Largest number upto T formed by combination of As and Bs

Given numbers T, A, and B, the task is to find the greatest number less than equal to T formed by the sum of any combination of A and B (A and B can be used any number of times) along with one special operation that is at any instance sum can be halved (integer division by 2) by its current value at most one time.

Examples:

Input: T = 8, A = 5, B = 6 
Output: 8
Explanation: There are two ways to reach the maximum value 8 

first way:

• Step 1: Choose 5 to add. the sum becomes 5.
• Step 2: Use special move and halve sum, sum becomes 2 (5÷2 == 2 as this is integer division).
• Step 3: Choose 6 to add, and the sum becomes 8.

second way: 

• Step 1: Choose 6 to add. the sum becomes 6.
• Step 2: Use a special move and halve sum, sum becomes 3 (6÷2 == 3).
• Step 3: Choose 5 to add, and the sum becomes 8.

8 is the maximum sum possible less than equal to T = 8

Input: T = 11, A = 5, B = 8
Output: 10

There is only one way to reach maximum value11

• Step 1: Choose 5 to add, Sum becomes 5.
• Step 2: Choose again 5 to add, Sum becomes 10.

10 is the maximum sum possible less than equal to T = 11

Naive approach: The basic way to solve the problem is as follows:

The basic way to solve this problem is to generate all possible combinations by using a recursive approach.

Time Complexity: O(2^N)
Auxiliary Space: O(1)

Efficient Approach:  The above approach can be optimized based on the following idea:

Dynamic programming can be used to solve this problem

• dp[i][j] represents maximum sum less than equal to T with j special operations left.
• recurrence relation: dp[i][j] = max(dp[i – A][j], dp[i – B][j]) = dp[i / 2][0]

it can be observed that the recursive function is called exponential times. That means that some states are called repeatedly. So the idea is to store the value of each state. This can be done using by store the value of a state and whenever the function is called, return the stored value without computing again.

Follow the steps below to solve the problem:

• Create a recursive function that takes i representing the current sum and j representing the number of special operations left to perform.
• Call recursive function for both choosing A into sum and choosing B into sum.
• Create a 2d array of dp[N][2] with initially filled with -1.
• If the answer for a particular state is computed then save it in dp[i][j].
• If the answer for a particular state is already computed then just return dp[i][j].

Below is the implementation of the above approach.

C++

// C++ code to implement the approach
#include <bits/stdc++.h>
using namespace std;

// DP table initialized with -1
int dp[1000001][2];

// Recursive Function to calculate maximum
// sum less than equal to T can be achieved
// by adding numbers A or B repeatedly
int recur(int i, int j, int T, int A, int B)
{

    // If answer for current state is already
    // calculated then just return dp[i][j]
    if (dp[i][j] != -1)
        return dp[i][j];

    int ans = INT_MIN;

    // Calling recursive function for
    // adding A
    if (i + A <= T)
        ans = max(ans, recur(i + A, j, T, A, B));

    // Else return value i
    else
        ans = i;

    // Calling recursive function for
    // adding B
    if (i + B <= T)
        ans = max(ans, recur(i + B, j, T, A, B));

    // Else return value ans = max(ans, i)
    else
        ans = max(ans, i);

    // Calling recursive function for special
    // operation of halving current sum
    if (j == 1)
        ans = max(ans, recur(i / 2, 0, T, A, B));

    // Save and return dp value
    return dp[i][j] = ans;
}

// Function to calculate maximum sum
// less than equal to T can be achieved
// by adding numbers A or B repeatedly
void maxTarLessThanEqualToT(int T, int A, int B)
{

    // Initializing dp array with - 1
    memset(dp, -1, sizeof(dp));

    cout << recur(0, 1, T, A, B) << endl;
}

// Driver Code
int main()
{

    // Input 1
    int T = 8, A = 5, B = 6;

    // Function Call
    maxTarLessThanEqualToT(T, A, B);

