Question Name:ROY & FLOWER FARM - SRM ELAB DSA

#include <bits/stdc++.h> 
#define int long long
#define fi first
#define se second
#define pb push_back
using namespace std; 
int mod=1e9+7;
int gcd(int a,int b){
 if(b==0)return a;
 return  gcd(b,a%b);
}
 
 
int32_t main(){
 ios_base::sync_with_stdio(false);
 int t;
 cin>>t;
 while(t--)
 {
  int n,p;
  cin>>n>>p;
  vector<pair<int,int> >v;
  v.pb({0,0});
  for(int i=1;i<=n;i++)
  {
      int x,y;
      cin>>x>>y;
      if(x-y>=0)
      {
          v.pb({x-y,y});
      }
  }
        n=v.size()-1;
  int dp[n+1][p+1];
  memset(dp,0,sizeof(dp));
 
  for(int i=1;i<=n;i++)
  {
   for(int j=1;j<=p;j++)
   {
    if(j>=v[i].se)
    {
     dp[i][j]=max(dp[i-1][j],dp[i-1][j-v[i].se]+v[i].fi);
    }
    else
        dp[i][j]=dp[i-1][j];
       //cout<<dp[i][j]<<" ";
   }//cout<<endl;
  }
  int spend=0,have=p;
  for(int i=0;i<=p;i++)
  {
      //cout<<dp[n][i]<<" ";
      if(dp[n][i]+p>have)
      {
          have=dp[n][i]+p;
          spend=i;
      }
  }//cout<<endl;
  cout<<spend<<" "<<have<<endl;
  //cout<<dp[n][p]<<endl;
  
 }
 
    return 0;
}

In-Depth Analysis of the Problem

This problem involves optimizing Roy’s spending on fertilizing plants to maximize his profits after selling them the next day. It is essentially a 0/1 Knapsack problem, where:

• Each plant can either be fertilized or not (binary decision).
• Fertilizing a plant comes with a cost, and selling a fertilized plant gives Roy profit.
• Roy wants to maximize his profit while minimizing the amount spent on fertilizing plants, under the constraint that he has only $P$ rupees available.

Problem Breakdown

Problem Description

• Roy has $N$ plants that need fertilization to be sold the next day.
• For each plant:
  • $X$ is the profit Roy earns by selling the plant.
  • $Y$ is the cost of fertilizing the plant.
• Roy starts with $P$ rupees, and we need to determine:
  • The minimum amount $A$ Roy should spend on fertilizing the plants.
  • The maximum amount of money $B$ Roy can have after selling the fertilized plants.

Input/Output

• Input:
  • $T$: Number of test cases.
  • For each test case:
    • $N$: Number of plants.
    • $P$: Amount of money Roy has.
    • $N$ pairs of integers $X$ and $Y$ (profit from selling and cost of fertilizing).
• Output:
  • For each test case, output the minimum money spent $A$ and the maximum money $B$ Roy will have after selling the fertilized plants.

Constraints:

• $1 \leq T \leq 10$
• $1 \leq N \leq 100$
• $1 \leq X, Y \leq 1000$
• $1 \leq P \leq 10000$

The constraints suggest that a brute force approach might not work efficiently, and we need a more structured approach, likely dynamic programming.

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Understanding the Knapsack Problem Structure

This problem is a variant of the 0/1 Knapsack problem:

• We have a set of items (plants), each with a weight (fertilization cost) and value (profit from selling).
• We want to select some items such that the total weight is within a given limit (Roy’s initial money $P$), and we maximize the value (profit) while minimizing the money spent.

Dynamic Programming Approach:

We can define the problem as a DP table:

• Let dp[i][j] represent the maximum profit Roy can achieve using the first $i$ plants and with $j$ rupees available to spend.

The state transitions are as follows:

1. If we don't fertilize the $i$-th plant, then: $$dp[i][j] = dp[i-1][j]$$
2. If we fertilize the $i$-th plant and we have enough money (i.e., $j \geq Y[i]$), then: $$dp[i][j] = \max(dp[i][j], dp[i-1][j - Y[i]] + X[i])$$ where $X[i]$ is the profit from selling the $i$-th plant and $Y[i]$ is the fertilization cost.

