Tip 1 : Be consistent in your practice
Tip 2 : Make notes of all the concepts you learn
Tip 3 : Keep revising the old concept
Tip 1 : Have two good live projects
Tip 2 : Participate in competitions such as hackathons, and coding competitions to show your coding talent



Create an auxiliary 2D dp[][3] array to store the minimum cost of previously colored houses.
Initialize dp[0][0], dp[0][1], and dp[0][2] as the cost of cost[i][0], cost[i][1], and cost[i][2] respectively.
Traverse the given array cost[][3] over the range [1, N] and update the cost of painting the current house with colors red, blue, and green with the minimum of the cost other two colors in dp[i][0], dp[i][1], and dp[i][2] respectively.
After completing the above steps, print the minimum of dp[N – 1][0], dp[N – 1][1], and dp[N – 1][2] as the minimum cost of painting all the houses with different adjacent colors.



The series is 1-based indexed.



The ‘ARR’ = [1, 2, -3, -4, 5], the subarray [5, 1, 2] has the maximum possible sum which is 8. This is possible as 5 is connected to 1 because ‘ARR’ is a circular array.
Create an integer array rightMax of length n.
Set rightMax[n - 1] to nums[n - 1], set suffixSum to nums[n - 1].
Iterate over i from n - 2 to 0
Increase suffixSum by nums[i]
Update rightMax[i] to max(rightMax[i + 1], suffixSum)
Calculate the normal sum maxSum using Kadane's algorithm.
Set specialSum to nums[0], set sum to 0.
Iterate over i from 0 to n - 2
Increase prefixSum by nums[i]
Update specialSum to max(specialSum, prefixSum + rightMax[i + 1]).
Return max(maxSum, specialSum)



The traversal should proceed from left to right according to the input adjacency list.
Adjacency list: { {1,2,3},{4}, {5}, {},{},{}}
The interpretation of this adjacency list is as follows:
Vertex 0 has directed edges towards vertices 1, 2, and 3.
Vertex 1 has a directed edge towards vertex 4.
Vertex 2 has a directed edge towards vertex 5.
Vertices 3, 4, and 5 have no outgoing edges.
We can also see this in the diagram below.
BFS traversal: 0 1 2 3 4 5

What is 2 tier and 3 tier architecture in DBMS? (Learn)
A two-tier DB architecture either buries the application logic within the server database, on the client (inside the UI), or both of them. A three-tier DB architecture buries the process or application logic in the middle-tier. Thus, it acts as a separate entity from the Client/ User Interface and the data Interface.
What is Kernel? (Learn)
The kernel is a computer program at the core of a computer's operating system and generally has complete control over everything in the system. It is the portion of the operating system code that is always resident in memory and facilitates interactions between hardware and software components.

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