I rarely use to comment on channels, but your channel made me to do so. This is the most amazing tutorial ever found + The Best UA-cam coding channel ever. Seriously! Thank You.... A Heart For You
First of really really nice explanation bhaiya and the dp approach is really 💥💥💥💥💯💯💯💯 But the reason for taking log base 2 in the time complexity is not that much clear can u explain it bhaiya And can u explain me how u are using dp here as far I know dp is using the past result so how are u using it here bhaiya otherwise very nice explanation 🥰🥰🥰
see dp[1] is stored initially right its basically bottom up type neeche se solution build karrahe hum dp[2] ke liye dp[1] usekiya but wo seeda nai dikra pointers use kiya humne and dp[n] depends on min(dp[i_2]*2,dp[i_3]*3,dp[i_5]*5) i_2 i_3 i_5 ye sab n ke less hi honga so we are using the values stored to build the higher solution so its Dp
Initially i2 i3 and i5 are 1 Then we checked 2*1 3*1 5*1 Among these 2*1 is the smallest hence we update the pointer for 2 i2++ which is (i2=2) Now array would have 0 1 2 Now we would check 2*2 3*1 5*1 As 3*1 is the smallest we would update the pointer for 3 i3++ so (i3=3) So array would have 0 1 2 3 Now i2=2 i3=2 i5=1 So according to this i3 is currently 2 So t[i3] should point to 2 and not 3
@@harshsuthar4947 Thanks for the reply, By the way i forget to remove this comment. I got intuition why i1, i2, i3 is just moving forward by one (We need to keep track of minimum ugly number for arr[i]) after watching complete algorithm.
May be that's why - The approach uses previously computed results (t[i2], t[i3], t[i5]) to avoid redundant calculations. The code efficiently handles the problem by storing the minimum values needed for the sequence, ensuring that each ugly number is generated in order.
May be that's why - The approach uses previously computed results (t[i2], t[i3], t[i5]) to avoid redundant calculations. The code efficiently handles the problem by storing the minimum values needed for the sequence, ensuring that each ugly number is generated in order.
class Solution { constructor() { this.mp = new Map(); } isUgly(n) { if (n 0) { if (this.isUgly(num)) { n--; } num++; } return num - 1; } } const solution = new Solution(); console.log(solution.nthUglyNumber(10)); // Example usage with n = 10
Bottom Up : class Solution { nthUglyNumber(n) { const t = new Array(n + 1); // t[i] = ith Ugly number; // we will return t[n] = nth ugly number t[1] = 1; // 1st Ugly number let i2, i3, i5; // i2th, i3rd, and i5th Ugly numbers initially point to 1, i.e., the first ugly number i2 = i3 = i5 = 1; for (let i = 2; i
I rarely use to comment on channels, but your channel made me to do so. This is the most amazing tutorial ever found + The Best UA-cam coding channel ever. Seriously! Thank You.... A Heart For You
You made my day. Thank you so much ❤️
You are the best tutor of DSA I have ever seen online.
In the middle of explanation, I realise we can use min-heap and i coded it and it works, thanks
Same
bhai usne intution min heap sol ki di or code dp wala krdiya sbka kat diya
@vikkasmaheshwari862 Apologies for any confusion. This seems quite an old video. In present, videos, I will ensure to avoid such confusion 😇🙏
no. 1 channel in programming 😍
man such a good lecture
very clear and concise explanation
After listening the explanation. I start to do coding myself. Thankyou for your explanations. They are always easy to understand.
Great Video mate as usual!!
You can assume I am binge watching /studying these whenever possible.
Thanks for the great content! Much appreciated
Means a lot to me 😇❤️🙏
Next Level Explanation sir........
Mazaa aaa gyaa
Thanks a lot man for listening to our requests.
Very helpful . Huge request to you❣
people like u r real gurus!
You, Me and all of us are genius.
Let’s all grow together and crack together ❤️❤️❤️
Big fan here of your teaching ❤
Excellent explanation. Thank you so much 🙏
Beautiful explanation
beautiful explanation
thanks
I did by binary search. BTW,amazing video.
1201. Ugly Number III
313. Super Ugly Number
I will try these today now..
btw u should cover it too if u can in your free time..
neat solution as always MIkbhaiya op ❤
Thanks a lot 💝
The real GENIUS .
Hats off to your teaching skill
Sir, please make video on Ugly Number III also
Awesome explanation, Thank you!
Thank you 🙏 😇
good tutorial. Looks like there is a lag between explanation and the screen update.
Sir , Aap contests ke solution bhi share kroge to bahut help milegi... ❤
Uploading today’s contest Qn-1 and Wn-4
They are good and will help us learn few things.