    // Input 2
    int T1 = 11, A1 = 5, B1 = 8;

    // Function Call
    maxTarLessThanEqualToT(T1, A1, B1);
    return 0;
}

Python3

# Python code to implement the approach

# DP table initialized with -1
dp = [[-1 for i in range(2)] for j in range(1000001)]

# Recursive Function to calculate maximum
# sum less than equal to T can be achieved
# by adding numbers A or B repeatedly
def recur(i, j, T, A, B):
    # If answer for current state is already
    # calculated then just return dp[i][j]
    if dp[i][j] != -1:
        return dp[i][j]
    ans = float('-inf')

    # Calling recursive function for
    # adding A
    if i + A <= T:
        ans = max(ans, recur(i + A, j, T, A, B))
    # Else return value i
    else:
        ans = i

    # Calling recursive function for
    # adding B
    if i + B <= T:
        ans = max(ans, recur(i + B, j, T, A, B))
    # Else return value ans = max(ans, i)
    else:
        ans = max(ans, i)

    # Calling recursive function for special
    # operation of halving current sum
    if j == 1:
        ans = max(ans, recur(i // 2, 0, T, A, B))

    # Save and return dp value
    dp[i][j] = ans
    return ans

# Function to calculate maximum sum
# less than equal to T can be achieved
# by adding numbers A or B repeatedly
def maxTarLessThanEqualToT(T, A, B):
    # Initializing dp array with - 1
    for i in range(1000001):
        for j in range(2):
            dp[i][j] = -1
    print(recur(0, 1, T, A, B))

# Driver Code
if __name__ == '__main__':
    # Input 1
    T = 8
    A = 5
    B = 6

    # Function Call
    maxTarLessThanEqualToT(T, A, B)

    # Input 2
    T1 = 11
    A1 = 5
    B1 = 8

    # Function Call
    maxTarLessThanEqualToT(T1, A1, B1)

# This code is contributed by shivamsharma215

C#

// C# code to implement the approach

using System;
using System.Linq;
using System.Collections.Generic;

class GFG
{

    // DP table initialized with -1
    static int[,] dp=new int[1000001,2];

    // Recursive Function to calculate maximum
    // sum less than equal to T can be achieved
    // by adding numbers A or B repeatedly
    static int recur(int i, int j, int T, int A, int B)
    {

        // If answer for current state is already
        // calculated then just return dp[i][j]
        if (dp[i,j] != -1)
            return dp[i,j];

        int ans = Int32.MinValue;

        // Calling recursive function for
        // adding A
        if (i + A <= T)
            ans = Math.Max(ans, recur(i + A, j, T, A, B));

        // Else return value i
        else
            ans = i;

        // Calling recursive function for
        // adding B
        if (i + B <= T)
            ans = Math.Max(ans, recur(i + B, j, T, A, B));

        // Else return value ans = max(ans, i)
        else
            ans = Math.Max(ans, i);

        // Calling recursive function for special
        // operation of halving current sum
        if (j == 1)
            ans = Math.Max(ans, recur(i / 2, 0, T, A, B));

        // Save and return dp value
        return dp[i,j] = ans;
    }

    // Function to calculate maximum sum
    // less than equal to T can be achieved
    // by adding numbers A or B repeatedly
    static void maxTarLessThanEqualToT(int T, int A, int B)
    {

        // Initializing dp array with - 1
        for(int i=0; i<1000001; i++)
            for(int j=0; j<2; j++)
                dp[i,j]=-1;

        Console.WriteLine(recur(0, 1, T, A, B));
    }

    // Driver Code
    static public void Main()
    {

        // Input 1
        int T = 8, A = 5, B = 6;

        // Function Call
        maxTarLessThanEqualToT(T, A, B);

        // Input 2
        int T1 = 11, A1 = 5, B1 = 8;

