9. Palindrome Number | Solve without String Conversion

Hey everyone, wanted to go a little light today on a friday and show you a solution to an easy problem. This solution will focus on using MATH! Hope you enjoy it.

[@sarasalvatierra7198]

i’ve searched so many videos to find the explanation of the math and you are the only one whose has explained it so well.

[@angeloloria6544]

great video, thank you for this breaking down the math for my own sake reversing a number, ex. 123: 1) use the modulo operator to divide by 10 and return the last decimal space, which will be 3 in this case (123/10 == 12.3) 2) use Math.floor, which will round down and return the nearest whole number, which will be 12 3) use the modulo operator in the same manner as step 1, which will return a 2 (12/10 == 1.2) 4) multiply the first number we extracted (3) times 10, to put it ahead of the second number we extracted (2) 5) add our multiplication result (30) to our second extracted number (2) to get 32. we've now reversed the last two digits of our original number (123) 6) use Math.floor on our last number, which gives us 1 7) use the modulo on that number again, which gives us 1 8) multiply our current reversed number (32) times 10 to put it ahead of our last extracted number (1) 9) add our multiplication result (320) to our last extracted number (1) to get 321, our reversed number

[@athirahasri9703]

I love this one. Thank you so much!

[@pinkhairblackman8141]

since the number will be a palindrome return reverse===x; will confirm if the number is the same since its a palindrome when reversed. math.floor(i/10) decrements until i>=1 is false in loop

[@UlianaSarkisova]

Thank you for such an easy to understand explanation!

[@Valer0n07]

What if an input = 10 or 1000 or 200? This solution will not generate a correct reversed output. But technically it will return a corect answer - false. This input (1234567899) also breaks this algorithm.

[@iharisyousaf]

What do you suggest for these cases

[@josey700]

Good day! Great content though the font size is just perfect for coding not so much for presenting

[@sayeesreeram]

Great Video :)

[@pweezy001]

Awesome video, thanks for sharing!

[@zherebetskyi4097]

Great and very helpfull video! Thank's man!

[@sarasalvatierra7198]

you are my savior

[@felipegarcialopez3701]

Great video!

[@rawstrife9762]

GOAT!

[@dshl21]

thankyou

[@jianqiuwang8798]

why i >= 1 ? can someone explain it ?

[@dannycho2287]

on line 8, we have the following: for(let i = x; i >= 1; i = Math.floor(i/10)) {...} Now, take an input of 121 as an example. The first iteration of the for loop takes us to i = 121; i is still greater than or equal to 1; i becomes Math.floor(121/10), which is 12. The second iteration of the for loop takes us to i = 12; i is still greater than or equal to 1; i becomes Math.floor(12/10), which is 1. The third iteration of the for loop takes us to i = 1; i is still greater than or equal to 1; i becomes Math.floor(1/10), which is now 0. Since i is now NOT greater than or equal to 1, we come to an end of the for loop. By this time, we already have a reverse version of the number 121.

[@jianqiuwang8798]

@@dannycho2287 hello Danny, thank you so much for the explanation, I really appreciate it!

[@parthgoswami3221]

@@dannycho2287 great explanation! thank you

[@rajbannasa7662]

thank you so much sir

[@iharisyousaf]

It fails in case of 10

[@Mikeishere1234]

make more of these please

[@athraafaraj8387]

Very help

[@RichardYoung11]

bring back the lofi!

[@ThinkFWD]

I just brought it back ! :), release a new vid with it. let me know if you have any question you want solved/explained