Thanks a lot bhaiya ❤❤
for 263. Ugly Number
the solution
is the TC also log n ?
Constraints:
-2^31
sir please make video on heap and trie topic videos concept and question playlist plzz plz🙏🙏🙏
do dp fundamentals video and upload as soon as possible bro
Check my DP Concepts playlist.
Sir can u please make a video on Super Ugly Number of DP Leetcode 313, Its a humble request please make it
Thank you Bhaiya 🙏🙏🙏
very nice explanation
sir plzz make video on ugly number 3 video
Thanks !
Why does i3++ gives the correct next ugly number? Any proof for it?
First of really really nice explanation bhaiya and the dp approach is really 💥💥💥💥💯💯💯💯
But the reason for taking log base 2 in the time complexity is not that much clear can u explain it bhaiya
And can u explain me how u are using dp here as far I know dp is using the past result so how are u using it here bhaiya otherwise very nice explanation 🥰🥰🥰
see dp[1] is stored initially right its basically bottom up type
neeche se solution build karrahe hum
dp[2] ke liye dp[1] usekiya but wo seeda nai dikra pointers use kiya humne
and dp[n] depends on min(dp[i_2]*2,dp[i_3]*3,dp[i_5]*5) i_2 i_3 i_5 ye sab n ke less hi honga so we are using the values stored to build the higher solution so its Dp
@@vishalplayzz2580 Thanks a lot
Awesome
Thanks
aaj k contest ka 2nd answer memoise ni ho rha tha ,pls do it(maximum energy boost from 2 drinks)
Uploading today’s contest Qn-1 and Qn-4
They are good and will help us learn few things.
thankyou sir
Thank you!
bhiaya Super ugly number ka v solution dedo n please
how are you taking t in 21st line without initializing it?
it should be arr
@19:47 why i3 is pointing to t[2] instead of t[3], can someone explain ? please!
Initially i2 i3 and i5 are 1
Then we checked
2*1
3*1
5*1
Among these 2*1 is the smallest hence we update the pointer for 2 i2++ which is (i2=2)
Now array would have 0 1 2
Now we would check
2*2
3*1
5*1
As 3*1 is the smallest we would update the pointer for 3
i3++ so (i3=3)
So array would have 0 1 2 3
Now
i2=2
i3=2
i5=1
So according to this i3 is currently 2
So t[i3] should point to 2 and not 3
@@harshsuthar4947 Thanks for the reply, By the way i forget to remove this comment. I got intuition why i1, i2, i3 is just moving forward by one (We need to keep track of minimum ugly number for arr[i]) after watching complete algorithm.
🔥❤️
what is dp here?
where we are using dp btw
May be that's why -
The approach uses previously computed results (t[i2], t[i3], t[i5]) to avoid redundant calculations.
The code efficiently handles the problem by storing the minimum values needed for the sequence, ensuring that each ugly number is generated in order.
ye dp kaise hua yr this looks greedy to me
May be that's why -
The approach uses previously computed results (t[i2], t[i3], t[i5]) to avoid redundant calculations.
The code efficiently handles the problem by storing the minimum values needed for the sequence, ensuring that each ugly number is generated in order.
@@gui-codesredundant how
@@gui-codesi think majorly it's Greedy
hii from 2024
another solution(priority queue)
class Solution {
public:
int nthUglyNumber(int n) {
vectorarr = {2,3,5};
long long count = 0;
priority_queuepq;
pq.push(1);
pq.push(3);
pq.push(5);
setst;
while(!pq.empty()){
long long top = pq.top();
pq.pop();
if(st.find(top)!= st.end()){
continue;
}
count++;
st.insert(top);
if(count == n){
cout
Wonderful. Thanks a lot for sharing 😇❤️
Thank you
Sir javascript se bhi kra do please
class Solution {
constructor() {
this.mp = new Map();
}
isUgly(n) {
if (n 0) {
if (this.isUgly(num)) {
n--;
}
num++;
}
return num - 1;
}
}
const solution = new Solution();
console.log(solution.nthUglyNumber(10)); // Example usage with n = 10
Bottom Up :
class Solution {
nthUglyNumber(n) {
const t = new Array(n + 1);
// t[i] = ith Ugly number;
// we will return t[n] = nth ugly number
t[1] = 1; // 1st Ugly number
let i2, i3, i5;
// i2th, i3rd, and i5th Ugly numbers initially point to 1, i.e., the first ugly number
i2 = i3 = i5 = 1;
for (let i = 2; i
java code
class Solution {
public int nthUglyNumber(int n) {
int[] arr=new int[n+1];
int i2;
int i3;
int i5;
i2=i3=i5=1;
arr[1]=1;
for(int i=2;i