Dynamic Programming

Prim's Minimum Spanning Tree is a Greedy Algorithm. All other are dynamic programming based.

http://www.geeksforgeeks.org/dynamic-programming-set-2-optimal-substructure-property/ Option (D) is incorrect because Greedy algorithms are generally faster than Dynamic programming. See http://www.geeksforgeeks.org/greedy-algorithms-set-1-activity-selection-problem/

Please see http://www.geeksforgeeks.org/dynamic-programming-set-3-longest-increasing-subsequence/

Kadane algorithm is used to find the maximum sum subarray in an array. It runs in O(n) time complexity. See details of the algorithm here.

It is basically matrix chain multiplication problem. We get minimum number of multiplications using ((M1 X (M2 X M3)) X M4). Total number of multiplications = 100x20x5 (for M2 x M3) + 10x100x5 + 10x5x80 = 19000.

X[I, j] (2 <= i <= n and ai <= j <= W), is true if any of the following is true 1) Sum of weights excluding ai is equal to j, i.e., if X[i-1, j] is true. 2) Sum of weights including ai is equal to j, i.e., if X[i-1, j-ai] is true so that we get (j – ai) + ai as j See http://www.geeksforgeeks.org/dynamic-programming-subset-sum-problem/ for details.

If we get the entry X[n, W] as true then there is a subset of {a1, a2, .. an} that has sum as W. Reference: http://en.wikipedia.org/wiki/Subset_sum_problem

In Longest common subsequence problem, there are two cases for X[0..i] and Y[0..j]

1) The last characters of two strings match. 
   The length of lcs is length of lcs of X[0..i-1] and Y[0..j-1]
2) The last characters don't match.
   The length of lcs is max of following two lcs values
   a) LCS of X[0..i-1] and Y[0..j]
   b) LCS of X[0..i] and Y[0..j-1]

//The LCS is of length 4. There are 3 LCS of length 4 "qprr", "pqrr" and qpqr   A subsequence is a sequence that can be derived from another sequence by selecting zero or more elements from it, without changing the order of the remaining elements. Subsequence need not be contiguous. Since the length of given strings A = “qpqrr” and B = “pqprqrp” are very small, we don’t need to build a 5x7 matrix and solve it using dynamic programming. Rather we can solve it manually just by brute force. We will first check whether there exist a subsequence  of length 5 since min_length(A,B) = 5. Since there is no subsequence , we will now check for length 4. “qprr”, “pqrr” and “qpqr” are common in both strings. X = 4 and Y = 3 X + 10Y = 34   This solution is contributed by Pranjul Ahuja

We have many ways to do matrix chain multiplication because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result of the matrix chain multiplication obtained will remain the same. Here we have four matrices A1, A2, A3, and A4, we would have: ((A1A2)A3)A4 = ((A1(A2A3))A4) = (A1A2)(A3A4) = A1((A2A3)A4) = A1(A2(A3A4)). However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product, or the efficiency. Here, A1 is a 10 × 5 matrix, A2 is a 5 x 20 matrix, and A3 is a 20 x 10 matrix, and A4 is 10 x 5. If we multiply two matrices A and B of order l x m and m x n respectively,then the number of scalar multiplications in the multiplication of A and B will be lxmxn. Then, The number of scalar multiplications required in the following sequence of matrices will be : A1((A2A3)A4) = (5 x 20 x 10) + (5 x 10 x 5) + (10 x 5 x 5) = 1000 + 250 + 250 = 1500. All other parenthesized options will require number of multiplications more than 1500.