Letter combinations of a phone number

Link: https://leetcode.com/problems/letter-combinations-of-a-phone-number/

Solution:

Topics: back tracking, DFS, grow

Intuition
Build out the combinations recursively and append them to the result. May be tempting to use for loops but the code is more complicated for no real gain…just keep it simple.

Implementation

def letter_combinations(digits):
	if len(digits) == 0:
		return []
	mapping = {
		'2': 'abc',
		'3': 'def',
		'4': 'ghi',
		'5': 'jkl',
		'6': 'mno',
		'7': 'pqrs',
		'8': 'tuv',
		'9': 'wxyz'
	}
	res = []
	def dfs(i, s):
		if i == len(digits):
			res.append(s)
			return
		for char in mapping[digits[i]]:
			dfs(i+1, s+char)
	dfs(0, '')
	return res
			
 
#time: o(n(4**n)) 
#NOTE: because there are at most 4**n combinations and each of them takes n time to build
 
#memory: o(n) stack space

Visual

Review 1
The iterative solution is probably more intuitive:

Implementation (iterative)

def letter_combos(digits):
	if len(digits) == 0:
		return []
	mapping = {
		'2': 'abc',
		'3': 'def',
		'4': 'ghi',
		'5': 'jkl',
		'6': 'mno',
		'7': 'pqrs',
		'8': 'tuv',
		'9': 'wxyz'
	}
	
	res = ['']
	for num in digits:
		new_res = []
		for char in mapping[num]:
			for combo in res:
				new_res.append(combo + char)
		res = new_res
	return res

Note on the time complexity: Each digit can have at most 4 possibilities therefore the complexity is o(4**n) where n is the number of digits. Think in bits!

review