Pratica di Shadowing: Kth Smallest element in a matrix | Leetcode #378 - Impara a parlare inglese con i video

Hello guys welcome back to TechDose

and in this video we will see the kth smallest element in a sorted matrix

which is from lead code number 378

and it is based on a searching algorithm let us now

look at the problem statement in this problem given an n by n matrix where each of the rows

and columns are sorted in ascending order return the kth smallest element in the matrix note

that they are asking about the kth smallest element in the sorted order

and not about the kth distinct element so you must find a solution

which has a memory complexity of better than order of n square

so let's take an example for better understanding in our case we are given a row wise

and column wise sorted matrix with non -decreasing order

and the matrix size is n by n our goal is

to find the k smallest element let's consider a one dimensional matrix it is not an n by n matrix

but i'm just giving you an example that

if you have a single dimensional array then in this case

if they are asking about the third smallest element then it will be two

and not three because they are not talking about the distinct element they are talking about even

if you have duplicates you can count them okay now

if you have an n by n matrix like we have

on the right hand side you see we have taken four

by four matrix every row is sorted in ascending order

and every column is also sorted in ascending order you see therefore in this case

if you have to find the kth smallest element

and let's find out the fourth smallest element what you can

do is you can convert these two dimensional matrix into one dimensional array

and simply get all the elements sorted

and you can find out the fourth smallest element

and in this case

if you enumerate everything in a non -decreasing order then the fourth element will be whatever fits at index 3

so that will be 2 in our case okay.

Now one technique can be to just flatten this entire matrix into a single dimensional array

and apply the sorting algorithm which will be n square log of n square

but a better technique could be to just push all the first elements into a heap

and then pop out the element which is the minimum out of this

and then push the next element of

that row this technique was already explained in the sliding window

maximum problem you can follow this the video link will be in the i button

or you can check out the description as well this can be solved using heap

and the time complexity will be n square log n

and the space complexity will be n square

but this is not the best approach in this case this

is just one of the approaches now in this case we can observe

that since we already know

that all the rows are sorted in non -decreasing order all

the columns are also sorted in non -decreasing order therefore whatever

will be our answer will be in the range of the smallest

and the largest one in this case the smallest one is 1

and the largest number is 11 so our answer cannot lie below 1

or beyond 11 right therefore the kth smallest element must be within the range of 1 to 11

and it must definitely be present in the matrix therefore what

we can do is we can search in our answer range okay we know the range of answer

so the minimum possible answer you can say this is lowest possible answer is 1 and the highest possible answer is 11. so

if we want to search in this answer range then one approach we can take is to take n pointers

and simply we can apply each of the pointer to the

first element just like we did in the heap previously it is a similar approach

but this is kind of a more simpler one in this

case you just compare all the n pointers like you can take p q r s

and you compare all these pointers p q r s

and see which one is smaller whichever is smaller you just have a counter

so you can have a count value c equals to zero now

when you skip this you move it to the right and the count will increase by 1.

Again, compare 2, 2, 5, 7.

Whichever is smaller, you just skip there.

You could have skipped the second 2 as well.

When two values are same, you can skip any one.

Now, the count of elements you have skipped is 2.

Now, let's say I was talking about the third smallest element.

Now, since you have skipped two elements, so out of these, whatever we are pointing, one of these will be the third smallest element

because two elements have already been skipped so

which one do you think will be the third smallest the

smallest of all these pqrs pointers right the smallest one will be the next one

which is skipped and if you find out the smallest it will be two okay

so the answer in this case will be equals to two this approach is also valid

and you see that for finding out each element like for skipping

which element do I need to skip you have to compare all the PQRS pointers.

so it is an n by n matrix

so there are n rows here right

so for each element skip you have to make n comparisons find out the minimum out of n elements

and you know that the highest k value can be the last one

which is n square itself therefore the time complexity in this case will be n square

which is the highest k value multiplied by for skipping each

element you need to do n comparisons therefore order of n cube will be the time complexity

but the space complexity will just be order of 1 okay

because we are not taking any extra space

so this is a simple approach not as good as the heap approach

but the heap approach used extra space in this case you

are optimizing space by taking more time to solve the same problem

so heap approach can be considered better since we already know

that all the rows and columns are already sorted therefore we also know

that our answer will be in the range of 1 to 11 on a sorted array

or on a sorted matrix we can always apply binary search.

So let's see how we can apply binary search in order to find the kth smallest element.

Let's say that our kth smallest element k value is equal to 3.

So we want to find the third smallest element.

So we already know the range of answer.

What is the lowest answer possible?

The lowest answer possible is one which is given by the top left corner.

The highest value which can be possible in the answer is the bottom right corner

which is 11 okay and we can say this since it is row wise column wise sorted

100

so we have to take a candidate so let's take the candidate as mid

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which will be low plus high minus low by 2 this

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is the technique i already told you in the binary partition how to do this

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and why we do this

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so the answer will be 1 plus 11 minus 1 by 2 which comes out to be 6.

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So, let's try this out like this becomes the candidate

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that what if I take 6 is this 6 the kth smallest element that is what I am trying to find.

