Add knapsack problem (#2330)

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Rajat Jain 2021-09-17 22:32:57 +05:30 committed by GitHub
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package DynamicProgramming;
/* A Naive recursive implementation
of 0-1 Knapsack problem */
public class BruteForceKnapsack {
// A utility function that returns
// maximum of two integers
static int max(int a, int b) {
return (a > b) ? a : b;
}
// Returns the maximum value that
// can be put in a knapsack of
// capacity W
static int knapSack(int W, int wt[], int val[], int n) {
// Base Case
if (n == 0 || W == 0)
return 0;
// If weight of the nth item is
// more than Knapsack capacity W,
// then this item cannot be included
// in the optimal solution
if (wt[n - 1] > W)
return knapSack(W, wt, val, n - 1);
// Return the maximum of two cases:
// (1) nth item included
// (2) not included
else
return max(val[n - 1] + knapSack(W - wt[n - 1], wt, val, n - 1), knapSack(W, wt, val, n - 1));
}
// Driver code
public static void main(String args[]) {
int val[] = new int[] { 60, 100, 120 };
int wt[] = new int[] { 10, 20, 30 };
int W = 50;
int n = val.length;
System.out.println(knapSack(W, wt, val, n));
}
}

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package DynamicProgramming;
// A Dynamic Programming based solution
// for 0-1 Knapsack problem
public class DyanamicProgrammingKnapsack {
static int max(int a, int b) {
return (a > b) ? a : b;
}
// Returns the maximum value that can
// be put in a knapsack of capacity W
static int knapSack(int W, int wt[], int val[], int n) {
int i, w;
int K[][] = new int[n + 1][W + 1];
// Build table K[][] in bottom up manner
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(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];
}
// Driver code
public static void main(String args[]) {
int val[] = new int[] { 60, 100, 120 };
int wt[] = new int[] { 10, 20, 30 };
int W = 50;
int n = val.length;
System.out.println(knapSack(W, wt, val, n));
}
}

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package DynamicProgramming;
// Here is the top-down approach of
// dynamic programming
public class MemoizationTechniqueKnapsack {
//A utility function that returns
//maximum of two integers
static int max(int a, int b) {
return (a > b) ? a : b;
}
//Returns the value of maximum profit
static int knapSackRec(int W, int wt[], int val[], int n, int[][] dp) {
// Base condition
if (n == 0 || W == 0)
return 0;
if (dp[n][W] != -1)
return dp[n][W];
if (wt[n - 1] > W)
// Store the value of function call
// stack in table before return
return dp[n][W] = knapSackRec(W, wt, val, n - 1, dp);
else
// Return value of table after storing
return dp[n][W] = max((val[n - 1] + knapSackRec(W - wt[n - 1], wt, val, n - 1, dp)),
knapSackRec(W, wt, val, n - 1, dp));
}
static int knapSack(int W, int wt[], int val[], int N) {
// Declare the table dynamically
int dp[][] = new int[N + 1][W + 1];
// Loop to initially filled the
// table with -1
for (int i = 0; i < N + 1; i++)
for (int j = 0; j < W + 1; j++)
dp[i][j] = -1;
return knapSackRec(W, wt, val, N, dp);
}
//Driver Code
public static void main(String[] args) {
int val[] = { 60, 100, 120 };
int wt[] = { 10, 20, 30 };
int W = 50;
int N = val.length;
System.out.println(knapSack(W, wt, val, N));
}
}