Choice of Area

Last Updated : 20 Jan, 2023

Consider a game, in which you have two types of powers, A and B and there are 3 types of Areas X, Y and Z. Every second you have to switch between these areas, and each area has specific properties by which your power A and power B increase or decrease. We need to keep choosing areas in such a way that our survival time is maximized. Survival time ends when any of the powers, A or B reaches less than 0.

Examples:

Initial value of Power A = 20        
Initial value of Power B = 8

Area X (3, 2) : If you step into Area X, 
                A increases by 3, 
                B increases by 2

Area Y (-5, -10) : If you step into Area Y, 
                   A decreases by 5, 
                   B decreases by 10

Area Z (-20, 5) : If you step into Area Z, 
                  A decreases by 20, 
                  B increases by 5

It is possible to choose any area in our first step.
We can survive at max 5 unit of time by following 
these choice of areas :
X -> Z -> X -> Y -> X

This problem can be solved using recursion, after each time unit we can go to any of the area but we will choose that area which ultimately leads to maximum survival time. As recursion can lead to solving same subproblem many time, we will memoize the result on basis of power A and B, if we reach to same pair of power A and B, we won’t solve it again instead we will take the previously calculated result.

Given below is the simple implementation of above approach.

CPP

//  C++ code to get maximum survival time
#include <bits/stdc++.h>
using namespace std;

//  structure to represent an area
struct area
{
    //  increment or decrement in A and B
    int a, b;
    area(int a, int b) : a(a), b(b)
    {}
};

//  Utility method to get maximum of 3 integers
int max(int a, int b, int c)
{
    return max(a, max(b, c));
}

//  Utility method to get maximum survival time
int maxSurvival(int A, int B, area X, area Y, area Z,
                int last, map<pair<int, int>, int>& memo)
{
    //  if any of A or B is less than 0, return 0
    if (A <= 0 || B <= 0)
        return 0;
    pair<int, int> cur = make_pair(A, B);

    //  if already calculated, return calculated value
    if (memo.find(cur) != memo.end())
        return memo[cur];

    int temp;

    //  step to areas on basis of last choose area
    switch(last)
    {
    case 1:
        temp = 1 + max(maxSurvival(A + Y.a, B + Y.b,
                                   X, Y, Z, 2, memo),
                       maxSurvival(A + Z.a, B + Z.b,
                                  X, Y, Z, 3, memo));
        break;
    case 2:
        temp = 1 + max(maxSurvival(A + X.a, B + X.b,
                                  X, Y, Z, 1, memo),
                       maxSurvival(A + Z.a, B + Z.b,
                                  X, Y, Z, 3, memo));
        break;
    case 3:
        temp = 1 + max(maxSurvival(A + X.a, B + X.b,
                                  X, Y, Z, 1, memo),
                       maxSurvival(A + Y.a, B + Y.b,
                                  X, Y, Z, 2, memo));
        break;
    }

    //  store the result into map
    memo[cur] = temp;

    return temp;
}

//  method returns maximum survival time
int getMaxSurvivalTime(int A, int B, area X, area Y, area Z)
{
    if (A <= 0 || B <= 0)
        return 0;
    map< pair<int, int>, int > memo;

    //  At first, we can step into any of the area
    return
        max(maxSurvival(A + X.a, B + X.b, X, Y, Z, 1, memo),
            maxSurvival(A + Y.a, B + Y.b, X, Y, Z, 2, memo),
            maxSurvival(A + Z.a, B + Z.b, X, Y, Z, 3, memo));
}

//  Driver code to test above method
int main()
{
    area X(3, 2);
    area Y(-5, -10);
    area Z(-20, 5);

    int A = 20;
    int B = 8;
    cout << getMaxSurvivalTime(A, B, X, Y, Z);

    return 0;
}

Java

/*package whatever //do not write package name here */
import java.util.*;
import java.io.*;

class GFG {
  // Java code to get maximum survival time

  // class to represent an area
  static class area
  {
    // increment or decrement in A and B
    public int a, b;
    public area(int a, int b){
      this.a = a;
      this.b = b;
    }
  };

  // class to represent pair
  static class Pair{
    public int first,second;
    public Pair(int first,int second){
      this.first = first;
      this.second = second;
    }
  }


  // Utility method to get maximum of 3 integers
  static int max(int a, int b, int c)
  {
    return Math.max(a, Math.max(b, c));
  }

  // Utility method to get maximum survival time
  static int maxSurvival(int A, int B, area X, area Y, area Z,
                         int last, HashMap<Pair, Integer> memo)
  {
    // if any of A or B is less than 0, return 0
    if (A <= 0 || B <= 0)
      return 0;
    Pair cur = new Pair(A, B);

