LWC 53：691. Stickers to Spell Word
LWC 53：691. Stickers to Spell Word
LWC 53：691. Stickers to Spell Word
传送门：691. Stickers to Spell Word
Problem:
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 target string 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. What is the minimum number of stickers that you need to spell out the target? If the task is impossible, return -1.
Example 1:
Input: [“with”, “example”, “science”], “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: [“notice”, “possible”], “basicbasic” Output: -1 Explanation: We can’t form the target “basicbasic” from cutting letters from the given stickers.
Note:
stickers has length in the range [1, 50]. stickers consists of lowercase English words (without apostrophes). target has length in the range [1, 15], and consists of lowercase English letters. In all test cases, all words were chosen randomly from the 1000 most common US English words, and the target was chosen as a concatenation of two random words. The time limit may be more challenging than usual. It is expected that a 50 sticker test case can be solved within 35ms on average.
思路： 因为字母顺序无关，所以只需要统计target中每个字母出现的次数，接着对于给定的stickers，如果选择了某个sticker（当然你也可以不选），那么就从target中减去共同出现的字母频次（以sticker为准），这样一来，该问题就变成了重复子问题。
代码如下：
static final int INF = 0x3f3f3f3f;
Map<String, Integer> mem = new HashMap<>();
int f(String target, int n) {
if (mem.containsKey(target)) return mem.get(target);
int[] tar = preprocess(target);
int ans = INF;
for (int i = 0; i < n; ++i) {
StringBuilder sb = new StringBuilder();
for (int j = 0; j < 32; ++j) {
if (tar[j] > 0) {
for (int k = 1; k <= Math.max(0, tar[j] - sticker_map[i][j]); ++k) {
sb.append((char)(j + 'a'));
}
}
}
String sub = sb.toString();
if (sub.length() != target.length()) {
ans = Math.min(ans, f(sub, n) + 1);
}
}
mem.put(target, ans);
return ans;
}
int[][] sticker_map;
public int minStickers(String[] stickers, String target) {
if (judge(stickers, target)) return -1;
int n = stickers.length;
sticker_map = new int[n][32];
for (int i = 0; i < n; ++i) {
sticker_map[i] = preprocess(stickers[i]);
}
mem.put("", 0);
return f(target, n);
}
boolean judge(String[] stickers, String target) {
int[] map = new int[32];
for (char c : target.toCharArray()) {
if (map[c - 'a'] == 0)
map[c - 'a']++;
}
for (String s : stickers) {
for (char c : s.toCharArray()) map[c - 'a'] --;
}
for (int i = 0; i < 32; ++i) {
if (map[i] >= 1) return true;
}
return false;
}
int[] preprocess(String target) {
int[] map = new int[32];
for (char c : target.toCharArray()) map[c - 'a']++;
return map;
}实际上是状态压缩DP，采用BOTTOM-UP，代码如下：
public int minStickers(String[] stickers, String target) {
int N = target.length();
int[] dp = new int[1 << N];
for (int i = 1; i < 1 << N; ++i) dp[i] = -1;
for (int state = 0; state < 1 << N; ++state) {
if (dp[state] == -1) continue;
for (String sticker : stickers) {
int now = state;
for (char c : sticker.toCharArray()) {
for (int i = 0; i < N; ++i) {
if (((now >> i) & 1) == 1) continue;
if (c == target.charAt(i)) {
now |= 1 << i;
break;
}
}
}
if (dp[now] == -1 || dp[now] > dp[state] + 1) {
dp[now] = dp[state] + 1;
}
}
}
return dp[(1 << N) - 1];
}从递归的版本可以看出，前后状态发生变化的实际是target，HashMap记录的也是target各字符出现频次的【综合状态】，因此容易想到子问题出现与否只与target中每一位是否能被sticker构造相关（顺序无关），而target长度最大只有15位，因此可以用int数组表示当前位是否被选择，消耗内存O(1 << N).