# Maximum items that can be filled in K Knapsacks of given Capacity

Given an integer array W[] consisting of weights of items and ‘K’ knapsacks of capacity ‘C’, find maximum weight we can put in the knapsacks if breaking of an item is not allowed.

Examples:

Input : w[] = {3, 9, 8}, k = 1, c = 11
Output : 11
The required subset will be {3, 8}
where 3+8 = 11

Input : w[] = {3, 9, 8}, k = 1, c = 10
Output : 9

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We will use Dynamic programming to solve this problem.
We will use two variables to represent the states of DP.

1. ‘i’ – The current index we are working on.
2. ‘R’ – It contains the remaining capacity of each and every knapsack.

Now, how will a single variable store the remaining capacity of every knapsack?

For that, we will initialise ‘R’ as R = C + C*(C+1) + C*(C+1)^2 + C*(C+1)^3 ..+ C*(C+1)^(k-1)
This initialises all the ‘k’ knapsacks with capacity ‘C’.

Now, we need to perform two queries:

• Reading remaining space of jth knapsack: (r/(c+1)^(j-1))%(c+1).
• Decreasing remaining space of jth knapsack by x: set r = r – x*(c+1)^(j-1).

Now, at each step, we will have k+1 choices.

1. Reject index ‘i’.
2. Put item ‘i’ in knapsack 1.
3. Put item ‘i’ in knapsack 2.
4. Put item ‘i’ in knapsack 3.

5. k+1) Put item ‘i’ in knapsack k.

We will choose the path that maximizes the result.

Below is the implementation of above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// 2-d array to store states of DP ` `vector > dp; ` ` `  `// 2-d array to store if a state ` `// has been solved ` `vector > v; ` ` `  `// Vector to store power of variable 'C'. ` `vector<``int``> exp_c; ` ` `  `// function to compute the states ` `int` `FindMax(``int` `i, ``int` `r, ``int` `w[], ` `            ``int` `n, ``int` `c, ``int` `k) ` `{ ` ` `  `    ``// Base case ` `    ``if` `(i >= n) ` `        ``return` `0; ` ` `  `    ``// Checking if a state has been solved ` `    ``if` `(v[i][r]) ` `        ``return` `dp[i][r]; ` ` `  `    ``// Setting a state as solved ` `    ``v[i][r] = 1; ` `    ``dp[i][r] = FindMax(i + 1, r, w, n, c, k); ` ` `  `    ``// Recurrence relation ` `    ``for` `(``int` `j = 0; j < k; j++) { ` `        ``int` `x = (r / exp_c[j]) % (c + 1); ` `        ``if` `(x - w[i] >= 0) ` `            ``dp[i][r] = max(dp[i][r], w[i] +  ` `            ``FindMax(i + 1, r - w[i] * exp_c[j], w, n, c, k)); ` `    ``} ` ` `  `    ``// Returning the solved state ` `    ``return` `dp[i][r]; ` `} ` ` `  `// Function to initialize global variables ` `// and find the initial value of 'R' ` `int` `PreCompute(``int` `n, ``int` `c, ``int` `k) ` `{ ` ` `  `    ``// Resizing the variables ` `    ``exp_c.resize(k); ` `    ``exp_c[0] = 1; ` ` `  `    ``for` `(``int` `i = 1; i < k; i++){ ` `        ``exp_c[i] = (exp_c[i - 1] * (c + 1)); ` `    ``} ` `    ``dp.resize(n); ` `    ``for` `(``int` `i = 0; i < n; i++){ ` `        ``dp[i].resize(exp_c[k - 1] * (c + 1), 0); ` `    ``} ` `    ``v.resize(n); ` `    ``for` `(``int` `i = 0; i < n; i++){ ` `        ``v[i].resize(exp_c[k - 1] * (c + 1), 0); ` `    ``} ` ` `  `    ``// Variable to store the initial value of R ` `    ``int` `R = 0; ` `    ``for` `(``int` `i = 0; i < k; i++){ ` `        ``R += exp_c[i] * c; ` `    ``} ` `    ``return` `R; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Input array ` `    ``int` `w[] = { 3, 8, 9 }; ` ` `  `    ``// number of knapsacks and capacity ` `    ``int` `k = 1, c = 11; ` ` `  `    ``int` `n = ``sizeof``(w) / ``sizeof``(``int``); ` ` `  `    ``// Performing required pre-computation ` `    ``int` `r = PreCompute(n, c, k); ` ` `  `    ``// finding the required answer ` `    ``cout << FindMax(0, r, w, n, c, k); ` ` `  `    ``return` `0; ` `} `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# 2-d array to store states of DP ` `x ``=` `100` `dp ``=` `[[``0` `for` `i ``in` `range``(x)]  ` `         ``for` `i ``in` `range``(x)] ` ` `  `# 2-d array to store if a state ` `# has been solved ` `v ``=` `[[``0` `for` `i ``in` `range``(x)]   ` `        ``for` `i ``in` `range``(x)] ` ` `  `# Vector to store power of variable 'C'. ` `exp_c ``=` `[] ` ` `  `# function to compute the states ` `def` `FindMax(i, r, w, n, c, k): ` ` `  `    ``# Base case ` `    ``if` `(i >``=` `n): ` `        ``return` `0` ` `  `    ``# Checking if a state has been solved ` `    ``if` `(v[i][r]): ` `        ``return` `dp[i][r] ` ` `  `    ``# Setting a state as solved ` `    ``v[i][r] ``=` `1` `    ``dp[i][r] ``=` `FindMax(i ``+` `1``, r, w, n, c, k) ` ` `  `    ``# Recurrence relation ` `    ``for` `j ``in` `range``(k): ` `        ``x ``=` `(r ``/``/` `exp_c[j]) ``%` `(c ``+` `1``) ` `        ``if` `(x ``-` `w[i] >``=` `0``): ` `            ``dp[i][r] ``=` `max``(dp[i][r], w[i] ``+` `            ``FindMax(i ``+` `1``, r ``-` `w[i] ``*` `exp_c[j], ` `                                   ``w, n, c, k)) ` ` `  `    ``# Returning the solved state ` `    ``return` `dp[i][r] ` ` `  `# Function to initialize global variables ` `# and find the initial value of 'R' ` `def` `PreCompute(n, c, k): ` ` `  ` `  `    ``# Resizing the variables ` `    ``exp_c.append(``1``) ` ` `  `    ``for` `i ``in` `range``(``1``, k): ` `        ``exp_c[i] ``=` `(exp_c[i ``-` `1``] ``*` `(c ``+` `1``)) ` ` `  `    ``# Variable to store the initial value of R ` `    ``R ``=` `0` `    ``for` `i ``in` `range``(k): ` `        ``R ``+``=` `exp_c[i] ``*` `c ` ` `  `    ``return` `R ` ` `  `# Driver Code ` ` `  `# Input array ` `w ``=``[``3``, ``8``, ``9``] ` ` `  `# number of knapsacks and capacity ` `k ``=` `1` `c ``=` `11` ` `  `n ``=` `len``(w) ` ` `  `# Performing required pre-computation ` `r ``=` `PreCompute(n, c, k) ` ` `  `# finding the required answer ` `print``(FindMax(``0``, r, w, n, c, k)) ` ` `  `# This code is contributed by Mohit Kumar `

Output:

```11
```

Time complexity : O(N*k*C^k).

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : mohit kumar 29