Question

Recall that in the "Knapsack Problem", there are n items having respective values V1..n) and weights W1..n), all greater than 0 and one needs to maximize the total value of the subset of the items placed in the knapsack limited by a weight capacity of W In the 0-1 Knapsack Problem, each item must be either be included or excluded in its entirety, in light of the fact that this problem is

Answer 1

knapsack 0-1 broblem Om TA, D E LONG :: It contains overlapped subproblems. So, we can follow DP apperbach. wt[] = Array containing weight of every element. W Mabimum weight val [ ] == Array containing values of every element. ne Total number of element AR

it will J. Knapsack (al, wt[], val[],"): of 2. aus [nti][k+1] l1 To store answers. for lizo sn) ♡ jne for (j zo sW). yliczo or j=20). u austil ljjzo else if (wthi-1] <=j) austellj] = may valtistan [i-1] [aji wtri) aust 2-11251) 8. 9. 10 alle also pus[i][j] = kie-1(j); return aus En J LWS

cbtnes is au NP problem in a first loop runs on) times windows 10 4. Second loop rouns o (n) times. - Total complexity o (nW).

fractional knapsack greedy solution in A. Ju dus we will tirst calculate valve weight a ratio and then sost items. Cast item will be the one with most value in least weight. So, we gather all the elemente from end till we reach the capacit

2222222PLNO i i oj jj o toj os diam struct item & int value; eut weight: double ratio. Sarr [n]; Fract knapsack CW, wth? vall], n) sort Carr, arren, end) Il sort on basis of ratio ent weight.co. double udle for Cirn-1-20); if (weight tarril. weight <= N) weight t= arril. Loeight name t= amil.name else م (ع) علمهار int x=W-weight value += arr olvame x (remo arrtil weight return value return value

Gu this problem, sorting can be done oln logn) o lim Other loop runs ocn) times.) So, final complexity 50 (n logn),