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Commit d1b1de9

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feat: add solutions to lc problems: No.1813,1814,1816 (#1758)
1 parent f847c0f commit d1b1de9

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10 files changed

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10 files changed

+191
-4
lines changed

‎solution/1800-1899/1813.Sentence Similarity III/README.md‎

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@@ -168,6 +168,27 @@ func areSentencesSimilar(sentence1 string, sentence2 string) bool {
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}
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```
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### **TypeScript**
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```ts
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function areSentencesSimilar(sentence1: string, sentence2: string): boolean {
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const words1 = sentence1.split(' ');
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const words2 = sentence2.split(' ');
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if (words1.length < words2.length) {
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return areSentencesSimilar(sentence2, sentence1);
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}
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const [m, n] = [words1.length, words2.length];
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let [i, j] = [0, 0];
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while (i < n && words1[i] === words2[i]) {
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++i;
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}
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while (j < n && words1[m - 1 - j] === words2[n - 1 - j]) {
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++j;
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}
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return i + j >= n;
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}
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```
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### **...**
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```

‎solution/1800-1899/1813.Sentence Similarity III/README_EN.md‎

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@@ -46,6 +46,16 @@
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## Solutions
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**Solution 1: Two Pointers**
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We split the two sentences into two word arrays `words1` and `words2` by spaces. Let the lengths of `words1` and `words2` be $m$ and $n,ドル respectively, and assume that $m \ge nn.
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We use two pointers $i$ and $j,ドル initially $i = j = 0$. Next, we loop to check whether `words1[i]` is equal to `words2[i]`, and if so, pointer $i$ continues to move right; then we loop to check whether `words1[m - 1 - j]` is equal to `words2[n - 1 - j]`, and if so, pointer $j$ continues to move right.
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After the loop, if $i + j \ge n,ドル it means that the two sentences are similar, and we return `true`; otherwise, we return `false`.
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The time complexity is $O(L),ドル and the space complexity is $O(L),ドル where $L$ is the sum of the lengths of the two sentences.
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<!-- tabs:start -->
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### **Python3**
@@ -145,6 +155,27 @@ func areSentencesSimilar(sentence1 string, sentence2 string) bool {
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}
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```
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### **TypeScript**
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```ts
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function areSentencesSimilar(sentence1: string, sentence2: string): boolean {
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const words1 = sentence1.split(' ');
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const words2 = sentence2.split(' ');
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if (words1.length < words2.length) {
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return areSentencesSimilar(sentence2, sentence1);
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}
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const [m, n] = [words1.length, words2.length];
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let [i, j] = [0, 0];
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while (i < n && words1[i] === words2[i]) {
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++i;
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}
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while (j < n && words1[m - 1 - j] === words2[n - 1 - j]) {
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++j;
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}
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return i + j >= n;
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}
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```
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### **...**
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```
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function areSentencesSimilar(sentence1: string, sentence2: string): boolean {
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const words1 = sentence1.split(' ');
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const words2 = sentence2.split(' ');
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if (words1.length < words2.length) {
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return areSentencesSimilar(sentence2, sentence1);
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}
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const [m, n] = [words1.length, words2.length];
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let [i, j] = [0, 0];
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while (i < n && words1[i] === words2[i]) {
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++i;
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}
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while (j < n && words1[m - 1 - j] === words2[n - 1 - j]) {
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++j;
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}
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return i + j >= n;
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}

‎solution/1800-1899/1814.Count Nice Pairs in an Array/README.md‎

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@@ -298,6 +298,30 @@ var countNicePairs = function (nums) {
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};
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```
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### **TypeScript**
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```ts
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function countNicePairs(nums: number[]): number {
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const rev = (x: number): number => {
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let y = 0;
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while (x) {
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y = y * 10 + (x % 10);
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x = Math.floor(x / 10);
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}
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return y;
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};
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const mod = 10 ** 9 + 7;
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const cnt = new Map<number, number>();
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let ans = 0;
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for (const x of nums) {
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const y = x - rev(x);
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ans = (ans + (cnt.get(y) ?? 0)) % mod;
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cnt.set(y, (cnt.get(y) ?? 0) + 1);
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}
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return ans;
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}
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```
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### **...**
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```

