Shadowing Practice: Two Sum - Leetcode 1 - HashMap - Python - Learn English Speaking with Video
Les maken...
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Let's solve leak code 1 to sum, the most popular leak code question.
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So we're given an input array and some target, in this case 9.
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And we want to find the two values in this input array that sum to 9.
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So in this case, it's 2 and 7.
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Now we want to return the indices of these two values.
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So the index of 2 is 0, the index of 7 is 1.
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So we return 0 and 1.
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We're guaranteed that there's exactly one solution, so we don't have to worry about not finding a solution, and we don't have to worry about multiple solutions.
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Now, the most intuitive way to solve this problem is basically just check every combination of two values
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and see if they can sum up to our target.
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So we start at two, we check every combination we can make that includes two.
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So we scan through the remainder of the array, one, five, three, and check if any of those numbers added to two sums to our target four.
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In this case, none of them do.
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So next we can repeat the process.
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Let's check every combination including one that sums up to target four.
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So we scan through every element that comes after it, five and three, and we find that one added with three sums up to our target four.
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Notice that we didn't have to check the values
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that came before one because we already checked the combination two and one when we were up over here.
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Remember when we checked every combination with 2.
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So we didn't have to repeat that work down here.
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We only had to check the numbers that came after 1.
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So the runtime of this algorithm isn't super efficient.
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This is basically brute force.
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We're going through the entire array of length n and we're going to do that worst case n times for each number.
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This means that overall worst case time complexity will be O of n squared.
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So can we do better?
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Now the thing to notice is that for each number,
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for example, 1, the value we're looking for is the difference between the target and this value 1.
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So we're looking for 4 minus 1, which is equal to 3.
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So that means this is the only value we can add to 1 that'll equal the target.
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So we don't have to check every number, we just want to know if 3 exists.
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Now the easiest way we can do this, the most efficient, is by making a hash map of every value in our input array
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so we can instantly check if the value 3 exists.
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Now let's try the same problem except let's use a hash map this time.
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Now in our hash map, we're going to be mapping each value to the index of each value.
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So the index of 2 is 0, the index of 1 is 1, the index of 5 is 2, the index of 3 is 3.
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So in our hash map, we're going to be mapping the value to the index.
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now we could add every value in this array into the hash map before we start iterating through it
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but there's actually an easier way to do it
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if we added the entire array into the hash map initially
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then we would get to the value 2 first right we
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would want to check does the difference between target 4 minus this value 2
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which is equal to 2 exists in our hash map and we would find that 2 does exist in our hash map, but we're not allowed to reuse the same one, right?
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Because they're both at the same index.
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We can't use the same value twice, so we would have to compare the index of our current
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two with the index of the two that's in our hash map.
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There's actually an easier way to do this though, and it's a little clever, and let me show you how to do it that way.
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So doing it this clever way, initially we say our hash map is empty.
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So we get to the value two first of all, right?
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And we want to look for the difference 4 minus 2 in our hash map.
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Our hash map is empty, so we don't find 2.
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So then, after we've visited this element, then we can add it to our hash map.
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So now that I'm done visiting it, I'm going to move to the second element 1.
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And before I do that, I'm going to add this value 2 to our hash map, and the index of this value is going to be 0.
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Now I'm at 1.
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I'm looking for 4 minus 1, which is 3.
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I see 3 isn't in our hash map, but it actually is in our array.
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So what's the problem?
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Well, for now, we're going to say we don't find a 3.
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So we add 1 to our hash map.
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The index of this 1 is 1.
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And now we move to the next element, 5.
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we check does 4 minus 5 uh it's 4 minus 5 exist in our hash map that's negative 1
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so no it does not then we add this 5 to our hash map
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and it's index which is 2
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and we move to the last value in the array 3
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we check does 4 minus 3 exist in our hash map now that's 1
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so we see it does exist right over here.
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The value exists and its index is one.
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So now we found our two values that sum to the target and we want to return their indexes,
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their indices, which are going to be one and three.
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So with this algorithm, we don't have to initialize our hash map.
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It can be initially empty and then we can just iterate through this array in one pass.
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Now the reason the algorithm can work in that way with just one pass is this.
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So let's say we had a giant array, right?
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We know for sure that there are two elements in this array that sum to our target, right?
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We don't know where they are.
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They're at some arbitrary location.
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When we visit the first one of those elements, our hash map is only gonna be this portion of the array.
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It's only gonna have the values that came before the first value.
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So we're going to notice that the second value
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that can sum to the target is not going to be in our hash map yet.
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But once we get to the second value, our hash map is going to be this portion.
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So every value that comes before this, right?
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So we're going to be guaranteed that once we visit the second element that sums up to the target, we're going to be guaranteed that the first one is already in our hash map.
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So we're guaranteed to find the solution.
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Now, since we only have to iterate through the array once, and we're adding each value to our hash map, which is a constant time operation,
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and we're checking if a value exists in our hash map, which is also a constant time operation, the time complexity is going to be big O of n.
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We are using extra memory, right?
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That hash map isn't free, so the memory complexity is also going to be O of n
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because we could potentially add every value to the hash map.
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So now let's code the solution.
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So remember we need a hash map, right?
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I'm going to call this previous map because it's basically every element that comes before the current element.
