By Dr Antonio Gulli

ISBN-10: 1495320480

ISBN-13: 9781495320484

This e-book offers a suite of Dynamic programming difficulties, their answer, and the C++ code relating to them.

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Extra resources for A Collection of Dynamic Programming Interview Questions Solved in C++

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Longest Palindrome String – Given a string, compute the longest palindromic substring 14. String Palindromes -– Given a string, find the minimum number of characters to be inserted for converting the string into a palindrome 15. LCS -– Given two strings S1 and S2, find the longest common substring 16. LCS – Given two strings, find the longest common subsequence 17. LIS – Given an array of integers, find the longest increasing subsequence 18. Bridge Matching – n cities are on the northern bank of a river and n cities are on the southern bank.

Ship Battle – Given a matrix M and an array V, match the array in the matrix Appendix Dynamic programming: the art of solving simple problems first Dynamic programming (DP) solves intricate problems by breaking them down into simpler components. DP can be applied to problems exhibiting two properties: overlapping subproblems and optimal substructure. Overlapping subproblems means that any recursive algorithm solving the problem should solve the same subproblems over and over, rather than generating new subproblems.

If then. dim2(); j++) if (size < s(i, j)) { size = s(i, j); maxI = i; maxJ = j; } } Complexity Complexity is 27. Matrix Parenthesization -- Given a set of m Matrices find the most efficient way of multiplying them Solution Matrix multiplication is associative so there are multiple ways of performing the multiplication and therefore the number of operations (sum and multiplications) performed on scalars is different. We can scan the vector and find a solution recursively. size(); Matrix m(n, n); for (unsigned int i = 0; i < n; ++i) m(i, i) = 0; unsigned int j, cost; // len = 2, 1 <= i < n - 1 ; j = i + 1 // len = 3, 1 <= i < n - 2; j = i + 2 // // len = n-2, i=1, i = 2 , j = i + n - 3 // len = n-1, i=1 , j = n - 1 for (unsigned int len = 2; len < n; ++len) for (unsigned i = 1; i < n - len + 1; ++i) { j = i + len - 1; m(i, j) = std::numeric_limits::max(); for (unsigned int k = i; k <= j - 1; ++k) { cost = m(i, k) + m(k + 1, j) + p[i - 1] * p[k] * p[j]; if (cost < m(i, j)) { m(i, j) = cost; } } } return m(1, n - 1); } Complexity Complexity is 28.

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A Collection of Dynamic Programming Interview Questions Solved in C++ by Dr Antonio Gulli

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