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# How to solve a 0/1 Knapsack Problem using Dynamic Programming?

**A knapsack problem is a problem in combinatorial optimization in which the goal is to find the most efficient way to fill a knapsack with a given set of items. Each item has a specific weight and value, and the goal is to fill the knapsack with as much value as possible while keeping the total weight below a given limit.**

The problem can be solved using a variety of algorithms, such as **dynamic programming**, **greedy algorithms**, and **branch and bound**. However, the most efficient algorithm for solving the problem depends on the specific instance of the problem and the amount of time and computational resources available.

This problem is a classic example of a combinatorial optimization problem and has been studied extensively by mathematicians and computer scientists. Many real-world problems can be formulated as a Knapsack problem, such as **resource allocation**, **packaging**, and **manufacturing**. Unfortunately, **it is also NP-complete**, meaning that no known algorithm can solve all instances of the problem in polynomial time.

There are mainly two types of Knapsack problems.

**0/1 Knapsack****Problem****Fractional Knapsack Problem**

In this part, we’ll solve the 0/1 Knapsack and in the next part, we’ll solve the fractional knapsack problem.

## 0-1 Knapsack Problem:

**In the 0-1 Knapsack problem, the main statement remains the same, i.e. we’ve to maximise the value within the given weight. But in this case, we have a condition that we can pick either the complete element or nothing. We can’t pick a fraction of the element.**

For example, see this image below-

Now solving this problem using Dynamic programming is a more general approach. This approach typically involves solving a series of sub-problems and then combining the solutions to these sub-problems to find the solution to the original problem.

There are two main steps in the dynamic programming approach:

- Determine the order in which the sub-problems should be solved.
- Solve the sub-problems and combine the solutions to find the solution to the original problem.

The order in which the sub-problems are solved is important, as it will determine the amount of work that needs to be done. In the 0-1 Knapsack Problem, the sub-problems can be solved in either a top-down or bottom-up fashion.

Once the order in which the sub-problems should be solved has been determined, the sub-problems can be solved using a recursive or iterative approach.

A recursive approach involves solving the sub-problem for the first item, then solving the sub-problem for the second item, and so on until the solution to the original problem is found.

An iterative approach would start by solving the sub-problem for the small items, then the medium items, and finally the large items.

Let’s understand the problem with the code and explanation.

### Code:

```
// solution for 0-1 Knapsack problem
#include <stdio.h>
int max(int a, int b)
{
return (a > b) ? a : b;
}
/* Returns the maximum value that can be put in the bag */
int knapSack(int W, int wt[], int prof[], int n)
{
int i, w;
int K[n + 1][W + 1];
for (i = 0; i <= n; i++)
{
for (w = 0; w <= W; w++)
{
if (i == 0 || w == 0)
K[i][w] = 0;
else if (wt[i - 1] <= w)
K[i][w] = max(prof[i - 1] + K[i - 1][w - wt[i - 1]],
K[i - 1][w]);
else
K[i][w] = K[i - 1][w];
}
}
return K[n][W];
}
int main(void)
{
int W, n, i;
float prof[100], wt[100];
float wp[100];
printf("Weight Limit: ");
scanf("%d", &W);
printf("Total number of stones-\n");
scanf("%d", &n);
for (i = 0; i < n; i++)
{
printf("Weight & Corresponding profit-\n");
scanf("%d%d", &wt[i], &prof[i]);
}
printf("Maximum Value is- %d", knapSack(W, wt, prof, n));
}
```

### Input:

For input, let’s take the same example of bag and stone.

Weight Limit: 8

Total number of stones-

4

Weight & Corresponding profit-1200

Weight & Corresponding profit-

2

400

Weight & Corresponding profit-

4

100

Weight & Corresponding profit-

3

500

### Output:

Maximum Value is- 1100

### Explanation:

To solve the 0/1 Knapsack Problem using Dynamic Programming, we need to create a table first containing the values and the weights.

- First, create a table with the required weight and number of items.
- Then make another table of weights with their corresponding values.
- Then put values according to the weights.

Follow these steps to find out the solution to this problem.

#### Time Complexity:

The 0/1 knapsack problem is a classic dynamic programming problem that has a time complexity of **O(nW)**, where **n** is the number of items and **W** is the capacity of the knapsack.

See the video-

So that’s all for this blog.

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