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Every previous element is going to be stored in this map.
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We're going to be mapping the value to the index of that value.
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So now let's iterate through every value in this array.
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We need the index as well as the actual number.
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So let's do it like this in Python.
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Before we add this to our map, let's check if the difference, which is equal to target minus n.
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Now let's check if this difference is already in the hash map.
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if it is, then we can return the solution, which is going to be a pair of the indices.
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So I can get the first index like this, and the second index is just i.
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Now if we don't find the solution, then we have to update our hash map.
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So for this value n, I'm going to say the index is i, and then we're going to continue.
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Since we're guaranteed that a solution exists, we don't have to return anything out here, right?
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But I'll just put a return for no reason.
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Now let's see if it works. And it works perfectly.
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So with this kind of neat little trick with just doing it in one pass, you can reduce the amount of code you have to write
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and not have to worry about like edge cases and comparisons and things like that.
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Woordenschat en spreektips bij deze les
Deze video bevat 110 zinnen en 1461 woorden om na te spreken. Het gesproken deel duurt 8:26. De spreker praat in een natuurlijk tempo van ongeveer 173 woorden per minuut, dicht bij een alledaags gesprek. 88% van de woorden hoort bij de 3.000 meest gebruikte Engelse woorden; de rest kun je beter vooraf bekijken.
Belangrijke woorden in deze video
15 woorden uit de video die het leren waard zijn, met uitspraak en betekenis:
| Woord | Uitspraak | Betekenis |
|---|---|---|
| hash zelfstandig naamwoord | /ˈhæʃ/ | hachee |
| element zelfstandig naamwoord | /ˈɛləmənt/ | element |
| minus | /ˈmaɪ.nəs/ | min |
| guarantee zelfstandig naamwoord | /ˌɡæɹ.ənˈtiː/ | garantie |
| sum zelfstandig naamwoord | /sʌm/ | som |
| algorithm zelfstandig naamwoord | /ˈælɡəɹɪðm̩/ | algoritme |
| complexity zelfstandig naamwoord | /kəmˈplɛk.sɪ.ti/ | ingewikkeldheid, complexiteit |
| efficient bijvoeglijk naamwoord | /ɪˈfɪʃənt/ | efficiënt |
| solve werkwoord | /sɒlv/ | oplossen |
| leak zelfstandig naamwoord | /liːk/ | lek |
| scan werkwoord | /skæn/ | afspeuren |
| compare werkwoord | /kəmˈpɛɚ/ | vergelijken |
| potentially bijwoord | /pəˈtɛnʃ(ə)li/ | potentieel |
| remainder zelfstandig naamwoord | /ɹəˈmeɪndɚ/ | rest, restant |
| arbitrary bijvoeglijk naamwoord | /ˈɑː.bɪ.tɹə.ɹi/ | willekeurig, arbitrair |
Uitspraak om op te letten
De spreker gebruikt 48 samentrekkingen en verkorte vormen, zoals we're, don't, I'm. Spreek ze kort uit, zoals je ze hoort.
- De klanken “sh” en “zh”: hash /ˈhæʃ/, efficient /ɪˈfɪʃənt/, potentially /pəˈtɛnʃ(ə)li/, initialize /ɪˈnɪʃəlaɪz/
- Lange woorden — let op de klemtoon: complexity /kəmˈplɛk.sɪ.ti/, potentially /pəˈtɛnʃ(ə)li/, arbitrary /ˈɑː.bɪ.tɹə.ɹi/, intuitive /ɪnˈtjuːɪtɪv/, initialize /ɪˈnɪʃəlaɪz/
Zo oefen je met deze video
- Luister de hele video één keer zonder te spreken en noteer de woorden die je niet kent.
- Begin op 0,75× snelheid, spreek zin voor zin na en ga terug naar normale snelheid zodra het makkelijk gaat.
- Neem jezelf op en vergelijk met het origineel; let daarbij op woorden als hash, element, minus.
Grammatica in deze video
De structuren die de spreker het meest gebruikt, met de exacte woorden uit de video:
| Structuur | In de video |
|---|---|
| Betrekkelijke bijzinnen who / which + zin — extra informatie over een persoon of ding | index which is · indices, which are · map, which is |
| Lijdende vorm be + voltooid deelwoord — het gaat om wat er gebeurt, niet om wie het doet | be guaranteed · be stored |
Wat is de Shadowing-techniek?
Shadowing is een wetenschappelijk onderbouwde taalleermethode die oorspronkelijk is ontwikkeld voor professionele tolkentraining en gepopulariseerd door polyglot Dr. Alexander Arguelles. De methode is eenvoudig maar krachtig: je luistert naar native Engelse audio en herhaalt het onmiddellijk hardop — als een schaduw die de spreker volgt met slechts 1–2 seconden vertraging. In tegenstelling tot passief luisteren of grammaticadrills, dwingt shadowing je hersenen en mondspieren om echte spraakpatronen tegelijkertijd te verwerken en te reproduceren. Onderzoek toont aan dat het de uitspraaknauwkeurigheid, intonatie, ritme, verbonden spraak, luisterbegrip en spreekvaardigheid aanzienlijk verbetert — waardoor het een van de meest effectieve methoden is voor IELTS Speaking-voorbereiding en echte Engelse communicatie.