Problem List

Group Anagrams

May 30, 2025 • Go • hash table, string, sorting, array • medium

Problem

Group all anagrams from a slice of strings. An anagram is formed by rearranging the letters of a word.

Approach

1. Use a helper function to check if two strings are anagrams.
2. Iterate through the list of words, placing each into an existing group if it matches the first word in that group.
3. If no group matches, start a new one.
4. Not the most efficient approach, but it gets the job done.

Edge Cases

["a", "b", "c", "a"] → duplicates must be grouped together.

Reflections

This brute force solution works for small inputs but doesn't scale, but I am happy that I've gone from no mediums to a few in just the last couple days. Soon my easy solutions will be fully polished, my mediums will improve, and I'll start working on hards. :)

Performance

Runtime beats: 5%
Memory beats: 46%

Complexity

Time: O(n^2 * k) Check every pair of strings in the worst case, and isAnagram takes O(k) time per comparison.
Space: O(n * k)Store the groups and use a map for character counting, which takes up O(n * k) space.

Go Solution

func groupAnagrams(strs []string) [][]string {
  res := [][]string{{strs[0]}}
  if len(strs) == 1 {
    return res
  }

  for i := 1; i < len(strs); i++ {
    placed := false
    for j := 0; j < len(res); j++ {
      if isAnagram(strs[i], res[j][0]) {
        placed = true
        res[j] = append(res[j], strs[i])
        break
      }
    }
    if !placed {
      res = append(res, []string{strs[i]})
    }
  }

  return res
}

func isAnagram(a, b string) bool {
  if len(a) != len(b) {
    return false
  }

  m := make(map[rune]int)
  for _, l := range a {
    m[l]++
  }
  for _, l := range b {
    m[l]--
    if m[l] < 0 {
      return false
    }
  }

  return true
}

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