    // if already calculated, return calculated value
    if (memo.containsKey(cur))
      return memo.get(cur);

    int temp = 0;

    // step to areas on basis of last choose area
    switch(last)
    {
      case 1:
        temp = 1 + Math.max(maxSurvival(A + Y.a, B + Y.b,
                                        X, Y, Z, 2, memo),
                            maxSurvival(A + Z.a, B + Z.b,
                                        X, Y, Z, 3, memo));
        break;
      case 2:
        temp = 1 + Math.max(maxSurvival(A + X.a, B + X.b,
                                        X, Y, Z, 1, memo),
                            maxSurvival(A + Z.a, B + Z.b,
                                        X, Y, Z, 3, memo));
        break;
      case 3:
        temp = 1 + Math.max(maxSurvival(A + X.a, B + X.b,
                                        X, Y, Z, 1, memo),
                            maxSurvival(A + Y.a, B + Y.b,
                                        X, Y, Z, 2, memo));
        break;
    }

    // store the result into map
    memo.put(cur,temp);

    return temp;
  }

  // method returns maximum survival time
  static int getMaxSurvivalTime(int A, int B, area X, area Y, area Z)
  {
    if (A <= 0 || B <= 0)
      return 0;
    HashMap<Pair,Integer> memo = new HashMap<>();

    // At first, we can step into any of the area
    return
      max(maxSurvival(A + X.a, B + X.b, X, Y, Z, 1, memo),
          maxSurvival(A + Y.a, B + Y.b, X, Y, Z, 2, memo),
          maxSurvival(A + Z.a, B + Z.b, X, Y, Z, 3, memo));
  }

  // Driver Code
  public static void main(String args[])
  {
    area X = new area(3, 2);
    area Y = new area(-5, -10);
    area Z = new area(-20, 5);

    int A = 20;
    int B = 8;
    System.out.println(getMaxSurvivalTime(A, B, X, Y, Z));
  }
}

// This code is contributed by shinjanpatra

Python3

# Python code to get maximum survival time

# Class to represent an area


class area:
    def __init__(self, a, b):
        self.a = a
        self.b = b

# Utility method to get maximum survival time


def maxSurvival(A, B, X, Y, Z, last, memo):
    # if any of A or B is less than 0, return 0
    if (A <= 0 or B <= 0):
        return 0
    cur = area(A, B)

    # if already calculated, return calculated value
    for ele in memo.keys():
        if (cur.a == ele.a and cur.b == ele.b):
            return memo[ele]

    # step to areas on basis of last chosen area
    if (last == 1):
        temp = 1 + max(maxSurvival(A + Y.a, B + Y.b,
                                   X, Y, Z, 2, memo),
                       maxSurvival(A + Z.a, B + Z.b,
                                   X, Y, Z, 3, memo))
    elif (last == 2):
        temp = 1 + max(maxSurvival(A + X.a, B + X.b,
                                   X, Y, Z, 1, memo),
                       maxSurvival(A + Z.a, B + Z.b,
                                   X, Y, Z, 3, memo))
    elif (last == 3):
        temp = 1 + max(maxSurvival(A + X.a, B + X.b,
                                   X, Y, Z, 1, memo),
                       maxSurvival(A + Y.a, B + Y.b,
                                   X, Y, Z, 2, memo))

    # store the result into map
    memo[cur] = temp

    return temp

# method returns maximum survival time


def getMaxSurvivalTime(A, B, X, Y, Z):
    if (A <= 0 or B <= 0):
        return 0
    memo = dict()

    # At first, we can step into any of the area
    return max(maxSurvival(A + X.a, B + X.b, X, Y, Z, 1, memo),
               maxSurvival(A + Y.a, B + Y.b, X, Y, Z, 2, memo),
               maxSurvival(A + Z.a, B + Z.b, X, Y, Z, 3, memo))


# Driver code to test above method
X = area(3, 2)
Y = area(-5, -10)
Z = area(-20, 5)

A = 20
B = 8
print(getMaxSurvivalTime(A, B, X, Y, Z))

# This code is contributed by Soumen Ghosh.

// C# code to get maximum survival time
using System;
using System.Collections.Generic;

class GFG {

  // class to represent an area
  class area {
    // increment or decrement in A and B
    public int a, b;
    public area(int a, int b)
    {
      this.a = a;
      this.b = b;
    }
  };

  // class to represent pair
  class Pair {
    public int first, second;
    public Pair(int first, int second)
    {
      this.first = first;
      this.second = second;
    }
  }

  // Utility method to get maximum of 3 integers
  static int max(int a, int b, int c)
  {
    return Math.Max(a, Math.Max(b, c));
  }

