# 0-1 Knapsack Problem in C?

A knapsack is a bag. And the knapsack problem deals with the putting items to the bag based on the value of the items. It aim is to maximise the value inside the bag. In 0-1 Knapsack you can either put the item or discard it, there is no concept of putting some part of item in the knapsack.

## Sample Problem

Value of items = {20, 25,40}
Weights of items = {25, 20, 30}
Capacity of the bag = 50

## Weight distribution

25,20{1,2}
20,30 {2,3}
If we use {1,3} the weight will be above the max allowed value.
For {1,2} : weight= 20+25=45 Value = 20+25 = 45
For {2,3}: weight=20+30=50 Value = 25+40=65

The maximum value is 65 so we will put the item 2 and 3 in the knapsack.

## PROGRAM FOR 0-1 KNAPSACK PROBLEM

#include<stdio.h>
int max(int a, int b) {
if(a>b){
return a;
} else {
return b;
}
}
int knapsack(int W, int wt[], int val[], int n) {
int i, w;
int knap[n+1][W+1];
for (i = 0; i <= n; i++) {
for (w = 0; w <= W; w++) {
if (i==0 || w==0)
knap[i][w] = 0;
else if (wt[i-1] <= w)
knap[i][w] = max(val[i-1] + knap[i-1][w-wt[i-1]], knap[i-1][w]);
else
knap[i][w] = knap[i-1][w];
}
}
return knap[n][W];
}
int main() {
int val[] = {20, 25, 40};
int wt[] = {25, 20, 30};
int W = 50;
int n = sizeof(val)/sizeof(val);
printf("The solution is : %d", knapsack(W, wt, val, n));
return 0;
}

## Output

The solution is : 65