diff --git a/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README.md b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README.md index 16eadec79aa72..e4be1bb9e729e 100644 --- a/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README.md +++ b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README.md @@ -73,17 +73,17 @@ tags: ### 方法一：贪心 + 哈希表 -我们先用哈希表 `cnt` 统计每个单词出现的次数。 +我们先用一个哈希表 $\textit{cnt}$ 统计每个单词出现的次数。 -遍历 `cnt` 中的每个单词 $k$ 以及其出现次数 $v$： +遍历 $\textit{cnt}$ 中的每个单词 $k$ 以及其出现次数 $v$： -如果 $k$ 中两个字母相同，那么我们可以将 $\left \lfloor \frac{v}{2} \right \rfloor \times 2$ 个 $k$ 连接到回文串的前后，此时如果 $k$ 还剩余一个，那么我们可以先记录到 $x$ 中。 +- 如果 $k$ 中两个字母相同，那么我们可以将 $\left \lfloor \frac{v}{2} \right \rfloor \times 2$ 个 $k$ 连接到回文串的前后，此时如果 $k$ 还剩余一个，那么我们可以先记录到 $x$ 中。 -如果 $k$ 中两个字母不同，那么我们要找到一个单词 $k'$，使得 $k'$ 中的两个字母与 $k$ 相反，即 $k' = k[1] + k[0]$。如果 $k'$ 存在，那么我们可以将 $\min(v, cnt[k'])$ 个 $k$ 连接到回文串的前后。 +- 如果 $k$ 中两个字母不同，那么我们要找到一个单词 $k'$，使得 $k'$ 中的两个字母与 $k$ 相反，即 $k' = k[1] + k[0]$。如果 $k'$ 存在，那么我们可以将 $\min(v, \textit{cnt}[k'])$ 个 $k$ 连接到回文串的前后。 遍历结束后，如果 $x$ 不为空，那么我们还可以将一个单词连接到回文串的中间。 -时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为 `words` 的长度。 +时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为单词的数量。 @@ -111,7 +111,7 @@ class Solution { public int longestPalindrome(String[] words) { Map cnt = new HashMap<>(); for (var w : words) { - cnt.put(w, cnt.getOrDefault(w, 0) + 1); + cnt.merge(w, 1, Integer::sum); } int ans = 0, x = 0; for (var e : cnt.entrySet()) { @@ -183,6 +183,26 @@ func longestPalindrome(words []string) int { } ``` +#### TypeScript + +```ts +function longestPalindrome(words: string[]): number { + const cnt = new Map(); + for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1); + let [ans, x] = [0, 0]; + for (const [k, v] of cnt.entries()) { + if (k[0] === k[1]) { + x += v & 1; + ans += Math.floor(v / 2) * 2 * 2; + } else { + ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2; + } + } + ans += x ? 2 : 0; + return ans; +} +``` + diff --git a/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README_EN.md b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README_EN.md index e947717112683..029e61dd8d93e 100644 --- a/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README_EN.md +++ b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/README_EN.md @@ -73,7 +73,19 @@ Note that "ll" is another longest palindrome that can be created, and -### Solution 1 +### Solution 1: Greedy + Hash Table + +First, we use a hash table $\textit{cnt}$ to count the occurrences of each word. + +Iterate through each word $k$ and its count $v$ in $\textit{cnt}$: + +- If the two letters in $k$ are the same, we can concatenate $\left \lfloor \frac{v}{2} \right \rfloor \times 2$ copies of $k$ to the front and back of the palindrome. If there is one $k$ left, we can record it in $x$ for now. + +- If the two letters in $k$ are different, we need to find a word $k'$ such that the two letters in $k'$ are the reverse of $k$, i.e., $k' = k[1] + k[0]$. If $k'$ exists, we can concatenate $\min(v, \textit{cnt}[k'])$ copies of $k$ to the front and back of the palindrome. + +After the iteration, if $x$ is not empty, we can also place one word in the middle of the palindrome. + +The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the number of words. @@ -101,7 +113,7 @@ class Solution { public int longestPalindrome(String[] words) { Map cnt = new HashMap<>(); for (var w : words) { - cnt.put(w, cnt.getOrDefault(w, 0) + 1); + cnt.merge(w, 1, Integer::sum); } int ans = 0, x = 0; for (var e : cnt.entrySet()) { @@ -173,6 +185,26 @@ func longestPalindrome(words []string) int { } ``` +#### TypeScript + +```ts +function longestPalindrome(words: string[]): number { + const cnt = new Map(); + for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1); + let [ans, x] = [0, 0]; + for (const [k, v] of cnt.entries()) { + if (k[0] === k[1]) { + x += v & 1; + ans += Math.floor(v / 2) * 2 * 2; + } else { + ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2; + } + } + ans += x ? 2 : 0; + return ans; +} +``` + diff --git a/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.java b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.java index 30dd228456061..228a887591207 100644 --- a/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.java +++ b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.java @@ -2,7 +2,7 @@ class Solution { public int longestPalindrome(String[] words) { Map cnt = new HashMap<>(); for (var w : words) { - cnt.put(w, cnt.getOrDefault(w, 0) + 1); + cnt.merge(w, 1, Integer::sum); } int ans = 0, x = 0; for (var e : cnt.entrySet()) { diff --git a/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.ts b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.ts new file mode 100644 index 0000000000000..04ba6edec8550 --- /dev/null +++ b/solution/2100-2199/2131.Longest Palindrome by Concatenating Two Letter Words/Solution.ts @@ -0,0 +1,15 @@ +function longestPalindrome(words: string[]): number { + const cnt = new Map(); + for (const w of words) cnt.set(w, (cnt.get(w) || 0) + 1); + let [ans, x] = [0, 0]; + for (const [k, v] of cnt.entries()) { + if (k[0] === k[1]) { + x += v & 1; + ans += Math.floor(v / 2) * 2 * 2; + } else { + ans += Math.min(v, cnt.get(k[1] + k[0]) || 0) * 2; + } + } + ans += x ? 2 : 0; + return ans; +}