A person wants to transfer bananas over to a destination A km away. He initially has B bananas and a camel. The camel cannot carry more than C bananas at a time and eats a banana every km it travels. Given three integers A, B, and C, the task is to find the maximum number of bananas the person can transfer to the destination using the camel.
Note: The given problem is a generalized version of the famous Camel-Banana puzzle.
Examples:
Input: A = 10, B = 30, C = 10
Output: 5Input: A = 1000, B = 3000, C = 1000
Output: 533
Approach: The given problem can be solved with the help of Dynamic Programming using Memoization using the following key points:
- It can be observed that the most effective way to transfer bananas is to divide the path (u, v) having A km into some smaller parts. Suppose x is a breakpoint in the path (u, v). The optimal choice is to transfer all the bananas from u to x and then from x to v.
- There can be any number of breakpoints in the path (u, v) such that the count of breakpoints < A.
- The total number of trips the camel which can carry C bananas at a time has to make in order to transfer X bananas over any distance can be calculated by the formula 2 * X / C – 1, if C is a factor of X (i.e, X % C = 0) otherwise 2 * X / C +1.
Using the above observations, the given problem can be solved by following the below steps:
- Consider a 2D array dp[][], where a state dp[A][B] represents the maximum number of bananas a camel can transfer over a distance of A km having B bananas initially. Initialize the dp[][] array with -1.
- Create a recursive function to iterate over the given path of A km and create a breakpoint at each valid index and recursively call the function for the remaining path.
- Memoize the maximum number of bananas for each state and return the memoized value if the current state is already calculated.
Below is the implementation of the above approach:
C++
// C++ program of the above approach#include <bits/stdc++.h>using namespace std;
// Stores the overlapping stateint dp[1001][3001];
// Recursive function to find the maximum// number of bananas that can be transferred// to A distance using memoizationint recBananaCnt(int A, int B, int C)
{ // Base Case where count of bananas
// is less that the given distance
if (B <= A) {
return 0;
}
// Base Case where count of bananas
// is less that camel's capacity
if (B <= C) {
return B - A;
}
// Base Case where distance = 0
if (A == 0) {
return B;
}
// If the current state is already
// calculated
if (dp[A][B] != -1) {
return dp[A][B];
}
// Stores the maximum count of bananas
int maxCount = INT_MIN;
// Stores the number of trips to transfer
// B bananas using a camel of capacity C
int tripCount = B % C == 0 ? ((2 * B) / C) - 1
: ((2 * B) / C) + 1;
// Loop to iterate over all the
// breakpoints in range [1, A]
for (int i = 1; i <= A; i++) {
// Recursive call over the
// remaining path
int curCount
= recBananaCnt(A - i,
B - tripCount * i, C);
// Update the maxCount
if (curCount > maxCount) {
maxCount = curCount;
// Memoize the current value
dp[A][B] = maxCount;
}
}
// Return answer
return maxCount;
}// Function to find the maximum number of// bananas that can be transferredint maxBananaCnt(int A, int B, int C)
{ // Initialize dp array with -1
memset(dp, -1, sizeof(dp));
// Function Call
return recBananaCnt(A, B, C);
}// Driver Codeint main()
{ int A = 1000;
int B = 3000;
int C = 1000;
cout << maxBananaCnt(A, B, C);
return 0;
} |
Java
// Java program of the above approachpublic class GFG {
// Stores the overlapping state
final static int dp[][] = new int[1001][3001];
// Recursive function to find the maximum
// number of bananas that can be transferred
// to A distance using memoization
static int recBananaCnt(int A, int B, int C)
{
// Base Case where count of bananas
// is less that the given distance
if (B <= A) {
return 0;
}
// Base Case where count of bananas
// is less that camel's capacity
if (B <= C) {
return B - A;
}
// Base Case where distance = 0
if (A == 0) {
return B;
}
// If the current state is already
// calculated
if (dp[A][B] != -1) {
return dp[A][B];
}
// Stores the maximum count of bananas
int maxCount = Integer.MIN_VALUE;
// Stores the number of trips to transfer
// B bananas using a camel of capacity C
int tripCount = B % C == 0 ? ((2 * B) / C) - 1 : ((2 * B) / C) + 1;
// Loop to iterate over all the
// breakpoints in range [1, A]
for (int i = 1; i <= A; i++) {
// Recursive call over the
// remaining path
int curCount
= recBananaCnt(A - i,
B - tripCount * i, C);
// Update the maxCount
if (curCount > maxCount) {
maxCount = curCount;
// Memoize the current value
dp[A][B] = maxCount;
}
}
// Return answer
return maxCount;
}
// Function to find the maximum number of
// bananas that can be transferred
static int maxBananaCnt(int A, int B, int C)
{
// Initialize dp array with -1
for(int i = 0; i < 1001; i++)
for (int j = 0; j < 3001; j++)
dp[i][j] = -1;
// Function Call
return recBananaCnt(A, B, C);
}
