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dynamic programming
Parshu gate
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Algorithms
Sep 17, 2017
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dynamic-programming
algorithms
Parshu gate
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Algorithms
Sep 17, 2017
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Parshu gate
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sandygate
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Nov 26, 2018
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is the answer 12??
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int max(int a, int b) { return (a > b) ? a : b; } // Returns the maximum value that can be // put in a knapsack of capacity W int knapSack(int W, int wt[], int val[], int n) { // Base Case if (n == 0 || W == 0) return 0; // If ... , wt, val, n - 1), knapSack(W, wt, val, n - 1)); } This statement implies that the max value return by the two different recursive problems right?
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Nandkishor3939
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Jan 22, 2019
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My answer came out to be 13: because when we will compute T(13) { as we are using Dynamic programming , it will have to compute value of T(12),T(11),…...T(2) only once(as it will store it and reuse it) so the stack size will be 1 (for T(13))+11 (for T(12),T(11),…...T(2)) = 12…...(48/4) } will any one help me out
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Given a text array $T[1…..n]$ and a pattern array $P[1….m]$ such that T and P are character taken from alphabet $\sum$, $\sum={a,b,c,…..z}$. String matching problem is to find all the occurence of P in T. A pattern occur with shift s in T if $P[1…..m]=T[s+1,…...s+m]$. Consider $T=bacacbaacacac$ $P=cac$ The sum of the value of all s is ________
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