‎solution/1800-1899/1814.Count Nice Pairs in an Array/README_EN.md‎

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## Solutions
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**Method 1: Equation Transformation + Hash Table**
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For the index pair $(i, j),ドル if it satisfies the condition, then we have $nums[i] + rev(nums[j]) = nums[j] + rev(nums[i]),ドル which means $nums[i] - nums[j] = rev(nums[j]) - rev(nums[i])$.
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Therefore, we can use $nums[i] - rev(nums[i])$ as the key of a hash table and count the number of occurrences of each key. Finally, we calculate the combination of values corresponding to each key, add them up, and get the final answer.
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Note that we need to perform modulo operation on the answer.
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The time complexity is $O(n \times \log M),ドル where $n$ and $M$ are the length of the $nums$ array and the maximum value in the $nums$ array, respectively. The space complexity is $O(n)$.
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<!-- tabs:start -->
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### **Python3**
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};
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```
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### **TypeScript**
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```ts
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function countNicePairs(nums: number[]): number {
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const rev = (x: number): number => {
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let y = 0;
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while (x) {
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y = y * 10 + (x % 10);
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x = Math.floor(x / 10);
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}
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return y;
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};
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const mod = 10 ** 9 + 7;
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const cnt = new Map<number, number>();
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let ans = 0;
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for (const x of nums) {
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const y = x - rev(x);
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ans = (ans + (cnt.get(y) ?? 0)) % mod;
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cnt.set(y, (cnt.get(y) ?? 0) + 1);
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}
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return ans;
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}
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```
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### **...**
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```
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function countNicePairs(nums: number[]): number {
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const rev = (x: number): number => {
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let y = 0;
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while (x) {
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y = y * 10 + (x % 10);
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x = Math.floor(x / 10);
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}
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return y;
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};
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const mod = 10 ** 9 + 7;
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const cnt = new Map<number, number>();
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let ans = 0;
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for (const x of nums) {
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const y = x - rev(x);
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ans = (ans + (cnt.get(y) ?? 0)) % mod;
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cnt.set(y, (cnt.get(y) ?? 0) + 1);
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}
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return ans;
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}

‎solution/1800-1899/1816.Truncate Sentence/README.md‎

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@@ -59,7 +59,7 @@ s 中的单词为 ["What", "is" "the", "solution", "to", "this", "problem"]
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**方法一:模拟**
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我们从前往后遍历字符串 $s,ドル对于当前遍历到的字符 $s[i],ドル如果 $s[i]$ 是空格,那么 $k$ 自减 1,当 $k$ 为 0 时,说明已经截取了 $k$ 个单词,截取字符串 $s[0:i]$ 返回即可。
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我们从前往后遍历字符串 $s,ドル对于当前遍历到的字符 $s[i],ドル如果 $s[i]$ 是空格,那么 $k$ 自减 1ドル$,当 $k$ 为 0ドル$ 时,说明已经截取了 $k$ 个单词,截取字符串 $s[0..i)$ 返回即可。
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遍历结束,返回 $s$ 即可。
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}
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```
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### **TypeScript**
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```ts
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function truncateSentence(s: string, k: number): string {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] === ' ' && --k === 0) {
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return s.slice(0, i);
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}
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}
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return s;
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}
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```
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### **JavaScript**
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```js
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*/
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var truncateSentence = function (s, k) {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] == ' ' && --k == 0) {
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if (s[i] === ' ' && --k === 0) {
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return s.slice(0, i);
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}
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}

‎solution/1800-1899/1816.Truncate Sentence/README_EN.md‎

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@@ -54,6 +54,14 @@ Hence, you should return &quot;What is the solution&quot;.</pre>
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## Solutions
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**Method 1: Simulation**
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We traverse the string $s$ from the beginning. For the current character $s[i],ドル if it is a space, we decrement $k$. When $k$ becomes 0ドル,ドル it means that we have extracted $k$ words, so we return the substring $s[0..i)$.
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After the traversal, we return $s$.
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The time complexity is $O(n),ドル where $n$ is the length of the string $s$. Ignoring the space complexity of the answer, the space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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}
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```
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### **TypeScript**
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```ts
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function truncateSentence(s: string, k: number): string {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] === ' ' && --k === 0) {
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return s.slice(0, i);
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}
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}
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return s;
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}
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```
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### **JavaScript**
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```js
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*/
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var truncateSentence = function (s, k) {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] == ' ' && --k == 0) {
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if (s[i] === ' ' && --k === 0) {
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return s.slice(0, i);
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}
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}

‎solution/1800-1899/1816.Truncate Sentence/Solution.js‎

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*/
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var truncateSentence = function (s, k) {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] == ' ' && --k == 0) {
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if (s[i] === ' ' && --k === 0) {
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return s.slice(0, i);
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}
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}
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function truncateSentence(s: string, k: number): string {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] === ' ' && --k === 0) {
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return s.slice(0, i);
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}
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}
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return s;
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}

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