Maximize the number of boxes using red and yellow balls

We are given R red balls and Y yellow balls to fill in boxes. We can fill a box using one of the two ways:

• Using A red balls and B yellow balls
• Using B red balls and A yellow balls.

Maximize the number of boxes we can fill.

Note: One ball can belong to only one box

Examples:

Input: R = 10, Y = 12, A = 2, B = 5
Output: 3
Explanation: First and Second box will have 2 Red balls and 5 Yellow Balls. Third box will have 5 Red Balls and 2 Yellow Balls

Input: R = 1, Y = 1, A = 2, B = 2
Output: 0
Explanation: 0 box can be filled.

Approach: To solve the problem, follow the below idea:

The problem can be solved by using Ternary Search on the number of times we use the first way to fill a box, that is using A red balls and B yellow balls. We can use ternary search because if we keep on increasing the number of times to fill the box using first way, then the number of ways to fill the box increases initially, reaches the maximum point and then starts decreasing. Therefore, the number of ways to fill the box behaves as a Unimodal function and therefore we can use Ternary Search on it. Let’s say we filled X boxes using A red balls and B yellow balls, so for the remaining (R – A * X) red balls and (Y – B * Y) yellow balls, we can only use the second way to fill the boxes. Now, find the total number of boxes by summing up boxes of both types.

Step-by-step algorithm:

• Declare a function, say getBoxes(R, Y, A, B, mid) to return the number of boxes if we use the first way mid times to fill the boxes.
• Now, declare the range of the number of times we can use the first way to fill the number of boxes.
• Apply ternary search and reduce the search space.
• Finally, iterate over all every value between lo and hi to get the maximum number of boxes one can fill.

Below is the implementation of the algorithm:

C++

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

// function to find the total number of boxes if we use the
// first way mid times to fill boxes and use the second way
// to fill the remaining boxes
int getBoxes(int R, int Y, int A, int B, int mid)
{

    // Number of boxes filled using first way
    int firstWay = mid;

    // Remaining red balls after filling boxes using first
    // way
    int remainingRedBalls = (R - (A * mid));

    // Remaining Yellow balls after filling boxes using
    // second way
    int remainingYellowBalls = (Y - (B * mid));

    // Number of boxes filled using second way
    int secondWay = min(remainingRedBalls / B,
                        remainingYellowBalls / A);

    // returning the total number of boxes filled
    return firstWay + secondWay;
}

// Function to find the maximum number of boxes
int solve(int R, int Y, int A, int B)
{

    // range for the number of times we have used type 1 way
    // to fill the boxes
    int lo = 0, hi = min(R / A, Y / B);

    // ternary search
    while (hi - lo > 2) {
        int mid1 = lo + (hi - lo) / 3;
        int mid2 = hi - (hi - lo) / 3;
        int res1 = getBoxes(R, Y, A, B, mid1);
        int res2 = getBoxes(R, Y, A, B, mid2);
        if (res1 == res2) {
            lo = mid1;
            hi = mid2;
        }
        else if (res1 < res2) {
            lo = mid1;
        }
        else {
            hi = mid2;
        }
    }
    int res = 0;
    while (lo <= hi) {
        res = max(res, getBoxes(R, Y, A, B, lo));
        lo += 1;
    }
    return res;
}

int main()
{

    // Sample Input
    int R = 10, Y = 12, A = 2, B = 5;

    cout << solve(R, Y, A, B) << "\n";
}

Java

class Main {

    // Function to find the total number of boxes
    // using the first way mid times to fill boxes
    // and using the second way to fill the remaining boxes
    static int getBoxes(int R, int Y, int A, int B, int mid) {

        // Number of boxes filled using the first way
        int firstWay = mid;

        // Remaining red balls after filling boxes using the first way
        int remainingRedBalls = (R - (A * mid));

        // Remaining yellow balls after filling boxes using the second way
        int remainingYellowBalls = (Y - (B * mid));

        // Number of boxes filled using the second way
        int secondWay = Math.min(remainingRedBalls / B,
                                 remainingYellowBalls / A);

        // Returning the total number of boxes filled
        return firstWay + secondWay;
    }

    // Function to find the maximum number of boxes
    static int solve(int R, int Y, int A, int B) {

        // Range for the number of times we have used the first way
        // to fill the boxes
        int lo = 0, hi = Math.min(R / A, Y / B);

        // Ternary search
        while (hi - lo > 2) {
            int mid1 = lo + (hi - lo) / 3;
            int mid2 = hi - (hi - lo) / 3;
            int res1 = getBoxes(R, Y, A, B, mid1);
            int res2 = getBoxes(R, Y, A, B, mid2);
            if (res1 == res2) {
                lo = mid1;
                hi = mid2;
            } else if (res1 < res2) {
                lo = mid1;
            } else {
                hi = mid2;
            }
        }
        int res = 0;
        while (lo <= hi) {
            res = Math.max(res, getBoxes(R, Y, A, B, lo));
            lo += 1;
        }
        return res;
    }

    public static void main(String[] args) {

        // Sample Input
        int R = 10, Y = 12, A = 2, B = 5;