        // Function Call
        maxTarLessThanEqualToT(T1, A1, B1);
    }
}

Javascript

// Javascript code to implement the approach

// DP table initialized with -1
let dp=new Array(1000001);
for(let i=0;i<1000001; i++)
    dp[i]=new Array(2);

// Recursive Function to calculate maximum
// sum less than equal to T can be achieved
// by adding numbers A or B repeatedly
function recur(i, j, T, A, B)
{

    // If answer for current state is already
    // calculated then just return dp[i][j]
    if (dp[i][j] != -1)
        return dp[i][j];

    let ans = Number.MIN_SAFE_INTEGER;

    // Calling recursive function for
    // adding A
    if (i + A <= T)
        ans = Math.max(ans, recur(i + A, j, T, A, B));

    // Else return value i
    else
        ans = i;

    // Calling recursive function for
    // adding B
    if (i + B <= T)
        ans = Math.max(ans, recur(i + B, j, T, A, B));

    // Else return value ans = max(ans, i)
    else
        ans = Math.max(ans, i);

    // Calling recursive function for special
    // operation of halving current sum
    if (j == 1)
        ans = Math.max(ans, recur(Math.floor(i / 2), 0, T, A, B));

    // Save and return dp value
    return dp[i][j] = ans;
}

// Function to calculate maximum sum
// less than equal to T can be achieved
// by adding numbers A or B repeatedly
function maxTarLessThanEqualToT(T, A, B)
{
    // Initializing dp array with - 1
    for(let i=0; i<1000001; i++)
    {
        for(let j=0; j<2; j++)
            dp[i][j]=-1;
    }
    console.log(recur(0, 1, T, A, B));
}

// Driver Code

// Input 1
let T = 8, A = 5, B = 6;

// Function Call
maxTarLessThanEqualToT(T, A, B);

// Input 2
let T1 = 11, A1 = 5, B1 = 8;

// Function Call
maxTarLessThanEqualToT(T1, A1, B1);

Java

// Java code to implement the approach

import java.io.*;

class GFG {

    // DP table initialized with -1
    static int[][] dp = new int[1000001][2];

    // Recursive Function to calculate maximum
    // sum less than equal to T can be achieved
    // by adding numbers A or B repeatedly
    static int recur(int i, int j, int T, int A, int B)
    {

        // If answer for current state is already
        // calculated then just return dp[i][j]
        if (dp[i][j] != -1)
            return dp[i][j];

        int ans = Integer.MIN_VALUE;

        // Calling recursive function for
        // adding A
        if (i + A <= T)
            ans = Math.max(ans, recur(i + A, j, T, A, B));

        // Else return value i
        else
            ans = i;

        // Calling recursive function for
        // adding B
        if (i + B <= T)
            ans = Math.max(ans, recur(i + B, j, T, A, B));

        // Else return value ans = max(ans, i)
        else
            ans = Math.max(ans, i);

        // Calling recursive function for special
        // operation of halving current sum
        if (j == 1)
            ans = Math.max(ans, recur(i / 2, 0, T, A, B));

        // Save and return dp value
        return dp[i][j] = ans;
    }

    // Function to calculate maximum sum
    // less than equal to T can be achieved
    // by adding numbers A or B repeatedly
    static void maxTarLessThanEqualToT(int T, int A, int B)
    {

        // Initializing dp array with - 1
        for(int i=0; i<1000001; i++)
            for(int j=0; j<2; j++)
                dp[i][j]=-1;

        System.out.println(recur(0, 1, T, A, B));
    }

    // Driver Code
    public static void main (String[] args) {
        // Input 1
        int T = 8, A = 5, B = 6;

        // Function Call
        maxTarLessThanEqualToT(T, A, B);

        // Input 2
        int T1 = 11, A1 = 5, B1 = 8;

        // Function Call
        maxTarLessThanEqualToT(T1, A1, B1);
    }
}

Output

8
10

Time Complexity: O(N)
Auxiliary Space: O(N)