ++j;
if (j - i >= 3) ans.push_back({ i, j - 1 });
i = j;
}
return ans;
}
};

• class Solution:
def largeGroupPositions(self, s: str) -> List[List[int]]:
i, n = 0, len(s)
ans = []
while i < n:
j = i
while j < n and s[j] == s[i]:
j += 1
if j - i >= 3:
ans.append([i, j - 1])
i = j
return ans

############

class Solution:
def largeGroupPositions(self, S):
"""
:type S: str
:rtype: List[List[int]]
"""
groups = []
before_index, before_char = 0, S[0]
for i, s in enumerate(S):
if s != before_char:
if i - before_index >= 3:
groups.append([before_index, i - 1])
before_index = i
before_char = s
if i - before_index >= 2:
groups.append([before_index, i])
return groups

• func largeGroupPositions(s string) [][]int {
i, n := 0, len(s)
ans := [][]int{}
for i < n {
j := i
for j < n && s[j] == s[i] {
j++
}
if j-i >= 3 {
ans = append(ans, []int{i, j - 1})
}
i = j
}
return ans
}