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So, in order to find

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if 6 is the kth smallest element thus in the simple

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approach what we could do is in every list we could

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count how many elements are less than 6 less than equals to 6 you can say.

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So, in the first row there are four such elements

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which are less than equals to six in the second row three elements in the third row two elements

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and in the fourth row there are no elements

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which are less than equals to six

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so total elements less than equals to six are four plus three plus two

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which is nine six is kind of the ninth smallest element

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but we are talking about the third smallest element right so

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if it is the third smallest element

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and six is actually the ninth smallest element then I should

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decrease the number like I will definitely get a number lower

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than 6 right therefore in order to decrease it my low value will remain the same

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but my high value will be equals to the mid value

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you will understand at the end like why I did not

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make high equals to mid minus 1 there are multiple variations of binary search

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and you have to adjust accordingly sometimes high will become equals

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to mid minus 1 sometimes it will not be in this

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case we should not take i equals to mid minus one by the end you will definitely understand

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so let's just take it and let's see what will be the new mid value

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that will be one plus six minus one by two

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so this comes out to be three let's see how many elements are less than equals to three

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now in the previous case we had applied i mean the simple linear search

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which was again very time consuming

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so what we should do is instead of applying like how

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many elements are less than equals to three on a given row, we should do that by using binary search

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and how many elements are less than equals to three can be counted

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or the first index which is having a value greater than three can be given by upper bound.

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So if you apply let's say upper bound of two on the first row, then it will stop at index three

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because index three is the first index where the elements are

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greater than two okay this is the first occurrence in this

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row therefore for 6 as well we should have applied binary search so

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if we apply binary search on 6 again then the upper

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bound will stop at index 4 here the upper bound will

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stop at index 3 here it will stop at index 2

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and here it will stop at index 0

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so you can add up all the indices 4 3 7 7 plus 2 9

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so again the answer is same

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but for applying binary search upper bound it takes order of log of the length of the single row right

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so that is log of n now again we are applying the same for three

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so for three the upper bound on the first row will

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stop at index three the upper bound on the on the in on the second row

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which is index one row will stop at one for this two it will stop at zero

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and for three it will stop at zero so how many elements did we see we saw three plus one, which is four elements, right.

153

So, this is saying that this is the fourth smallest.

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Okay, three is the fourth smallest probably.

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Therefore, what we should do, we should search for a lower number.

156

Therefore, your L will be equals to one and your H will be equals to three.

157

In this case, let's take the mid value and this will be one plus three minus one by two, which will be equals to two.

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Now let's find out the same for two.

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So for two in the first row, we will stop at index three in the second row will stop at index one.

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And here we will stop at index zero.

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And here as well we will stop at index zero.

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So for this as well, we got the count equals to four.

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So definitely I will keep on searching.

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Low will be one and high will be equals to two now.

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So mid will be equals to low plus high by two or you can say low plus high minus low by two, which will be equals to 1.

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Now for one, my search will stop here.

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And for all the other rows, it will stop at index zero.

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So here the count will be equals to one, which is less than the third smallest element, right.

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So we should go on the right hand side.

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So low will be mid plus one and low will be two, high will be two.

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And as soon as low and high are same, we will stop there.

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And whatever is the high value, we will return it as an answer, high value or low value whatever so in this case we can say

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that 2 is the answer 2 is the third smallest element you can see

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that if I had made high equals to mid minus 1 then in this step where

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the high value was 3 and we had taken the mid as 2

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and the 2's count had come out to be 4 in

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this update for the next value the high value could have been made equals to 1

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which is mid minus 1 right so that would not preserve the result

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so i think

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if you just do some dry runs you will yourself understand about should i update it to mid minus one

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or should i update it to equals to mid

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and the same can happen for multiple questions

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when you update low sometimes it can be just low equals to mid

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or sometimes it can become mid plus one

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so it can vary depending on the type of question you are solving

186

so let's discuss about the time complexity how many times do

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we need to do this binary search here like binary search on the range of answer right

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so what can be the range of answer it will be

189

the maximum minus minimum you can take the mod of this this will be the given range of answer

190

and since we are doing binary search on the range

191

so it will be log of this

192

which we can write as log of r

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which is the binary search on the range of numbers the

194

range will be equals to max minus mid okay now having

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seen this for each of this binary search on the range

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of answer we are applying binary search on all the rows how many rows are there n rows

197

and what is the size of each row the size of

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each row is n therefore log of n operation is required to find upper bound on each row

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and there are n rows therefore n log n time is

200

spent for each verification of I mean whenever you find a mid value you need to count how many

201

elements are less than equals to the given mid value right

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so it will take n log n time on the entire matrix

203

and we are repeating it log r times therefore the time complexity will be n log n log r

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which is very very less as compared to order of n square

205

and the space complexity is order of one

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so this is one of the best approach for this problem

207

let's look at the code i would like to announce about our live training programs data structures

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and i will help you clear all your doubts in the

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one -on -one sessions you can know more about this by querying us on the whatsapp number

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or you can also visit our website techdose .co .in in this code you are given a matrix and a value k.