        System.out.println(solve(R, Y, A, B));
    }
}


// This code is contributed by rambabuguphka

Python3

# Function to find the total number of boxes if we use the
# first way mid times to fill boxes and use the second way
# to fill the remaining boxes
def get_boxes(R, Y, A, B, mid):
    # Number of boxes filled using first way
    first_way = mid

    # Remaining red balls after filling boxes using first way
    remaining_red_balls = R - (A * mid)

    # Remaining Yellow balls after filling boxes using
    # second way
    remaining_yellow_balls = Y - (B * mid)

    # Number of boxes filled using second way
    second_way = min(remaining_red_balls // B,
                     remaining_yellow_balls // A)

    # Returning the total number of boxes filled
    return first_way + second_way

# Function to find the maximum number of boxes
def solve(R, Y, A, B):
    # Range for the number of times we have used type 1 way
    # to fill the boxes
    lo, hi = 0, min(R // A, Y // B)

    # Ternary search
    while hi - lo > 2:
        mid1 = lo + (hi - lo) // 3
        mid2 = hi - (hi - lo) // 3
        res1 = get_boxes(R, Y, A, B, mid1)
        res2 = get_boxes(R, Y, A, B, mid2)
        if res1 == res2:
            lo = mid1
            hi = mid2
        elif res1 < res2:
            lo = mid1
        else:
            hi = mid2

    res = 0
    while lo <= hi:
        res = max(res, get_boxes(R, Y, A, B, lo))
        lo += 1
    return res

if __name__ == "__main__":
    # Sample Input
    R, Y, A, B = 10, 12, 2, 5

    # Calling the function and printing the result
    print(solve(R, Y, A, B))

C#

using System;

public class Program
{
    // function to find the total number of boxes if we use the
    // first way mid times to fill boxes and use the second way
    // to fill the remaining boxes
    static int GetBoxes(int R, int Y, int A, int B, int mid)
    {
        // Number of boxes filled using first way
        int firstWay = mid;

        // Remaining red balls after filling boxes using first way
        int remainingRedBalls = R - (A * mid);

        // Remaining Yellow balls after filling boxes using second way
        int remainingYellowBalls = Y - (B * mid);

        // Number of boxes filled using second way
        int secondWay = Math.Min(remainingRedBalls / B, remainingYellowBalls / A);

        // returning the total number of boxes filled
        return firstWay + secondWay;
    }

    // Function to find the maximum number of boxes
    static int Solve(int R, int Y, int A, int B)
    {
        // range for the number of times we have used type 1 way
        // to fill the boxes
        int lo = 0, hi = Math.Min(R / A, Y / B);

        // ternary search
        while (hi - lo > 2)
        {
            int mid1 = lo + (hi - lo) / 3;
            int mid2 = hi - (hi - lo) / 3;
            int res1 = GetBoxes(R, Y, A, B, mid1);
            int res2 = GetBoxes(R, Y, A, B, mid2);
            if (res1 == res2)
            {
                lo = mid1;
                hi = mid2;
            }
            else if (res1 < res2)
            {
                lo = mid1;
            }
            else
            {
                hi = mid2;
            }
        }
        int res = 0;
        while (lo <= hi)
        {
            res = Math.Max(res, GetBoxes(R, Y, A, B, lo));
            lo += 1;
        }
        return res;
    }

    public static void Main(string[] args)
    {
        // Sample Input
        int R = 10, Y = 12, A = 2, B = 5;

        Console.WriteLine(Solve(R, Y, A, B));
    }
}

Javascript

// Function to find the total number of boxes if we use the
// first way mid times to fill boxes and use the second way
// to fill the remaining boxes
function getBoxes(R, Y, A, B, mid) {
    // Number of boxes filled using first way
    let firstWay = mid;

    // Remaining red balls after filling boxes using first way
    let remainingRedBalls = R - (A * mid);

    // Remaining Yellow balls after filling boxes using second way
    let remainingYellowBalls = Y - (B * mid);

    // Number of boxes filled using second way
    let secondWay = Math.min(Math.floor(remainingRedBalls / B), Math.floor(remainingYellowBalls / A));

    // Returning the total number of boxes filled
    return firstWay + secondWay;
}

// Function to find the maximum number of boxes
function solve(R, Y, A, B) {
    // Range for the number of times we have used type 1 way to fill the boxes
    let lo = 0, hi = Math.min(Math.floor(R / A), Math.floor(Y / B));

    // Ternary search
    while (hi - lo > 2) {
        let mid1 = lo + Math.floor((hi - lo) / 3);
        let mid2 = hi - Math.floor((hi - lo) / 3);
        let res1 = getBoxes(R, Y, A, B, mid1);
        let res2 = getBoxes(R, Y, A, B, mid2);
        if (res1 === res2) {
            lo = mid1;
            hi = mid2;
        } else if (res1 < res2) {
            lo = mid1;
        } else {
            hi = mid2;
        }
    }

    let res = 0;
    while (lo <= hi) {
        res = Math.max(res, getBoxes(R, Y, A, B, lo));
        lo += 1;
    }
    return res;
}

// Sample Input
let R = 10, Y = 12, A = 2, B = 5;
console.log(solve(R, Y, A, B));

Output

3

Time Complexity: O(2 * log₃(min(R/A, Y/B)), where R is the total number of Red Balls, Y is the total number of Yellow balls and A and B are the number of balls to fill a box.
Auxiliary Space: O(1)