After filling the table, we compute the minimum money $A$ that needs to be spent such that the total money $B$ (after selling the plants) is maximized.

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Detailed Code Implementation

#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second
#define pb push_back
using namespace std;

int mod=1e9+7;

// GCD function (unused in this solution)
int gcd(int a, int b){
    if(b == 0) return a;
    return gcd(b, a % b);
}

int32_t main(){
    ios_base::sync_with_stdio(false);  // Fast I/O
    int t;
    cin >> t;  // Number of test cases

    while(t--) {
        int n, p;
        cin >> n >> p;  // Number of plants and initial money Roy has

        vector<pair<int, int>> v;  // Vector to store (profit, cost) pairs for each plant
        v.pb({0, 0});  // A dummy value to simplify 1-based indexing
        
        for(int i = 1; i <= n; i++) {
            int x, y;
            cin >> x >> y;  // Input profit and cost for each plant
            if(x - y >= 0) {  // Only consider plants that yield non-negative profit
                v.pb({x - y, y});
            }
        }
        
        n = v.size() - 1;  // Updated number of valid plants after filtering
        int dp[n+1][p+1];
        memset(dp, 0, sizeof(dp));  // Initialize DP table with 0s
        
        // Fill the DP table
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= p; j++) {
                if(j >= v[i].se) {
                    // Option 1: Not fertilizing this plant
                    // Option 2: Fertilizing this plant
                    dp[i][j] = max(dp[i-1][j], dp[i-1][j - v[i].se] + v[i].fi);
                } else {
                    dp[i][j] = dp[i-1][j];
                }
            }
        }
        
        // Finding the optimal spending and final money
        int spend = 0, have = p;
        for(int i = 0; i <= p; i++) {
            if(dp[n][i] + p > have) {
                have = dp[n][i] + p;  // Update max money Roy will have after selling the plants
                spend = i;  // The money spent for this optimal solution
            }
        }
        
        // Output the result for this test case
        cout << spend << " " << have << endl;
    }

    return 0;
}

---

Explanation of Code

1. Input Parsing:

   • First, we read the number of test cases $t$.
   • For each test case, we read $n$ (number of plants) and $p$ (the money Roy has).

2. Filtering Valid Plants:

   • For each plant, we input the selling price $X$ and fertilization cost $Y$.
   • We only keep the plants where $X - Y \geq 0$, i.e., where the plant yields non-negative profit.

3. Dynamic Programming Table:

   • We define a DP table dp[i][j] where:
     • $i$ represents the first $i$ plants.
     • $j$ represents the amount of money Roy can spend (ranging from 0 to $p$).
   • The transitions are:
     • If we don't fertilize the $i$-th plant, $dp[i][j] = dp[i-1][j]$.
     • If we fertilize the $i$-th plant and can afford it, $dp[i][j] = \max(dp[i-1][j], dp[i-1][j - Y[i]] + X[i])$.

4. Final Calculation:

   • After filling the DP table, we look for the solution by checking all possible spending amounts and tracking the maximum profit Roy can achieve.
   • We track the minimum money spent and the maximum money Roy will have after selling the fertilized plants.

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Time Complexity

• For each test case:
  • Building the DP table takes $O(n \times p)$, where $n$ is the number of valid plants (filtered) and $p$ is the maximum money Roy has.
  • The time complexity for each test case is $O(n \times p)$, and given that $T \leq 10$, $n \leq 100$, and $p \leq 10000$, this approach is efficient.

Space Complexity

• The space complexity is $O(n \times p)$ for storing the DP table, where $n$ is the number of plants and $p$ is the money Roy has.

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Edge Cases Considered

• Zero Profit Plants: Plants with negative profit are discarded.
• Boundary Cases: All plants have a profit-to-cost ratio close to zero or the entire budget is spent fertilizing the plants.