تدريب Shadowing: Two Sum - Leetcode 1 - HashMap - Python - تعلم التحدث بالإنجليزية عبر الفيديو
اختر وضع التدريب
كل وضع يفتح غرفة تدريب بملء الشاشة.
110 جمل
إذا كانت الجمل قصيرة أو طويلة جدًا، انقر على Edit لتعديلها.
1
Let's solve leak code 1 to sum, the most popular leak code question.
2
So we're given an input array and some target, in this case 9.
3
And we want to find the two values in this input array that sum to 9.
4
So in this case, it's 2 and 7.
5
Now we want to return the indices of these two values.
6
So the index of 2 is 0, the index of 7 is 1.
7
So we return 0 and 1.
8
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.
9
Now, the most intuitive way to solve this problem is basically just check every combination of two values
10
and see if they can sum up to our target.
11
So we start at two, we check every combination we can make that includes two.
12
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.
13
In this case, none of them do.
14
So next we can repeat the process.
15
Let's check every combination including one that sums up to target four.
16
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.
17
Notice that we didn't have to check the values
18
that came before one because we already checked the combination two and one when we were up over here.
19
Remember when we checked every combination with 2.
20
So we didn't have to repeat that work down here.
21
We only had to check the numbers that came after 1.
22
So the runtime of this algorithm isn't super efficient.
23
This is basically brute force.
24
We're going through the entire array of length n and we're going to do that worst case n times for each number.
25
This means that overall worst case time complexity will be O of n squared.
26
So can we do better?
27
Now the thing to notice is that for each number,
28
for example, 1, the value we're looking for is the difference between the target and this value 1.
29
So we're looking for 4 minus 1, which is equal to 3.
30
So that means this is the only value we can add to 1 that'll equal the target.
31
So we don't have to check every number, we just want to know if 3 exists.
32
Now the easiest way we can do this, the most efficient, is by making a hash map of every value in our input array
33
so we can instantly check if the value 3 exists.
34
Now let's try the same problem except let's use a hash map this time.
35
Now in our hash map, we're going to be mapping each value to the index of each value.
36
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.
37
So in our hash map, we're going to be mapping the value to the index.
38
now we could add every value in this array into the hash map before we start iterating through it
39
but there's actually an easier way to do it
40
if we added the entire array into the hash map initially
41
then we would get to the value 2 first right we
42
would want to check does the difference between target 4 minus this value 2
43
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?
44
Because they're both at the same index.
45
We can't use the same value twice, so we would have to compare the index of our current
46
two with the index of the two that's in our hash map.
47
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.
48
So doing it this clever way, initially we say our hash map is empty.
49
So we get to the value two first of all, right?
50
And we want to look for the difference 4 minus 2 in our hash map.
51
Our hash map is empty, so we don't find 2.
52
So then, after we've visited this element, then we can add it to our hash map.
53
So now that I'm done visiting it, I'm going to move to the second element 1.
54
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.
55
Now I'm at 1.
56
I'm looking for 4 minus 1, which is 3.
57
I see 3 isn't in our hash map, but it actually is in our array.
58
So what's the problem?
59
Well, for now, we're going to say we don't find a 3.
60
So we add 1 to our hash map.
61
The index of this 1 is 1.
62
And now we move to the next element, 5.
63
we check does 4 minus 5 uh it's 4 minus 5 exist in our hash map that's negative 1
64
so no it does not then we add this 5 to our hash map
65
and it's index which is 2
66
and we move to the last value in the array 3
67
we check does 4 minus 3 exist in our hash map now that's 1
68
so we see it does exist right over here.
69
The value exists and its index is one.
70
So now we found our two values that sum to the target and we want to return their indexes,
71
their indices, which are going to be one and three.
72
So with this algorithm, we don't have to initialize our hash map.
73
It can be initially empty and then we can just iterate through this array in one pass.
74
Now the reason the algorithm can work in that way with just one pass is this.
75
So let's say we had a giant array, right?
76
We know for sure that there are two elements in this array that sum to our target, right?
77
We don't know where they are.
78
They're at some arbitrary location.
79
When we visit the first one of those elements, our hash map is only gonna be this portion of the array.
80
It's only gonna have the values that came before the first value.
81
So we're going to notice that the second value
82
that can sum to the target is not going to be in our hash map yet.
83
But once we get to the second value, our hash map is going to be this portion.
84
So every value that comes before this, right?
85
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.
86
So we're guaranteed to find the solution.
87
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,
88
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.
89
We are using extra memory, right?
90
That hash map isn't free, so the memory complexity is also going to be O of n
91
because we could potentially add every value to the hash map.
92
So now let's code the solution.
93
So remember we need a hash map, right?
94
I'm going to call this previous map because it's basically every element that comes before the current element.
95
Every previous element is going to be stored in this map.
96
We're going to be mapping the value to the index of that value.
97
So now let's iterate through every value in this array.
98
We need the index as well as the actual number.
99
So let's do it like this in Python.
100
Before we add this to our map, let's check if the difference, which is equal to target minus n.
101
Now let's check if this difference is already in the hash map.
102
if it is, then we can return the solution, which is going to be a pair of the indices.
103
So I can get the first index like this, and the second index is just i.
104
Now if we don't find the solution, then we have to update our hash map.
105
So for this value n, I'm going to say the index is i, and then we're going to continue.
106
Since we're guaranteed that a solution exists, we don't have to return anything out here, right?
107
But I'll just put a return for no reason.
108
Now let's see if it works. And it works perfectly.
109
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
110
and not have to worry about like edge cases and comparisons and things like that.
