27_RodCuttin.cpp

// https://www.codingninjas.com/codestudio/problems/rod-cutting-problem_800284

// Given a rod of length ‘N’ units. The rod can be cut into different sizes and each
// size has a cost associated with it. Determine the maximum cost obtained by cutting
// the rod and selling its pieces.

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

class Solution5
{
    // Tabulation: Space Optimised: Single Array 
public:
    int cutRod(vector<int> &price, int n)
    {
        vector<vector<int>> dp(n, vector<int>(n + 1));
        vector<int> cp(n + 1);
        for (int len = 1; len <= n; ++len)
        {
            cp[len] = price[0] * len;
        }

        for (int i = 1; i < n; ++i)
        {
            for (int j = 0; j <= n; ++j)
            {
                int cut = INT_MIN;
                if (j >= i + 1)
                    cut = price[i] + cp[j - 1 - i];
                int noCut = cp[j];
                cp[j] = max(cut, noCut);
            }
        }

        return cp[n];
    }
};

class Solution4
{
    // Tabulation: Space Optimised
public:
    int cutRod(vector<int> &price, int n)
    {
        vector<vector<int>> dp(n, vector<int>(n + 1));
        vector<int> cp(n + 1), xp(n + 1);
        for (int len = 1; len <= n; ++len)
        {
            cp[len] = price[0] * len;
        }

        for (int i = 1; i < n; ++i)
        {
            for (int j = 0; j <= n; ++j)
            {
                int cut = INT_MIN;
                if (j >= i + 1)
                    cut = price[i] + xp[j - 1 - i];
                int noCut = cp[j];
                xp[j] = max(cut, noCut);
            }
            cp = xp;
        }

        return cp[n];
    }
};

class Solution3
{
    // Tabulation
public:
    int cutRod(vector<int> &price, int n)
    {
        vector<vector<int>> dp(n, vector<int>(n + 1));
        for (int len = 1; len <= n; ++len)
        {
            dp[0][len] = price[0] * len;
        }

        for (int i = 1; i < n; ++i)
        {
            for (int j = 0; j <= n; ++j)
            {
                int cut = INT_MIN;
                if (j >= i + 1)
                    cut = price[i] + dp[i][j - 1 - i];
                int noCut = dp[i - 1][j];
                dp[i][j] = max(cut, noCut);
            }
        }

        return dp[n - 1][n];
    }
};

class Solution2
{
    // Recursion: Memoization
public:
    int solve(vector<int> &price, vector<vector<int>> &dp, int len, int n)
    {
        if (n == -1)
            return 0;

        if (dp[n][len] != -1)
            return dp[n][len];

        int cut = 0;
        if (len >= n + 1)
            cut = price[n] + solve(price, dp, len - (n + 1), n);
        int noCut = solve(price, dp, len, n - 1);
        return dp[n][len] = max(cut, noCut);
    }

    int cutRod(vector<int> &price, int n)
    {
        vector<vector<int>> dp(n, vector<int>(n + 1, -1));
        return solve(price, dp, n, n - 1);
    }
};

class Solution1
{
    // Recursion: Optimised
public:
    int solve(vector<int> &price, int len, int n)
    {
        if (n == -1)
            return 0;
        int cut = 0;
        if (len >= n + 1)
            cut = price[n] + solve(price, len - (n + 1), n);
        int noCut = solve(price, len, n - 1);
        return max(cut, noCut);
    }

    int cutRod(vector<int> &price, int n)
    {
        return solve(price, n, n - 1);
    }
};

class Solution
{
    // BruteForce: Recursion
public:
    void solve(vector<int> &price, int &ans, int &cost, int &len, int n)
    {
        // base condition
        if (n == 0)
        {
            ans = max(cost, ans);
            return;
        }
        // Cut
        if (len >= n)
        {
            cost += price[n - 1];
            len -= n;
            solve(price, ans, cost, len, n);
            len += n;
            cost -= price[n - 1];
        }
        // Dont cut
        solve(price, ans, cost, len, n - 1);
    }

    int cutRod(vector<int> &price, int n)
    {
        int ans = INT_MIN, cost = 0;
        solve(price, ans, cost, n, n);
        return ans;
    }
};

int main()
{
    int n = 6;
    vector<int> price = {3, 5, 6, 7, 10, 12};

    Solution3 Obj1;
    cout << Obj1.cutRod(price, n);

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    return 0;
}