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groupAnagrams.md

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Problem statement: Given an array of strings strs, group the anagrams together into sublists. You can return the output in any order.

Note: An anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Examples:

Example 1:

Input: strs = ['eat', 'tea', 'tan', 'ate', 'nat', 'bat'] Output: [ [ 'eat', 'tea', 'ate' ], [ 'tan', 'nat' ], [ 'bat' ] ]

Example 2:

Input: strs = ['Eat', 'Tea', 'Tan', 'Ate', 'Nat', 'Bat'] Output: [ [ 'Eat' ], [ 'Tea' ], [ 'Tan' ], [ 'Ate' ], [ 'Nat' ], [ 'Bat' ] ]

Example 3:

Input: strs = ['apple', 'orange', 'banana'] Output: [ [ 'apple' ], [ 'orange' ], [ 'banana' ] ]

Algorithmic Steps This problem is solved with the help of hashtable using sorted strings as keys and list of anagrams as values. The algorithmic approach can be summarized as follows:

Approach 1

  1. Create an empty map(anagramGroups) to store the list of anagrams.

  2. Iterate an input array using for-of loop and find the sorted string for each string in an array. Thereafter, fetch the angram group from the map based on sorted string as a key.

  3. If the group has sorted string as a key, push the current string as a new anagram to the map. Otherwise, set the anagram map with a new group of anagram.

  4. Return anagramGroups values as array of anagram groups.

Approach 2

  1. Create an empty object(anagramGroups) to store the list of anagrams.

  2. Iterate an input array using for-of loop and find the sorted string for each string in an array. Thereafter, fetch the angram group from an object based on sorted string as a property.

  3. If the group contains sorted string as a object property, push the current string as a new anagram to the object. Otherwise, set the anagram map with a new group of anagram.

  4. Return anagramGroups values as array of anagram groups.

Time and Space complexity: This algorithm has a time complexity of O(K N*log N), where N is the length of longest string and K is the number of strings in an array. This is because sorting an each string takes O(N*log N) and it needs to be repeated for K strings. Hence, the overall complexity is O(K N*log N).

Here, we use map datastructure which is used to store upto K strings and each string with average length of N. Hence, the space complexity will be O(K * N).