Leetcode: Positions of Large Groups

Positions of Large Groups



Similar Problems:


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

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


  • Solution: Special treatment for last group
// Blog link: https://code.dennyzhang.com/positions-of-large-groups
// Basic Ideas: two pointers
//   groups: [start, end]
// Complexity: Time O(n), Space O(n)
func largeGroupPositions(S string) [][]int {
    res := [][]int{}
    start := 0
    for end := 0; end<len(S); end++ {
        // find a new group
        if S[end] != S[start] {
            if end-start>=3 {
                res = append(res, []int{start, end-1})
            }
            start = end
        }
        // last group
        if S[end] == S[start] && end == len(S)-1 {
            if end-start>=2 {
                res = append(res, []int{start, end})
            }
        }
    }
    return res
}

  • Solution: padding for last group
// Blog link: https://code.dennyzhang.com/positions-of-large-groups
// Basic Ideas: two pointers
//   groups: [start, end]
// Complexity: Time O(n), Space O(n)
func largeGroupPositions(S string) [][]int {
    res := [][]int{}
    start := 0
    // Note: ";" won't happen in S. 
    for end, ch := range S+";" {
        if ch != rune(S[start]) {
            if end-start >=3 {
                res = append(res, []int{start, end-1})
            }
            start = end
        }
    }
    return res
}
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