  // Utility method to get maximum survival time
  static int maxSurvival(int A, int B, area X, area Y,
                         area Z, int last,
                         Dictionary<Pair, int> memo)
  {
    // if any of A or B is less than 0, return 0
    if (A <= 0 || B <= 0)
      return 0;
    Pair cur = new Pair(A, B);

    // if already calculated, return calculated value
    if (memo.ContainsKey(cur))
      return memo[cur];

    int temp = 0;

    // step to areas on basis of last choose area
    switch (last) {
      case 1:
        temp
          = 1
          + Math.Max(maxSurvival(A + Y.a, B + Y.b,
                                 X, Y, Z, 2, memo),
                     maxSurvival(A + Z.a, B + Z.b,
                                 X, Y, Z, 3, memo));
        break;
      case 2:
        temp
          = 1
          + Math.Max(maxSurvival(A + X.a, B + X.b,
                                 X, Y, Z, 1, memo),
                     maxSurvival(A + Z.a, B + Z.b,
                                 X, Y, Z, 3, memo));
        break;
      case 3:
        temp
          = 1
          + Math.Max(maxSurvival(A + X.a, B + X.b,
                                 X, Y, Z, 1, memo),
                     maxSurvival(A + Y.a, B + Y.b,
                                 X, Y, Z, 2, memo));
        break;
    }

    // store the result into map
    memo[cur] = temp;

    return temp;
  }

  // method returns maximum survival time
  static int getMaxSurvivalTime(int A, int B, area X,
                                area Y, area Z)
  {
    if (A <= 0 || B <= 0)
      return 0;
    Dictionary<Pair, int> memo
      = new Dictionary<Pair, int>();

    // At first, we can step into any of the area
    return max(
      maxSurvival(A + X.a, B + X.b, X, Y, Z, 1, memo),
      maxSurvival(A + Y.a, B + Y.b, X, Y, Z, 2, memo),
      maxSurvival(A + Z.a, B + Z.b, X, Y, Z, 3,
                  memo));
  }

  // Driver Code
  public static void Main(String[] args)
  {
    area X = new area(3, 2);
    area Y = new area(-5, -10);
    area Z = new area(-20, 5);

    int A = 20;
    int B = 8;
    Console.WriteLine(
      getMaxSurvivalTime(A, B, X, Y, Z));
  }
}

// This code is contributed by lokeshpotta20.

JavaScript

<script>

// JavaScript code to get maximum survival time

// Class to represent an area
class area{
    constructor(a, b){
        this.a = a
        this.b = b
    }
}

// Utility method to get maximum survival time
function maxSurvival(A, B, X, Y, Z, last, memo){
    // if any of A or B is less than 0, return 0
    if (A <= 0 || B <= 0)
        return 0
    let cur = new area(A, B)

    // if already calculated, return calculated value
    for(let [key,value] of memo){
        if (cur.a == key.a && cur.b == key.b)
            return memo.get(key)
    }

    let temp;

    // step to areas on basis of last chosen area
    if (last == 1){
        temp = 1 + Math.max(maxSurvival(A + Y.a, B + Y.b,
                                   X, Y, Z, 2, memo),
                       maxSurvival(A + Z.a, B + Z.b,
                                   X, Y, Z, 3, memo))
    }
    else if (last == 2){
        temp = 1 + Math.max(maxSurvival(A + X.a, B + X.b,
                                   X, Y, Z, 1, memo),
               maxSurvival(A + Z.a, B + Z.b,
                   X, Y, Z, 3, memo))
    }
    else if (last == 3){
        temp = 1 + Math.max(maxSurvival(A + X.a, B + X.b,
                   X, Y, Z, 1, memo),
               maxSurvival(A + Y.a, B + Y.b,
                   X, Y, Z, 2, memo))
    }

    // store the result into map
    memo.set(cur , temp)

    return temp
}

// method returns maximum survival time
function getMaxSurvivalTime(A, B, X, Y, Z){
    if (A <= 0 || B <= 0)
        return 0
    let memo = new Map()

    // At first, we can step into any of the area
    return Math.max(maxSurvival(A + X.a, B + X.b, X, Y, Z, 1, memo),
           maxSurvival(A + Y.a, B + Y.b, X, Y, Z, 2, memo),
           maxSurvival(A + Z.a, B + Z.b, X, Y, Z, 3, memo))
}

// Driver code to test above method
let X = new area(3, 2)
let Y = new area(-5, -10)
let Z = new area(-20, 5)

let A = 20
let B = 8
document.write(getMaxSurvivalTime(A, B, X, Y, Z),"</br>")

  
// This code is contributed by shinjanpatra

</script>

Output

Utkarsh Trivedi

Improve

Article Tags :

Choice of Area

Explore

DSA Fundamentals

Data Structures

Algorithms

Advanced

Interview Preparation

Practice Problem

Thank You!

What kind of Experience do you want to share?