By Dr Antonio Gulli

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

**Read or Download A Collection of Dynamic Programming Interview Questions Solved in C++ PDF**

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

**Sample text**

Max sub-array problem - given an array of integers, compute the largest sum continuous sub array Solution This problem is a little gem. [1] The key intuition for this algorithm is illustrated by this picture represents the maximum value computed before analysing the current subsequence, while represents the maximum for the current subsequence. If , we violate the sequence continuity and therefore we can safely reset , a value which represents the empty string. size(); ++i) { sum += v[i]; if (sum > maxSum) { maxSum = sum; finish = right = i; start = left; } else if (sum < 0) { sum = 0; left = right = i+1; } } return maxSum; } Complexity Linear in the size of the input 4.

A jump cannot exceed the length contained in the current position. 11. Dice -– Given n dice, count how many ways to get sum s. 12. Coin Change – Given dollars, how many different ways there are to make the change into a set of coins? 13. 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.

Space occupancy is 3. Max sub-array problem - given an array of integers, compute the largest sum continuous sub array Solution This problem is a little gem. [1] The key intuition for this algorithm is illustrated by this picture represents the maximum value computed before analysing the current subsequence, while represents the maximum for the current subsequence. If , we violate the sequence continuity and therefore we can safely reset , a value which represents the empty string. size(); ++i) { sum += v[i]; if (sum > maxSum) { maxSum = sum; finish = right = i; start = left; } else if (sum < 0) { sum = 0; left = right = i+1; } } return maxSum; } Complexity Linear in the size of the input 4.