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feat: add typescript solution to lc problem: No.2291

No.2291.Maximum Profit From Trading Stocks

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Original file line number	Diff line number	Diff line change
`@@ -79,7 +79,16 @@ It can be shown that the maximum profit you can make is 0.`
`79`	`79`	`### TypeScript`
`80`	`80`
`81`	`81`	```ts
`82`		`-`
	`82`	`+function maximumProfit(present: number[], future: number[], budget: number): number {`
	`83`	`+ let packet = present.map((v, i) => [v, future[i] - v]);`
	`84`	`+ let dp = new Array(budget + 1).fill(0);`
	`85`	`+ for (let [v, w] of packet) {`
	`86`	`+ for (let j = budget; j >= v; j--) {`
	`87`	`+ dp[j] = Math.max(dp[j], dp[j - v] + w);`
	`88`	`+ }`
	`89`	`+ }`
	`90`	`+ return dp[budget];`
	`91`	`+};`
`83`	`92`	```
`84`	`93`
`85`	`94`	`### ...`

Lines changed: 10 additions & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -71,7 +71,16 @@ It can be shown that the maximum profit you can make is 0.`
`71`	`71`	`### TypeScript`
`72`	`72`
`73`	`73`	```ts
`74`		`-`
	`74`	`+function maximumProfit(present: number[], future: number[], budget: number): number {`
	`75`	`+ let packet = present.map((v, i) => [v, future[i] - v]);`
	`76`	`+ let dp = new Array(budget + 1).fill(0);`
	`77`	`+ for (let [v, w] of packet) {`
	`78`	`+ for (let j = budget; j >= v; j--) {`
	`79`	`+ dp[j] = Math.max(dp[j], dp[j - v] + w);`
	`80`	`+ }`
	`81`	`+ }`
	`82`	`+ return dp[budget];`
	`83`	`+};`
`75`	`84`	```
`76`	`85`
`77`	`86`	`### ...`

Lines changed: 10 additions & 0 deletions

Original file line number	Diff line number	Diff line change
`@@ -0,0 +1,10 @@`
	`1`	`+function maximumProfit(present: number[], future: number[], budget: number): number {`
	`2`	`+ let packet = present.map((v, i) => [v, future[i] - v]);`
	`3`	`+ let dp = new Array(budget + 1).fill(0);`
	`4`	`+ for (let [v, w] of packet) {`
	`5`	`+ for (let j = budget; j >= v; j--) {`
	`6`	`+ dp[j] = Math.max(dp[j], dp[j - v] + w);`
	`7`	`+ }`
	`8`	`+ }`
	`9`	`+ return dp[budget];`
	`10`	`+};`

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