// Driver Code
public static void main (String[] args) {
int A = 1000;
int B = 3000;
int C = 1000;
System.out.println(maxBananaCnt(A, B, C));
}
}// This code is contributed by AnkThon |
Python3
# Python program of the above approach# Stores the overlapping statedp = [[-1 for i in range(3001)] for j in range(1001)]
# Recursive function to find the maximum# number of bananas that can be transferred# to A distance using memoizationdef recBananaCnt(A, B, C):
# Base Case where count of bananas
# is less that the given distance
if (B <= A):
return 0
# Base Case where count of bananas
# is less that camel's capacity
if (B <= C):
return B - A
# Base Case where distance = 0
if (A == 0):
return B
# If the current state is already
# calculated
if (dp[A][B] != -1):
return dp[A][B]
# Stores the maximum count of bananas
maxCount = -2**32
# Stores the number of trips to transfer
# B bananas using a camel of capacity C
tripCount = ((2 * B) // C) - 1 if(B % C == 0 ) else ((2 * B) // C) + 1
# Loop to iterate over all the
# breakpoints in range [1, A]
for i in range(1,A+1):
# Recursive call over the
# remaining path
curCount = recBananaCnt(A - i, B - tripCount * i, C)
# Update the maxCount
if (curCount > maxCount):
maxCount = curCount
# Memoize the current value
dp[A][B] = maxCount
# Return answer
return maxCount
# Function to find the maximum number of# bananas that can be transferreddef maxBananaCnt(A, B, C):
# Function Call
return recBananaCnt(A, B, C)
# Driver CodeA = 1000
B = 3000
C = 1000
print(maxBananaCnt(A, B, C))
# This code is contributed by shivanisinghss2110 |
C#
// C# program of the above approachusing System;
public class GFG {
// Stores the overlapping state
static int[, ] dp = new int[1001, 3001];
// Recursive function to find the maximum
// number of bananas that can be transferred
// to A distance using memoization
static int recBananaCnt(int A, int B, int C)
{
// Base Case where count of bananas
// is less that the given distance
if (B <= A) {
return 0;
}
// Base Case where count of bananas
// is less that camel's capacity
if (B <= C) {
return B - A;
}
// Base Case where distance = 0
if (A == 0) {
return B;
}
// If the current state is already
// calculated
if (dp[A, B] != -1) {
return dp[A, B];
}
// Stores the maximum count of bananas
int maxCount = Int32.MinValue;
// Stores the number of trips to transfer
// B bananas using a camel of capacity C
int tripCount = B % C == 0 ? ((2 * B) / C) - 1
: ((2 * B) / C) + 1;
// Loop to iterate over all the
// breakpoints in range [1, A]
for (int i = 1; i <= A; i++) {
// Recursive call over the
// remaining path
int curCount
= recBananaCnt(A - i, B - tripCount * i, C);
// Update the maxCount
if (curCount > maxCount) {
maxCount = curCount;
// Memoize the current value
dp[A, B] = maxCount;
}
}
// Return answer
return maxCount;
}
// Function to find the maximum number of
// bananas that can be transferred
static int maxBananaCnt(int A, int B, int C)
{
// Initialize dp array with -1
for (int i = 0; i < 1001; i++)
for (int j = 0; j < 3001; j++)
dp[i, j] = -1;
// Function Call
return recBananaCnt(A, B, C);
}
// Driver Code
public static void Main(string[] args)
{
int A = 1000;
int B = 3000;
int C = 1000;
Console.WriteLine(maxBananaCnt(A, B, C));
}
}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript Program to implement
// the above approach
// Stores the overlapping state
// Initialize dp array with -1
let dp = new Array(1001);
for (let i = 0; i < dp.length; i++)
{
dp[i] = (new Array(3001).fill(-1))
}
// Recursive function to find the maximum
// number of bananas that can be transferred
// to A distance using memoization
function recBananaCnt(A, B, C)
{
// Base Case where count of bananas
// is less that the given distance
if (B <= A) {
return 0;
}
// Base Case where count of bananas
// is less that camel's capacity
if (B <= C) {
return B - A;
}
// Base Case where distance = 0
if (A == 0) {
return B;
}
// If the current state is already
// calculated
if (dp[A][B] != -1) {
return dp[A][B];
}
// Stores the maximum count of bananas
let maxCount = Number.MIN_VALUE;
// Stores the number of trips to transfer
// B bananas using a camel of capacity C
let tripCount = B % C == 0 ? Math.floor((2 * B) / C) - 1
: Math.floor((2 * B) / C) + 1;
// Loop to iterate over all the
// breakpoints in range [1, A]
for (let i = 1; i <= A; i++) {
// Recursive call over the
// remaining path
let curCount
= recBananaCnt(A - i,
B - tripCount * i, C);
// Update the maxCount
if (curCount > maxCount) {
maxCount = curCount;
// Memoize the current value
dp[A][B] = maxCount;
}
}
// Return answer
return maxCount;
}
// Function to find the maximum number of
// bananas that can be transferred
function maxBananaCnt(A, B, C) {
// Function Call
return recBananaCnt(A, B, C);
}
// Driver Code
let A = 1000;
let B = 3000;
let C = 1000;
document.write(maxBananaCnt(A, B, C));
// This code is contributed by Potta Lokesh
</script>
|
Output:
533
Time Complexity: O(A*A*B)
Auxiliary Space: O(A*B)