Sri Hari: Batch-3/Neetcode-150/Added hints #3740
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hints/anagram-groups.md
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| <details class="hint-accordion"> | ||
| <summary>Hint 3</summary> | ||
| <p> | ||
| We can simply use an <code>O(26)</code> size array to count the frequency of each character in a string. Then, we can use this array as the key in a hash map to group the strings. |
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This is minor, but might be worth explicitly mentioning that we can use an array since the character set is a through z (26 continuous characters)
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Yeah, I will be more beginner-friendly and specific towards the details.
hints/duplicate-integer.md
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| <summary>Hint 1</summary> | ||
| <p> | ||
| A brute force solution would be to check every element against every other element in the array. This would be an <code>O(n^2)</code> solution. Can you think of a better way? | ||
| A brute force solution would be to check every element against every other element in the array. This would be an <code>O(n<sup>2</sup>)</code> solution. Can you think of a better way? |
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Unfortunately, there's a limitation in the way my site renders markdown. The wont work as intended so I would leave this as-is for now. And do the same for all other files.
This isn't your fault. I appreciate you trying to improve this though!
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Ok, I will keep that as it is in all the files.
hints/is-palindrome.md
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| <details class="hint-accordion"> | ||
| <summary>Hint 1</summary> | ||
| <p> | ||
| A brute force solution would be to create a copy of the string, reverse it, and then check for equality. This would be an <code>O(n)</code> solution with extra space. Can you think of a better way? |
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Minor: We can be more specific by saying "Can you think of a way to do this without O(n) space?"
"Better" is a bit ambiguous.
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A great point to mention for a beginner user. I will be more specific.
| <details class="hint-accordion"> | ||
| <summary>Hint 2</summary> | ||
| <p> | ||
| Is there any repeated work we are doing? |
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Can be more specific, e.g.
Is there any way to identify the start of a sequence? For example, in [1,2,3,10,11,12], only 1 and 10 are the beginning of a sequence.
| <details class="hint-accordion"> | ||
| <summary>Hint 3</summary> | ||
| <p> | ||
| Yes, we should only start building the sequence with an element <code>num</code> such that <code>num - 1</code> is not present in the array. We can use a hash set for fast lookups. |
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Reword this one, after updating above.
| <details class="hint-accordion"> | ||
| <summary>Recommended Time & Space Complexity</summary> | ||
| <p> | ||
| You should aim for a solution with <code>O(n)</code> time and <code>O(1)</code> space, where <code>n</code> is the size of the input array. |
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For this one i think it's fine to have O(n) space since it's only a medium
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I thought it would be better to have a solution from your video, but fine, now I understood what you are trying to say. I will be more beginner friendly.
| <details class="hint-accordion"> | ||
| <summary>Hint 2</summary> | ||
| <p> | ||
| Is there a way to avoid the repeated work? Maybe there is a technique with linear time? |
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I would replace "Maybe there is a technique with linear time?" with "Maybe we can store the results of the repeated work in an array."
hints/three-integer-sum.md
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| <details class="hint-accordion"> | ||
| <summary>Hint 4</summary> | ||
| <p> | ||
| To efficiently find the <code>j</code> and <code>k</code> pairs, we can use the two pointer algorithm. |
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Might be worth mentioning we run the two pointer approach on all elements to the right of i.
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Yes, I will do that. But I have added 5 hints only for this problem as we don't want to ask a user a question and give answer in the same hint.
| <details class="hint-accordion"> | ||
| <summary>Hint 2</summary> | ||
| <p> | ||
| Can you think of an algorithm by taking the advantage of array being sorted? |
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Maybe we can be more specific:
If nums[0] + nums[n-1] > target, than we know nums[n-1] can not possibly be included in any pairs. Why? Because nums[n-1] is the largest element in the array. Even by adding it with nums[0], which is the smallest element, we still exceed the target.
This hints at a two pointer solution, and also gives some intuition of why it works.
hints/two-integer-sum.md
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| <details class="hint-accordion"> | ||
| <summary>Hint 3</summary> | ||
| <p> | ||
| We can fix index <code>i</code> and iterate on it. The equation becomes <code>nums[i] == target - nums[j]</code> when we rearrange it. Let <code>value = target - nums[j]</code> and check if <code>value</code> exists in the hash map as we iterate through the array, else store the current element in the hashmap with its index and continue. |
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replace value with difference.
We can fix index
iand iterate on it.
The above line seems unnecessary and a little bit meaningless.
Would it be better to remove that, and say something like: we can iterate through nums with index i.
We should also mention that we use a hashmap for O(1) lookups. I know it's implied, but since this is a beginner problem.
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Thanks @Srihari2222 Just curious what
is in reference to? |
@neetcode-gh |
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@Srihari2222 Thanks and great work! |
2 participants
@neetcode-gh
When we have no chance other than recommending the optimal solution, I used this line
"You have to aim for the solution with "XYZ" complexity".
But when I suggest sub-optimal solution, I used
"You have to aim for the solution as good or better than "XYZ" complexity".