Dynamic Programming Questions Long
Dynamic Programming can be used to solve the maximum subarray problem by utilizing the concept of subproblems and optimal substructure.
The maximum subarray problem aims to find the contiguous subarray within a given array of numbers that has the largest sum. To solve this problem using dynamic programming, we can follow the below steps:
1. Define the subproblem:
Let's consider an array A of size n. We define a subproblem as finding the maximum sum of a subarray ending at index i, denoted as maxEndingHere(i). This subproblem represents the maximum sum of a subarray that includes the element at index i.
2. Define the recurrence relation:
We can define the recurrence relation for the subproblem as follows:
maxEndingHere(i) = max(A[i], maxEndingHere(i-1) + A[i])
This relation states that the maximum sum of a subarray ending at index i is either the element at index i itself or the sum of the maximum subarray ending at index i-1 and the element at index i.
3. Define the base case:
The base case for the subproblem is when i = 0, i.e., the first element of the array. In this case, maxEndingHere(0) is simply A[0] since there is only one element.
4. Solve the subproblems iteratively:
We can solve the subproblems iteratively by starting from index 1 and calculating maxEndingHere(i) for each index i. At each step, we compare the current element with the sum of the previous maximum subarray and the current element, and update maxEndingHere(i) accordingly.
5. Track the maximum sum:
While solving the subproblems, we also keep track of the maximum sum found so far, denoted as maxSoFar. This variable stores the maximum sum of any subarray encountered during the iteration.
6. Return the maximum sum:
After iterating through all the elements, the maximum sum of a subarray will be stored in maxSoFar. We can return this value as the solution to the maximum subarray problem.
By following these steps, we can efficiently solve the maximum subarray problem using dynamic programming. The time complexity of this approach is O(n), where n is the size of the input array, as we only need to iterate through the array once.