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So, you can simply find out how many rows are there.

217

And since it is an n by n matrix, you know how many rows and columns will be there at the end, find out the lowest possible value, which is the first top left corner and the highest possible value,

218

which is the bottom right corner.

219

And then we have taken a mid variable.

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So, this is binary search on the range of answer and you can see how I have updated the high value,

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right so if the count is less than k then I will update the low to mid plus one

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and if it is higher than k then high will be equals to mid higher

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or equal okay and this is the function for upper bound right

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and you can like easily solve this you can read about

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upper bound how it is done now what can be a

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follow -up question here a follow -up question can be can

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you implement the upper bound another follow -up question is why have you updated this high equals to mid

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so you can give an example and

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uh try to make him understand like whatever example i have given the same example you can give

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and if you do high equals to mid minus one then my input will not give correct answer okay

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so you can take this input

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and try it out i hope you understood this approach please also practice the heap approach

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which was explained in the sliding window maximum problem link will be in the description below like

234

and share our video and subscribe to our channel in order to watch Watch more of this programming video, see you guys in the next video, thank you.

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Questo video contiene 234 frasi e 2984 parole da ripetere con lo shadowing. Il parlato dura 15:30. Chi parla va veloce, circa 192 parole al minuto: aspettati suoni legati e ridotti. Il 85% delle parole rientra nelle 3.000 più comuni dell’inglese; conviene studiare le altre prima di iniziare.

Vocaboli chiave di questo video

15 parole del video che vale la pena imparare, con pronuncia e significato:

Parola	Pronuncia	Significato
element sostantivo	/ˈɛləmənt/	elemento
binary aggettivo	/ˈbaɪ.nə.ɹi/	binario
minus	/ˈmaɪ.nəs/	meno
skip verbo	/skɪp/	balzare, saltellare
complexity sostantivo	/kəmˈplɛk.sɪ.ti/	complessità
decrease verbo	/dɪˈkɹiːs/	diminuire, calare
pointer sostantivo	/ˈpɔɪn.təɹ/	bacchetta
compare verbo	/kəmˈpɛɚ/	paragonare
dimensional aggettivo	/daɪˈmɛn.ʃə.nəl/	-dimensionale
technique sostantivo	/tɛkˈniːk/	tecnica
solve verbo	/sɒlv/	risolvere
array sostantivo	/əˈɹeɪ/	abbigliamento, agghindamento
column sostantivo	/ˈkɒləm/	colonna
algorithm sostantivo	/ˈælɡəɹɪðm̩/	algoritmo
input sostantivo	/ˈɪnpʊt/	immissione

I phrasal verb che sentirai

Parola	Significato
find out verbo	scoprire
add up verbo	sommarsi
check out verbo	investigare

La grammatica di questo video

Le strutture che chi parla usa di più, con le parole esatte del video:

Struttura	Nel video
Forma passiva be + participio passato — conta ciò che accade, non chi lo fa	is based · are sorted · are given
Frasi relative who / which + frase — un’informazione in più su una persona o una cosa	algorithm which will · element which is · one which is
Present perfect have/has + participio passato — un’azione passata che conta ancora adesso	have taken · have skipped · have already been skipped

Pronuncia a cui fare attenzione

I suoni “th”: algorithm /ˈælɡəɹɪðm̩/, ninth /naɪnθ/
I suoni “sh” e “zh”: dimensional /daɪˈmɛn.ʃə.nəl/, variation /ˌvɛəɹiˈeɪʃn̩/, verification /ˌvɛ.ɹə.fɪˈkeɪ.ʃən/, partition /pɑɹˈtɪ.ʃən/
Parole lunghe — attenzione all’accento: complexity /kəmˈplɛk.sɪ.ti/, dimensional /daɪˈmɛn.ʃə.nəl/, curriculum /kəˈɹɪk.jə.ləm/, accordingly /əˈkɔɹ.dɪŋ.li/, verification /ˌvɛ.ɹə.fɪˈkeɪ.ʃən/

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Registrati e confronta con l’originale, facendo attenzione a parole come element, binary, minus.

Cos'è la tecnica dello Shadowing?

Shadowing è una tecnica di apprendimento delle lingue supportata da studi scientifici, originariamente sviluppata per la formazione dei traduttori professionisti e resa popolare dal poliglotta Dr. Alexander Arguelles. Il metodo è semplice ma potente: ascolti un audio in inglese di madrelingua e lo ripeti immediatamente ad alta voce — come un'ombra che segue il parlante con un ritardo di solo 1–2 secondi. A differenza dell'ascolto passivo o degli esercizi di grammatica, lo shadowing costringe il tuo cervello e i muscoli della bocca a elaborare e riprodurre simultaneamente i modelli di discorso reale. La ricerca dimostra che migliora significativamente la precisione della pronuncia, l'intonazione, il ritmo, il discorso connesso, la comprensione dell'ascolto e la fluidità del parlato — rendendolo uno dei metodi più efficaci per la preparazione alla prova di speaking dell'IELTS e per la comunicazione reale in inglese.

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