LeetCode Solution - 121 Best time to Buy and Sell Stock | Javascript

[@david_yuzhakov]

Thanks, bro, I haven't found a simple and high-quality video like yours anywhere.

[@ThinkFWD]

Please feel free to leave a comment and let me know what you think !

[@ThinkFWD]

From user @ leetcode @dermatobia The solution above could be improved by not actually changing the input array to cache profits. we could actually remove a few lines of code and achieve the same output! .. we don't actually have to cache the profits in our prices array. Since the profit var does it indirectly already. :) var maxProfit = function(prices) { let buy = prices[0]; let profit = 0; for (let i = 0; i < prices.length; i++) { if (buy > prices[i]) { buy = prices[i]; } else { profit = Math.max(prices[i]-buy, profit); } } return profit; };

[@sawyerburnett8319]

Good call. I was about to say this. In this case it doesn't add any value to update the array, and could cause confusion or bugs in the solution.

[@jerryliu862]

I appreciate you sharing this video, It improves my coding skill.

[@edetmmekut809]

Hi really cool explaination ,but u dont need to mutate the array that extra over head cost

[@riturathinsharma5931]

This is a great solution mate , but the best solution is “divide and conquer”.

[@initialb811]

great stuff! small recommendation: can you add comments in your code as a way for viewers to follow along better? I find myself needing to scrub back a couple of times to review why something was written. This video is definitely easy enough to follow, but it's something I find myself needing to do since I'm binge watching your videos. =D Also, subscribed! Thanks for making these videos!

[@ibeeliot]

i binge on your videos. keep it up.

[@ThinkFWD]

will release one tonight :) stay tuned !

[@kathiateran5826]

Really appreciate your videos, you're great at explaining things!

[@ThinkFWD]

Thanks! Appreciate it

[@DmitriyMalayevProfile]

You're an awesome instructor. Can you please check why the volume is so low?

[@pinkhairblackman8141]

reason he set price[i]=0 is the edge case where you can never make a profit

[@nurlanikhsan3692]

Nice

[@ThinkFWD]

Thanks

[@kidoo1567]

U are best

[@rakeshrockie3213]

please solve more problems. love your videos

[@ThinkFWD]

will do ! planning on putting a few more stay tuned

[@OmarAhmed-xc7bs]

Nice work and explanation but please remove the music, it is a big distraction, tysm.

[@EshsUrbanPulseYT]

How this is dynamic programming, could anyone please explain?

[@KolaIL]

var maxProfit = function(prices) { let minprice = prices[0]; let profit = 0; for (let i = 0; i < prices.length; i++) { if (prices[i] < minprice) { minprice = prices[i]; } else if (prices[i] - minprice > profit) { profit = prices[i] - minprice; } } return profit };

[@user-zp1dv4yh5e]

Hey, I tried you appraoach , but it's not working function findMaxProfit2(prices){ let buy = [0] prices[0] = 0 let profit = 0 for(let i = 1; i < prices.length ;i++){ if(buy > prices[i] ){ buy = prices[i] price[i] = 0 }else{ profit = Math.max(prices[i]-buy, profit) } } return profit }