Answers for "Knapsack python"

python 0-1 kanpsack

#Returns the maximum value that can be stored by the bag

def knapSack(W, wt, val, n):
   # initial conditions
   if n == 0 or W == 0 :
      return 0
   # If weight is higher than capacity then it is not included
   if (wt[n-1] > W):
      return knapSack(W, wt, val, n-1)
   # return either nth item being included or not
   else:
      return max(val[n-1] + knapSack(W-wt[n-1], wt, val, n-1),
         knapSack(W, wt, val, n-1))
# To test above function
val = [50,100,150,200]
wt = [8,16,32,40]
W = 64
n = len(val)
print (knapSack(W, wt, val, n))

knapsack algorithm in python

# a dynamic approach
# Returns the maximum value that can be stored by the bag
def knapSack(W, wt, val, n):
   K = [[0 for x in range(W + 1)] for x in range(n + 1)]
   #Table in bottom up manner
   for i in range(n + 1):
      for w in range(W + 1):
         if i == 0 or w == 0:
            K[i][w] = 0
         elif wt[i-1] <= w:
            K[i][w] = max(val[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]
#Main
val = [50,100,150,200]
wt = [8,16,32,40]
W = 64
n = len(val)
print(knapSack(W, wt, val, n))

knapsack

#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> >a;
//dp table is full of zeros
int n,s,dp[1002][1002];
void ini(){
    for(int i=0;i<1002;i++)
        for(int j=0;j<1002;j++)
            dp[i][j]=-1;
}
int f(int x,int b){
	//base solution
	if(x>=n or b<=0)return 0;
	//if we calculate this before, we just return the answer (value diferente of 0)
	if(dp[x][b]!=-1)return dp[x][b];
	//calculate de answer for x (position) and b(empty space in knapsack)
	//we get max between take it or not and element, this gonna calculate all the
	//posible combinations, with dp we won't calculate what is already calculated.
	return dp[x][b]=max(f(x+1,b),b-a[x].second>=0?f(x+1,b-a[x].second)+a[x].first:INT_MIN);
}
int main(){
	//fast scan and print
	ios_base::sync_with_stdio(0);cin.tie(0);
	//we obtain quantity of elements and size of knapsack
	cin>>n>>s;
	a.resize(n);
	//we get value of elements
	for(int i=0;i<n;i++)
		cin>>a[i].first;
	//we get size of elements
	for(int i=0;i<n;i++)
		cin>>a[i].second;
	//initialize dp table
	ini();
	//print answer
	cout<<f(0,s);
	return 0;
}

Knapsack python

'''
Capacity of knapsack = W
weight list : wt = []
price list : pr = []
No. of items = N
'''
def kProfit(W,N,wt,pr,dp):
    # Base Condition
    if N==0 or W==0:
        return 0
    # If sub problem is previously solved tehn return it.
    if dp[N][W] is not None:
        return dp[N][W]
    if wt[N-1] <= W:
        dp[N][W] = max(pr[N-1]+kProfit(W-wt[N-1],N-1,wt,pr,dp), kProfit(W,N-1,wt,pr,dp))
        return dp[N][W]
    else:
        dp[N][W] = kProfit(W,N-1,wt,pr,dp)
        return dp[N][W]
if __name__ == '__main__':
    W = 11
    wt = [1, 2, 5, 6, 7]
    pr = [1, 6, 18, 22, 28]
    N = len(pr)
    # define DP array
    dp = [[None] * (W + 1) for _ in range(N + 1)]
    # Call for kProfit to calculate max profit
    maxProfit = kProfit(W,N,wt,pr,dp)
    print('Maximum Profit is : ',maxProfit)

knapsack problem

// memory efficient and iterative approach to the knapsack problem

#include <bits/stdc++.h>
using namespace std;

// n is the number of items
// w is the knapsack's capacity
int n, w;

int main() {
/*
input format:
n w
value_1 cost_1
value_2 cost_2
.
.
value_n cost_n
*/
    cin >> n >> w;
  	vector<long long> dp(w + 1, 0);

    for (int i = 0; i < n; ++i) {
        int value, cost;
        cin >> value >> cost;
        for (int j = w; j >= cost; --j)
            dp[j] = max(dp[j], value + dp[j - cost]);
    }

