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"
.
A group is identified by an interval [start, end]
, where start
and end
denote the start and end indices (inclusive) of the group. In the above example, "xxxx"
has the interval [3,6]
.
A group is considered large if it has 3 or more characters.
Return the intervals of every large group sorted in increasing order by start index.
Example 1:
Input: s = "abbxxxxzzy"
Output: [[3,6]]
Explanation: "xxxx" is the only
large group with start index 3 and end index 6.
Example 2:
Input: s = "abc" Output: [] Explanation: We have groups "a", "b", and "c", none of which are large groups.
Example 3:
Input: s = "abcdddeeeeaabbbcd" Output: [[3,5],[6,9],[12,14]] Explanation: The large groups are "ddd", "eeee", and "bbb".
Constraints:
1 <= s.length <= 1000
s
contains lowercase English letters only.
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 {
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;
}
}
class Solution {
public:
vector<vector<int>> largeGroupPositions(string s) {
int n = s.size();
int i = 0;
vector<vector<int>> ans;
while (i < n) {
int j = i;
while (j < n && s[j] == s[i]) {
++j;
}
if (j - i >= 3) {
ans.push_back({i, j - 1});
}
i = j;
}
return ans;
}
};
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
}