Problems
December 1, 2022
Evaluate the bracket pairs of a string
We will use a hashmap to store the key-value pairs. Then we will iterate over the string and if we encounter a (, we will start a new string. If we encounter a ), we will...
December 1, 2022
Determine if string halves are alike
We will itrerate over the string, for first half we increase the count and for second half we decrease the count. At the end if the count is 0, then the string halves are alike. Time complexity:...
ReadDecember 1, 2022
Majority element II
We will count the frequency of each element in the array and return the elements whose frequency is greater than n//3. We will use a hashmap to store the frequency of each element, and a...
November 30, 2022
Maximum average subtree
We will traverse the tree in post-order fashion. For each node, we will calculate the sum of all nodes in its subtree and the number of nodes in its subtree. Then we will check if the average of the...
ReadNovember 30, 2022
Find leaves of binary tree
We will traverse the tree in post-order fashion. For each node, we will check if it is a leaf node. If it is, we will append the value of the node in a list. Then we will check if the left and right...
ReadNovember 30, 2022
Recover a tree from preorder traversal
We save the construction path in a stack. In each loop, we get the number level of '-'. We get the value val of node to add. If the size of stack is bigger than the level of node, we pop the stack...
ReadNovember 30, 2022
Find all lonely numbers in the array
We will count the occurrences of each number in the array and return the numbers that have only one occurrence and no adjacent number is present in the list.
Time complexity: O(n), n...
November 30, 2022
Fizz buzz
We will iterate over the numbers from 1 to n and append the corresponding string to the result list. If the number is divisible by 3, we will append "Fizz" to the result list. If the number is...
ReadNovember 30, 2022
Remove duplicates from an unsorted linked list
We will use a hashmap to count the frequency of each node in the linked list. Then we will traverse the linked list again and remove the nodes that have frequency greater than 1. Time complexity:...
ReadNovember 30, 2022
Partitioning into minimum number of deci binary numbers
Assume max digit in n is x, because deci-binary only contains 0 and 1, we need at least x numbers to sum up a digit x. So we can just find the max digit in n and return it. Time complexity:...
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