Problems
September 6, 2022
Merge nodes in between zeros
We will create a dummy node for our result. Then we will start from the head, sum the values of the nodes until we reach another 0, then attach a node to our result list. We will repeat it until we...
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Query kth smallest trimmed number
We will take the nums, iterate over each query, and take the last trim values of the nums and store it as integer in a values array along with the index. Then we sort this values array and take the...
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Seat reservation manager
We will create a min heap with the capacity of the seat manager. Then when someone reserve a seat, we will pop it from our min heap, and when someone unreserve a seat, we will push back the value to...
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Sort an array
We will use divide and conquar to solve this sorting porblem. We will be using classic merge sort.
Time Complexity: O(nlog(n))
Space Complexity: O(nlog(n))
September 5, 2022
Path sum II
We will use DFS along with backtracking to get the paths of every root to leaf path that adds upto the target.
Time Complexity: O(n)
Space Complexity: O(n), for...
September 5, 2022
Sort list
We will use standard merge sort to sort the list, to get the middle node, we will use a fast and slow pointer approach, rest is standard merge sort stuffs.
Time Complexity: O(nlog(n))...
September 5, 2022
Count primes
We will use the algorithm Sieve of Eratosthenes to find the number of primes within a range.
Time Complexity: O(nlog(log(n)))
Space Complexity: O(n)
September 5, 2022
Restore ip addresses
We will create a decision tree based on the rule that each segment of the ip address can be between 0 and 255. Then we run our DFS to get all possible options and return the result. Time...
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Basic calculator II
We will take a stack. Initially our operation and operand will be 0 and + append each character to the stack. If the character is numeric, until we reach an operand we will calculate the...
September 5, 2022
Number of enclaves
We will run DFS from every bordering position of the grid and turn the 1's to 0's. Then we will just count the number of position where we still have 1's left, and return that count as result....
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