Problems
January 1, 2023
Count the number of k big indices
We will use binary search twice, one from left, one from right to find out how many elements are less than the current element.
Time complexity: O(nlog(n))
Space complexity:...
January 1, 2023
Distinct prime factors of product of array
First we calculate all the prime numbers up to n using the Sieve of Eratosthenes. Then we iterate over the array and for each element we will increment the counter for each prime factor...
January 1, 2023
Sum of all odd length subarrays
We will generate all possible subarrays of odd length and sum them.
Time complexity: O(n^3)
Space complexity: O(1)
We can do this in one line too-
January 1, 2023
Two sum bsts
We will traverse the tree and store the values in a set. Then we will iterate over the set and check if target - num is in the set.
Time complexity: O(n)
Space...
December 31, 2022
Create binary tree from descriptions
We will use a dictionary to store the nodes. We will iterate over the descriptions and for each description we will create a node and store it in the dictionary. Then we will iterate over the...
ReadDecember 31, 2022
Sum of all subset xor totals
We will use backtracking to solve this problem. We will start from the first element and then we will add it to the current subset and then we will call the function again with the next element. Then...
ReadDecember 31, 2022
Unique paths III
We first find the start squar and end square and path length which contains all squares except the blocked. Then we start the dfs from the start node to wards the end node. And we add the node to a...
ReadDecember 30, 2022
3sum smaller
First we will sort the numbers. Then we will use two pointers to find the number of triplets that sum to a smaller value than the target.
Time complexity: O(n^2)
Space complexity:...
December 30, 2022
adding-spaces-to-a-string
We will iterate over the string and add the spaces to the result.
Time complexity: O(n)
Space complexity: O(n)
December 30, 2022
Remove letter to equalize frequency
We will count the frequency of each letter. Then we will find the minimum and maximum frequency. If the minimum frequency is equal to the maximum frequency, then we can remove any letter. Otherwise,...
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