Omkar Gaikwad

Crafting digital dreams into reality 🌐 | Web Developer & Software Engineer

[Omkar Gaikwad @omkargdotcom]

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Day 15 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 14 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 12 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 11 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 10 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 9 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 8 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 7 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 6 of 21 Days POTD Challenge Powered By Deutsche Bank
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[Omkar Gaikwad @omkargdotcom]

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Day 5 of 21 Days POTD Challenge Powered By Deutsche Bank

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[Omkar Gaikwad @omkargdotcom]

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[Omkar Gaikwad @omkargdotcom]

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Day 4 of 21 Days POTD Challenge Powered By Deutsche Bank

Problem: Binary Tree to DLL

[Omkar Gaikwad @omkargdotcom]

A helper function (inorder) recursively traverses the tree and updates pointers. The first node encountered becomes the head of the DLL, and the final result is printed in both forward and reverse order.

[Omkar Gaikwad @omkargdotcom]

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[Omkar Gaikwad @omkargdotcom]

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Day 3 of 21 Days POTD Challenge Powered By Deutsche Bank

Problem: Alternate positive and negative numbers

Approach: Separate positive and negative values, then push it into the new array in alternate manner. Copy new array to original array.

[Omkar Gaikwad @omkargdotcom]

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[Omkar Gaikwad @omkargdotcom]

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Day 2 of 21 Days POTD Challenge Powered By Deutsche Bank

Problem - Given a Binary Tree, convert it into its mirror.

Approach - Use Recursion. Best way for solving Tree Problems.

[Omkar Gaikwad @omkargdotcom]

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[Omkar Gaikwad @omkargdotcom]

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Day 1 of 21 Days POTD Challenge Powered By Deutsche Bank

Problem - Given the head of a linked list, the task is to find the middle. Eg - The middle of 1-> 2->3->4->5 is 3. If there are two middle nodes (even count), return the second middle.

[Omkar Gaikwad @omkargdotcom]

My Approach - This approach is two-pointer technique. One pointer (fast) moves two steps at a time while the other (slow) moves one step. When the fast pointer reaches the end, the slow pointer is at the middle, returning its data. This method ensures a time complexity of O(n).