SDE - 1

to look for all positive contiguous segments of the array (max_ending_here is used for this). And keep track of maximum sum contiguous segment among all positive segments (max_so_far is used for this). Each time we get a positive sum compare it with max_so_far and update max_so_far if it is greater than max_so_far

Recursion, Backtracking, with memorization java

This problem can be solved using two stacks. Assume the two stacks are current: currentlevel and nextlevel. We would also need a variable to keep track of the current level order(whether it is left to right or right to left). We pop from the currentlevel stack and print the nodes value. Whenever the current level order is from left to right, push the nodes left child, then its right child to the stack nextlevel. Since a stack is a LIFO(Last-In-First_out) structure, next time when nodes are popped off nextlevel, it will be in the reverse order. On the other hand, when the current level order is from right to left, we would push the nodes right child first, then its left child. Finally, do-not forget to swap those two stacks at the end of each level(i.e., when current level is empty)

Keep three indices low = 1, mid = 1 and high = N and there are four ranges, 1 to low (the range containing 0), low to mid (the range containing 1), mid to high (the range containing unknown elements) and high to N (the range containing 2).
Traverse the array from start to end and mid is less than high. (Loop counter is i)
If the element is 0 then swap the element with the element at index low and update low = low + 1 and mid = mid + 1
If the element is 1 then update mid = mid + 1
If the element is 2 then swap the element with the element at index high and update high = high – 1 and update i = i – 1. As the swapped element is not processed
Print the array.

Tip 1 : know your strength weakness
Tip 2 : know about family, be active and responsible in family
Tip 3 : have passion about what you are doing