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[1] For example, for the array of values [−2, 1, −3, 4, −1, 2, 1, −5, 4], the contiguous subarray with the largest sum is [4, −1, 2, 1], with sum 6. Some properties of this problem are: If the array contains all non-negative numbers, then the problem is trivial; a maximum subarray is the entire array.
A multiple constrained problem could consider both the weight and volume of the books. (Solution: if any number of each book is available, then three yellow books and three grey books; if only the shown books are available, then all except for the green book.) The knapsack problem is the following problem in combinatorial optimization:
HackerRank's programming challenges can be solved in a variety of programming languages (including Java, C++, PHP, Python, SQL, and JavaScript) and span multiple computer science domains. [ 2 ] HackerRank categorizes most of their programming challenges into a number of core computer science domains, [ 3 ] including database management ...
Peterson's algorithm (or Peterson's solution) is a concurrent programming algorithm for mutual exclusion that allows two or more processes to share a single-use resource without conflict, using only shared memory for communication.
A schematic picture of the skip list data structure. Each box with an arrow represents a pointer and a row is a linked list giving a sparse subsequence; the numbered boxes (in yellow) at the bottom represent the ordered data sequence.
Max-sum MSSP is a special case of MKP in which the value of each item equals its weight. The knapsack problem is a special case of MKP in which m=1. The subset-sum problem is a special case of MKP in which both the value of each item equals its weight, and m=1. The MKP has a Polynomial-time approximation scheme. [6]
Unlike the unweighted version, there is no greedy solution to the weighted activity selection problem. However, a dynamic programming solution can readily be formed using the following approach: [1] Consider an optimal solution containing activity k. We now have non-overlapping activities on the left and right of k. We can recursively find ...
If the car is behind door 2 – with the player having picked door 1 – the host must open door 3, such the probability that the car is behind door 2 and the host opens door 3 is 1 / 3 × 1 = 1 / 3 . These are the only cases where the host opens door 3, so if the player has picked door 1 and the host opens door 3, the car is ...