A Collection of Dynamic Programming Interview Questions by Dr Antonio Gulli PDF

By Dr Antonio Gulli

ISBN-10: 1495320480

ISBN-13: 9781495320484

This publication provides a set of Dynamic programming difficulties, their resolution, and the C++ code on the topic of them.

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

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The code provided uses memorization for avoiding recomputation of the same values again and again. The solution is built bottom-up. size(); Matrix cost(n, n); Matrix sumFreq(n, n); unsigned int localCost; for (unsigned int i = 0; i < n; ++i) cost(i, i) = freq[i]; for (unsigned int i = 0; i < n; ++i) for (unsigned int j = i; j < n; ++j) { sumFreq(i, j) = 0; for (unsigned int s = i; s <= j && s < n; ++s) sumFreq(i, j) += freq[s]; } // range = 2 (0<=i j=i+1 1<=j<=n-1 // range = 3 (0<=i j=i+2 0<=j<=n-1 // range = n (i=0) -> j = n-1 for (unsigned int range = 2; range <= n; ++range) for (unsigned int i = 0; i < n - range + 1; ++i) { unsigned int j = i + range - 1; cost(i, j) = std::numeric_limits::max(); for (unsigned int k = i; k <= j; k++) { localCost = ((k > i) ?

Seg[n - i]) seg[n - i] = segmentDPhelper(n - i, seg); m = std::max(m, std::max(i * (n - i), seg[n - i] * i)); } return m; } unsigned int segmentDP(unsigned int n) { if (n == 0 || n == 1) return 0; std::vector seg(n, 0); return segmentDPhelper(n, seg); } Complexity Linear in time and space 24. Cutting a Rod – given a rod of length n and a vector of prices for different lengths, cut the rod for maximizing the gain Solution The idea is very similar to the one of the previous problem.

0) return false; return isSubsetSum(v, n, sum / 2); } bool isSubsetSumDP(const std::vector & v, const unsigned int n, unsigned int sum) { Matrix subset(sum + 1, n + 1); for (unsigned int j = 0; j <= n; j++) subset(0, j) = true; for (unsigned int i = 1; i <= sum; i++) subset(i, 0) = false; for (unsigned int i = 1; i <= sum; i++) for (unsigned int j = 1; j <= n; j++) { subset(i, j) = subset(i, j - 1); if (i >= v[j - 1]) { subset(i, j) = subset(i, j) || subset(i - v[j - 1], j - 1); } } return subset(sum, n); } Complexity Complexity is pseudo-polynomial 22.

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


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