Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. Items are indivisible; you either take an item or not (0-1 property).

Input:

value = [ 20, 5, 10, 40, 15, 25 ]

weight = [ 1, 2, 3, 8, 7, 4 ]

int W = 10

Output: Knapsack value is 60

(value = 20 + 40 = 60

weight = 1 + 8 = 9 < W)

The idea is to use **recursion** to solve this problem. For each item, there are two possibilities –

1. We include current item in knapSack and recurse for remaining items with decreased capacity of Knapsack. If the capacity becomes negative, do not recuse or return -INFINITY.

2. We exclude current item from knapSack and recurse for remaining items.

Finally, we return maximum value we get by including or excluding current item. The base case of the recursion would be when no items are left or capacity becomes 0.

Below solution finds the maximum value that can be attained with weight less than or equal to W recursively by using above relations.

**C++ implementation –**

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#include <bits/stdc++.h> using namespace std; // Values (stored in array v) // Weights (stored in array w) // Number of distinct items (n) // Knapsack capacity W int knapSack(int v[], int w[], int n, int W) { // base case: no items left or capacity becomes 0 if (n < 0 || W == 0) return 0; // base case: Negative capacity if (W < 0) return INT_MIN; // Case 1. include current item n in knapSack (v[n]) and recurse for // remaining items (n - 1) with decreased capacity (W - w[n]) int include = v[n] + knapSack(v, w, n - 1, W - w[n]); // Case 2. exclude current item n from knapSack and recurse for // remaining items (n - 1) int exclude = knapSack(v, w, n - 1, W); // return maximum value we get by including or excluding current item return max (include, exclude); } int main() { // Input: set of items each with a weight and a value int v[] = { 20, 5, 10, 40, 15, 25 }; int w[] = { 1, 2, 3, 8, 7, 4 }; // Knapsack capacity int W = 10; // number of items int n = sizeof(v) / sizeof(v[0]); cout << "Knapsack value is " << knapSack(v, w, n - 1, W); return 0; } |

**Output: **

Knapsack value is 60

The time complexity of above solution is exponential and auxiliary space used by the program is O(1).

The problem has an **optimal substructure**. That means the problem can be broken down into smaller, simple “subproblems”, which can further be divided into yet simpler, smaller subproblems until the solution becomes trivial. Above solution also exhibits **overlapping subproblems**. If we draw the recursion tree of the solution, we can see that the same sub-problems are getting computed again and again. We know that problems having optimal substructure and overlapping subproblems can be solved by using **Dynamic Programming**, in which subproblem solutions are *Memo*ized rather than computed again and again. Below *Memo*ized version follows the top-down approach, since we first break the problem into subproblems and then calculate and store values.

**Memoized C++ implementation –**

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#include <bits/stdc++.h> using namespace std; // create a map to store solutions of subproblems unordered_map<string, int> lookup; // Values (stored in array v) // Weights (stored in array w) // Number of distinct items (n) // Knapsack capacity W int knapSack(int v[], int w[], int n, int W) { // base case: no items left or capacity becomes 0 if (n < 0 || W == 0) return 0; // base case: Negative capacity if (W < 0) return INT_MIN; // construct a unique map key from dynamic elements of the input string key = to_string(n) + "|" + to_string(W); // if sub-problem is seen for the first time, solve it and // store its result in a map if (lookup.find(key) == lookup.end()) { // Case 1. include current item n in knapSack (v[n]) & recurse for // remaining items (n - 1) with decreased capacity (W - w[n]) int include = v[n] + knapSack(v, w, n - 1, W - w[n]); // Case 2. exclude current item n from knapSack and recurse for // remaining items (n - 1) int exclude = knapSack(v, w, n - 1, W); // assign max value we get by including or excluding current item lookup[key] = max (include, exclude); } // return solution to current sub-problem return lookup[key]; } int main() { // Input: set of items each with a weight and a value int v[] = { 20, 5, 10, 40, 15, 25 }; int w[] = { 1, 2, 3, 8, 7, 4 }; // Knapsack capacity int W = 10; // number of items int n = sizeof(v) / sizeof(v[0]); cout << "Knapsack value is " << knapSack(v, w, n - 1, W); return 0; } |

**Output: **

Knapsack value is 60

The time complexity of above solution is O(nW) where n is the number of items in the input and W is the Knapsack capacity. Auxiliary space used by the program is also O(nW).

We can also solve this problem in bottom-up manner. In the bottom-up approach, we solve smaller sub-problems first, then solve larger sub-problems from them. The following bottom-up approach computes T[i][j], for each 1 <= i <= n and 0 <= j <= W, which is maximum value that can be attained with weight less than or equal to j and using items up to first i items. It uses value of smaller values i and j already computed. It has the same asymptotic run-time as Memoization but no recursion overhead.

**C++ implementation –**

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#include <bits/stdc++.h> using namespace std; // Input: // Values (stored in array v) // Weights (stored in array w) // Number of distinct items (n) // Knapsack capacity W int knapSack(int v[], int w[], int n, int W) { // T[i][j] stores the maximum value that can be attained with // weight less than or equal to j using items up to first i items int T[n+1][W+1]; for (int j = 0; j <= W; j++) T[0][j] = 0; // memset (T[0], 0, sizeof T[0]); // do for ith item for (int i = 1; i <= n; i++) { // consider all weights from 0 to maximum capacity W for (int j = 0; j <= W; j++) { // don't include ith element if j-w[i-1] is negative if (w[i-1] > j) T[i][j] = T[i-1][j]; else // find max value by excluding or including the ith item T[i][j] = max(T[i-1][j], T[i-1][j-w[i-1]] + v[i-1]); } } // return maximum value return T[n][W]; } int main() { // Input: set of items each with a weight and a value int v[] = { 20, 5, 10, 40, 15, 25 }; int w[] = { 1, 2, 3, 8, 7, 4 }; // Knapsack capacity int W = 10; // number of items int n = sizeof(v) / sizeof(v[0]); cout << "Knapsack value is " << knapSack(v, w, n, W); return 0; } |

**Output: **

Knapsack value is 60

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