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This approach utilizes recursive backtracking in combination with memoization to solve the problem. The idea is to maintain a count of letters that are still needed to complete the target string and recursively explore combinations of stickers that reduce this count. Memoization helps in caching previously computed results for specific target states, avoiding redundant calculations.
The steps are as follows:
Time Complexity: O(T * 2^T * S), where T is the length of the target, and S is the number of stickers. This accounts for checking each subset of the target matched with each sticker.
Space Complexity: O(T * k), where k is the average word length in stickers, for the memoization storage.
1from collections import defaultdict
2import sys
3
4def minStickers(stickers, target):
5 n = len(stickers)
6 # Preprocess stickers into a list of letter-count dictionaries
7 sticker_count = [defaultdict(int) for _ in range(n)]
8 for i in range(n):
9 for char in stickers[i]:
10 sticker_count[i][char] += 1
11 memo = {}
12
13 def dfs(target):
14 if target in memo:
15 return memo[target]
16
17 target_count = defaultdict(int)
18 for char in target:
19 target_count[char] += 1
20
21 ans = sys.maxsize
22 # Try each sticker
23 for i in range(n):
24 if sticker_count[i][target[0]] == 0: # Cannot contribute to current first char
25 continue
26 # Construct a new target string after using this sticker
27 new_target = ''
28 for char in target_count:
29 if char in target_count:
30 needed = target_count[char] - sticker_count[i].get(char, 0)
31 if needed > 0:
32 new_target += char * needed
33
34 # Use one current sticker
35 if new_target != target: # If we made any progress
36 ans = min(ans, dfs(new_target) + 1)
37
38 memo[target] = ans if ans != sys.maxsize else -1
39 return memo[target]
40
41 result = dfs(target)
42 return result
43
44# Example Usage
45stickers = ["with", "example", "science"]
46target = "thehat"
47print(minStickers(stickers, target)) # Output: 3This Python solution utilizes a recursive function with memoization to determine the minimum number of stickers needed. It preprocesses each sticker to count letter frequencies and applies DFS to explore useful sticker combinations. Memoization caches results for recurring target patterns, efficiently solving overlapping subproblems.
In this approach, a depth-first search (DFS) strategy is employed, utilizing bitmasking to represent the state of the target string. Each bit corresponds to a character position in the target, and a bit set to 1 indicates that a letter is needed. This makes it efficient to track state transitions and reuse computed outcomes using a bitmask.
The bitmask representation simplifies target state management, especially with small problems like this one where all state combinations fit comfortably in integer bit representation.
Time Complexity: O(2^T * n * T), where T is the target length and n the number of stickers, reflecting operations per state per sticker.
Space Complexity: O(2^T + n), for the dp table managing the state combinations and sticker reading.
1import java.util.*;
2
3
A Depth First Search (DFS) approach combined with memoization can efficiently solve this problem. By recursively exploring each possible configuration of using stickers to form the target and caching results for already computed subproblems, we minimize redundant calculations.
The idea is to count the frequency of each character needed for the target and use it to explore different combinations of stickers. We process stickers in such a way that each branch of our recursive call attempts to use the result of one sticker to reduce the requirement for the remaining target. Memoization is used to store results for target states we've already calculated for.
Time Complexity: O(2^T * N * L), where T is the length of the target, N is the number of stickers, and L is the average length of a sticker.
Space Complexity: O(T), for the memoization cache.
1from collections import Counter
2
3
This approach uses dynamic programming with bit masking to keep track of which characters in the target have been fulfilled. The idea is to use a bit mask to represent the completeness of parts of the target. Each bit in the mask represents whether a character in the target has been used or not.
By iterating over each sticker, we apply the possible contribution of each sticker to partially fulfill the target (indicated by updating the bitmask). A DP array helps record the minimum sticker usage to reach any given bitmap state of the target.
Time Complexity: O(N * T * 2^T), where T is the length of the target, N is the number of stickers.
Space Complexity: O(2^T), for the DP array.
1#include <bits/stdc++.h>
2using namespace std;
3
4class Solution {
public:
int minStickers(vector<string>& stickers, string target) {
int m = target.size();
vector<int> dp(1 << m, -1);
dp[0] = 0;
for (auto& sticker : stickers) {
vector<int> cnt(26, 0);
for (char c : sticker) cnt[c - 'a']++;
for (int state = 0; state < (1 << m); state++) {
if (dp[state] == -1) continue;
int cur[26] = {0};
for (int i = 0; i < m; i++)
if (state & (1 << i)) cur[target[i] - 'a']++;
int new_state = state;
for (int i = 0; i < m; i++) {
if (! (state & (1 << i)) && cnt[target[i] - 'a'] > cur[target[i] - 'a']) {
new_state |= (1 << i);
}
}
dp[new_state] = min(dp[new_state] == -1 ? INT_MAX : dp[new_state], dp[state] + 1);
}
}
return dp[(1 << m) - 1];
}
};The solution implemented in Java leverages bitmasking along with a memoization HashMap to efficiently calculate the minimum number of stickers required. The helper function explores possible sticker applications iteratively, adjusting the target state and caching results to avoid redundant calculations.
This Python solution utilizes DFS with memoization to navigate possible sticker usage combinations recursively. The Counter from the collections module is used to manage and compare character frequencies efficiently. A nested function dfs() handles the recursions, checking if a given state of the target string has been computed before or if it can be completed with the current stickers. If a target can be formed, it records the minimum number of stickers required.
This C++ solution employs a dynamic programming approach with bitmasking. The DP array keeps track of the minimum number of stickers required to attain various states of completeness of the target string, which are represented via bit masks. Each state explores how stickers can contribute to completing the target, and bit masking efficiently labels which parts are completed or pending.