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added some solution between 3000-3010 #928
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added some solution between 3000-3010 #928
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Hi there! 👋 Thank you for submitting your pull request to CodeHarborHub. Great job on the contribution! 🎉 We appreciate your efforts and our team will review it soon. If you have any questions or need further assistance, feel free to reach out. Keep up the great work!
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Great job, @agarwalhimanshugaya! 🎉 Thank you for submitting your pull request to CodeHarborHub. We appreciate your contribution and enthusiasm! Our team will review it soon. If you have any questions or need further assistance, feel free to reach out. Thanks for contributing!
Here's the code health analysis summary for commits Analysis Summary
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Update all files code base on one files suggestion and can't delete dsa-solutions/lc-solutions/_category_.json
this file and you need write good format for all files based on suggestions
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- **Input:** $dimensions = [[9,3],[8,6]]$ | ||
- **Output:** $48$ | ||
- **Explanation:** $For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487. |
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replace
$For index = 0, length = 9 and width = 3. Diagonal length = sqrt(9 * 9 + 3 * 3) = sqrt(90) ≈ 9.487.
For index = 1, length = 8 and width = 6. Diagonal length = sqrt(8 * 8 + 6 * 6) = sqrt(100) = 10.
So, the rectangle at index 1 has a greater diagonal length therefore we return area = 8 * 6 = 48..$
to
For index = $0$, $\text{length} = 9$ and width = $3$. $\text{Diagonal length} = sqrt(9 \times 9 + 3 \times 3) = sqrt(90) ≈ 9.487$.
For $\text{index} = 1$, $length = 8$ and $\text{width} = 6$. $\text{Diagonal length} = sqrt(8 \times 8 + 6 \times 6) = sqrt(100) = 10$.
So, the rectangle at index 1 has a greater diagonal length therefore we return $area = 8 \times 6 = 48..$
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### Constraints | ||
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- $1 <= dimensions.length <= 100$ |
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replace $1 <= dimensions.length <= 100$
to $1 \leq dimensions.length \leq 100$
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### Conclusion | ||
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At last we can say that we can calculate maximum area of diagonal rectangle using timecomplexity of o(n) and spacecomplexity of o(1) using a simple for loop and some variable |
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replace o(n)
to $O(n)$
i apply all your given suggestions |
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now update code and focus on code because it is based on mdx
+ React.
dsa-solutions/lc-solutions/3000-3099/3000-Maximum area of diagonal rectangle.md
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dsa-solutions/lc-solutions/3000-3099/3006-find beautiful indices in the given array.md
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i remove all the error mention by you |
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update your code only one error
dsa-solutions/lc-solutions/3000-3099/3006-find beautiful indices in the given array.md
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Update code based on suggestions
dsa-solutions/lc-solutions/3000-3099/3005-count elements with maximum frequency.md
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dsa-solutions/lc-solutions/3000-3099/3010-divide an array into subarrays with minimum cost 1.md
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dsa-solutions/lc-solutions/3000-3099/3006-find beautiful indices in the given array.md
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mai push krrtha tha to nhi horha tha to muje merge krna pda origin k sath |
okay |
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Done
…a-of-diagonal-rectangle.md
…oves-to-capture-the-queen.md
…2-maximum-size-of-a-set-after-removals.md
…05-count-elements-with-maximum-frequency.md
…o 3006-find-beautiful-indices-in-the-given-array-i.md
…st 1.md to 3010-divide-an-array-into-subarrays-with-minimum-cost-i.md
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Related Issue
Fix: #835
Description
I add some of the solution between 3000-3010 which is solved by mine
Type of PR
Screenshots / Videos (if applicable)
[Attach any relevant screenshots or videos demonstrating the changes]
Checklist
Additional Context
[Include any additional information or context that might be helpful for reviewers.]
Resources for Guidance
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