تنزيل التطبيق
كل ما تحتاجه للتحدث بطلاقة
تقييم النطق بالذكاء الاصطناعيتقييم كل جملة
تدريب IPAأتقن كل صوت
المفرداتابنِ رصيدك من الكلمات
لعبة المفرداتتعلّم أثناء اللعب
نفس القناة
FOR YOU
فيديو موصى به
C12
182 جمل
We Let $200 GPT-Pro Think for 7 Days on Unsolved Math
Daily ConversationUniversity StudentsWorking Adults
C11
234 جمل
Kth Smallest element in a matrix | Leetcode #378
Daily ConversationUniversity Students
C11
27 جمل
You are wasting memory
Daily ConversationUniversity StudentsWorking Adults
C11
108 جمل
Cắt bánh - bài toán gây đau đầu nhiều thế kỷ
Daily ConversationUniversity StudentsWorking Adults
C11
146 جمل
🚫 Java Deadlock Explained: Why Your Code is Stuck 😱 (Beginners Must Watch!)
Daily ConversationUniversity StudentsWorking Adults
C11
575 جمل
Lec-6 Simplex Method | Maximization Problem | Mathematical Example Solution | Operations Research
Daily ConversationUniversity Students
C11
54 جمل
Traceroute: You're Probably Reading the Results Wrong
Daily ConversationUniversity StudentsWorking Adults
C181 جمل
Mathematician Collapses All Functions to One Weird Formula
Daily ConversationUniversity StudentsWorking Adults
C12
31 جمل
Heaps in 3 minutes — Intro
Daily ConversationUniversity Students
B2215 جمل
PTE SWT: One-Sentence Formula for 79+
Daily ConversationUniversity StudentsWorking Adults
C12
125 جمل
Quant Interview Questions #2 | Bayes Theorem
Interview EnglishUniversity StudentsWorking Adults
السياق والخلفية
في هذا الفيديو، يتم تناول مشكلة حسابية مشهورة تُعرف باسم "Two Sum" من منصة Leetcode. يتناول المتحدث كيفية حل هذه المشكلة البرمجية باستخدام خوارزمية فعالة تعتمد على HashMap. يتناول الفيديو كيفية تحديد قيمتين من مصفوفة معينة، تعد كل منهما بحيث يكون مجموع قيمتهما هو الهدف المحدد، وهو 9 في هذا المثال. هذا النوع من المشكلات لا يتطلب فقط المعرفة البرمجية، بل أيضًا الفهم العميق لكيفية تحسين الكود ليتناسب مع تجارب التعلم.
أفضل 5 عبارات للتواصل اليومي
- لنحسب مجموع قيمتين - تعبير يمكن استخدامه في المحادثات المتعلقة بالرياضيات أو البرمجة.
- هل يمكنك أن تخبرني عن الأرقام؟ - سؤال يسأل فيه الشخص عن الأرقام المعنية.
- أين يمكنني العثور على النتيجة؟ - يستخدم للاستفسار عن النتائج.
- علينا تجربة طرق مختلفة - عبارة تعبر عن ضرورة استكشاف أساليب جديدة.
- هيا نستخدم الخوارزميات - يشير إلى استخدام طرق محددة لحل المشكلات.
دليل الظل خطوة بخطوة
تعتبر طريقة التظليل في الإنجليزية وسيلة فعالة لتعزيز المهارات اللغوية. اتبع الخطوات التالية لمساعدتك في ممارسة ما تعلمته من الفيديو:
- استمع وانظر إلى الفيديو: ركز على طريقة عرض المتحدث للأفكار، وحاول استيعاب المفاهيم الأساسية.
- كرر العبارات: استخدم طريقة shadow speech لتكرار العبارات المهمة التي تسمعها. هذا سيساعدك على تحسين النطق والثقة.
- قم بتطبيق ما تعلمته: حاول استخدام الطرق والأفكار التي تم تقديمها في الفيديو أثناء حل مشكلات مشابهة بنفسك.
- اسجل صوتك: أثناء استخدام shadowspeaks، قم بتسجيل صوتك لتقييم تقدمك وتحسين أدائك.
- التفاعل مع الآخرين: قم بمشاركة ما تعلمته مع أصدقائك أو في مجموعة دراسية لمناقشة الأفكار ومحاولة حل تحديات جديدة.
بتطبيق هذه الخطوات، يمكنك تعزيز مهاراتك في تعلم الإنجليزية مع يوتيوب وتحقيق تحسين ملحوظ في التواصل باللغة الإنجليزية.
القواعد في هذا الفيديو
أكثر التراكيب التي يستخدمها المتحدث، مع العبارات نفسها من الفيديو:
| التركيب | في الفيديو |
|---|---|
| جمل الوصل who / which + جملة — معلومة إضافية عن شخص أو شيء | index which is · indices, which are · map, which is |
| المبني للمجهول be + اسم المفعول — التركيز على ما يحدث لا على من يفعله | be guaranteed · be stored |
ما هي تقنية التظليل الصوتي؟
التظليل الصوتي (Shadowing) تقنية تعلم لغة مدعومة علمياً، طُورت أصلاً لتدريب المترجمين الفوريين المحترفين. الطريقة بسيطة لكنها قوية: تستمع لصوت إنجليزي أصلي وتكرره فوراً بصوت عالٍ — كظل يتبع المتحدث بتأخير 1-2 ثانية. تُظهر الأبحاث تحسناً كبيراً في دقة النطق والتنغيم والإيقاع وربط الأصوات والاستماع والطلاقة.