    // the answer is dp[w]
    cout << dp[w];
}

python 0-1 kanpsack

#Returns the maximum value that can be stored by the bag

def knapSack(W, wt, val, n):
   # initial conditions
   if n == 0 or W == 0 :
      return 0
   # If weight is higher than capacity then it is not included
   if (wt[n-1] > W):
      return knapSack(W, wt, val, n-1)
   # return either nth item being included or not
   else:
      return max(val[n-1] + knapSack(W-wt[n-1], wt, val, n-1),
         knapSack(W, wt, val, n-1))
# To test above function
val = [50,100,150,200]
wt = [8,16,32,40]
W = 64
n = len(val)
print (knapSack(W, wt, val, n))

knapsack algorithm in python

# a dynamic approach
# Returns the maximum value that can be stored by the bag
def knapSack(W, wt, val, n):
   K = [[0 for x in range(W + 1)] for x in range(n + 1)]
   #Table in bottom up manner
   for i in range(n + 1):
      for w in range(W + 1):
         if i == 0 or w == 0:
            K[i][w] = 0
         elif wt[i-1] <= w:
            K[i][w] = max(val[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]
#Main
val = [50,100,150,200]
wt = [8,16,32,40]
W = 64
n = len(val)
print(knapSack(W, wt, val, n))

knapsack

#include<bits/stdc++.h>
using namespace std;
vector<pair<int,int> >a;
//dp table is full of zeros
int n,s,dp[1002][1002];
void ini(){
    for(int i=0;i<1002;i++)
        for(int j=0;j<1002;j++)
            dp[i][j]=-1;
}
int f(int x,int b){
	//base solution
	if(x>=n or b<=0)return 0;
	//if we calculate this before, we just return the answer (value diferente of 0)
	if(dp[x][b]!=-1)return dp[x][b];
	//calculate de answer for x (position) and b(empty space in knapsack)
	//we get max between take it or not and element, this gonna calculate all the
	//posible combinations, with dp we won't calculate what is already calculated.
	return dp[x][b]=max(f(x+1,b),b-a[x].second>=0?f(x+1,b-a[x].second)+a[x].first:INT_MIN);
}
int main(){
	//fast scan and print
	ios_base::sync_with_stdio(0);cin.tie(0);
	//we obtain quantity of elements and size of knapsack
	cin>>n>>s;
	a.resize(n);
	//we get value of elements
	for(int i=0;i<n;i++)
		cin>>a[i].first;
	//we get size of elements
	for(int i=0;i<n;i++)
		cin>>a[i].second;
	//initialize dp table
	ini();
	//print answer
	cout<<f(0,s);
	return 0;
}

Knapsack python

'''
Capacity of knapsack = W
weight list : wt = []
price list : pr = []
No. of items = N
'''
def kProfit(W,N,wt,pr,dp):
    # Base Condition
    if N==0 or W==0:
        return 0
    # If sub problem is previously solved tehn return it.
    if dp[N][W] is not None:
        return dp[N][W]
    if wt[N-1] <= W:
        dp[N][W] = max(pr[N-1]+kProfit(W-wt[N-1],N-1,wt,pr,dp), kProfit(W,N-1,wt,pr,dp))
        return dp[N][W]
    else:
        dp[N][W] = kProfit(W,N-1,wt,pr,dp)
        return dp[N][W]
if __name__ == '__main__':
    W = 11
    wt = [1, 2, 5, 6, 7]
    pr = [1, 6, 18, 22, 28]
    N = len(pr)
    # define DP array
    dp = [[None] * (W + 1) for _ in range(N + 1)]
    # Call for kProfit to calculate max profit
    maxProfit = kProfit(W,N,wt,pr,dp)
    print('Maximum Profit is : ',maxProfit)

knapsack problem

// memory efficient and iterative approach to the knapsack problem

#include <bits/stdc++.h>
using namespace std;

// n is the number of items
// w is the knapsack's capacity
int n, w;

int main() {
/*
input format:
n w
value_1 cost_1
value_2 cost_2
.
.
value_n cost_n
*/
    cin >> n >> w;
  	vector<long long> dp(w + 1, 0);

    for (int i = 0; i < n; ++i) {
        int value, cost;
        cin >> value >> cost;
        for (int j = w; j >= cost; --j)
            dp[j] = max(dp[j], value + dp[j - cost]);
    }

    // the answer is dp[w]
    cout << dp[w];
}