We are given n
different types of stickers
. Each sticker has a lowercase English word on it.
You would like to spell out the given string target
by cutting individual letters from your collection of stickers and rearranging them. You can use each sticker more than once if you want, and you have infinite quantities of each sticker.
Return the minimum number of stickers that you need to spell out target
. If the task is impossible, return -1
.
Example 1:
Input: stickers = ["with","example","science"], target = "thehat"
Output: 3
Explanation:
We can use 2 "with" stickers, and 1 "example" sticker.
After cutting and rearrange the letters of those stickers, we can form the target "thehat".
Also, this is the minimum number of stickers necessary to form the target string.
Example 2:
Input: stickers = ["notice","possible"], target = "basicbasic"
Output: -1
Explanation:
We cannot form the target "basicbasic" from cutting letters from the given stickers.
Constraints:
n == stickers.length
1 <= n <= 50
1 <= stickers[i].length <= 10
1 <= target.length <= 15
stickers[i]
and target
consist of lowercase English letters.How would you approach this problem? What data structures and algorithms would you use? Can you provide a detailed explanation of your solution and its time and space complexity?
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force strategy aims to discover the smallest number of stickers needed by exhaustively exploring all possible combinations of stickers. It's like trying every conceivable way to use the stickers until you spell out the target word. This involves checking every possible sticker, then every possible pair, then every possible group, and so on.
Here's how the algorithm would work step-by-step:
def stickers_to_spell_word_brute_force(stickers, target):
minimum_stickers = float('inf')
number_of_stickers = len(stickers)
# Iterate through all possible combinations of stickers
for i in range(1, 1 << number_of_stickers):
current_stickers = []
for j in range(number_of_stickers):
if (i >> j) & 1:
current_stickers.append(stickers[j])
# Check if current combination can form the target
combined_string = ''.join(current_stickers)
can_form = can_form_target(combined_string, target)
# If the combination spells the target and uses fewer stickers than current minimum, update
if can_form:
number_of_stickers_used = len(current_stickers)
minimum_stickers = min(minimum_stickers, number_of_stickers_used)
if minimum_stickers == float('inf'):
return -1
else:
return minimum_stickers
def can_form_target(combined_string, target):
target_counts = {}
for char in target:
target_counts[char] = target_counts.get(char, 0) + 1
combined_counts = {}
for char in combined_string:
combined_counts[char] = combined_counts.get(char, 0) + 1
# Check if all target characters are available in combined_string
for char, count in target_counts.items():
if char not in combined_counts or combined_counts[char] < count:
return False #not enough characters
return True
The most efficient way to solve this problem is to use a technique called dynamic programming, combined with some smart pre-processing of the available stickers. We aim to minimize the number of stickers used by intelligently selecting which sticker to use at each step, building up our target word in the fewest moves possible.
Here's how the algorithm would work step-by-step:
def stickers_to_spell_word(stickers, target):
target_length = len(target)
# Initialize DP table to store minimum stickers needed.
dp = [-1] * (1 << target_length)
dp[0] = 0
for mask in range(1 << target_length):
if dp[mask] == -1:
continue
for sticker in stickers:
new_mask = mask
sticker_letter_counts = {}
for letter in sticker:
sticker_letter_counts[letter] = sticker_letter_counts.get(letter, 0) + 1
for letter_index in range(target_length):
if (mask >> letter_index) & 1:
continue
letter = target[letter_index]
if letter in sticker_letter_counts and sticker_letter_counts[letter] > 0:
sticker_letter_counts[letter] -= 1
new_mask |= (1 << letter_index)
# If the new mask is unchanged, the sticker did not help
if new_mask > mask:
if dp[new_mask] == -1 or dp[new_mask] > dp[mask] + 1:
dp[new_mask] = dp[mask] + 1
return dp[(1 << target_length) - 1]
Case | How to Handle |
---|---|
Null or empty stickers array | Return -1 since the target cannot be formed from nothing. |
Null or empty target string | Return 0 since an empty target requires no stickers. |
Stickers array contains an empty string | Ignore empty strings as they contribute nothing towards forming the target. |
Target string contains characters not present in any sticker | Return -1 since the target can't be fully formed. |
Stickers contain overlapping characters that could be used more efficiently in combination | The solution explores all possible sticker combinations so the algorithm should naturally gravitate to the more efficient combination. |
Target string is very long, leading to deep recursion (if using a recursive solution) | Implement memoization to avoid recomputing solutions for the same target substring. |
Large number of stickers with significant character overlap | Precompute the frequency of each character in each sticker to improve the efficiency of the matching process. |
Integer overflow if tracking the minimum number of stickers requires a large integer | Use a 64-bit integer or handle overflows by checking if the number of stickers exceeds a maximum allowed value and return -1 if it does. |