Maximum number of mangoes that can be bought
Given two integers W and C, representing the number of watermelons and coins, the task is to find the maximum number of mangoes that can be bought given that each mango costs 1 watermelon and X coins and y coins can be earned selling a watermelon.
Examples:
Input: W = 10, C = 10, X = 1, Y = 1
Output: 10
Explanation: The most optimal way is to use 10 watermelons and 10 coins to buy 10 mangoes. Hence, the maximum number of mangoes that can be bought is 10.Input: W = 4, C = 8, X = 4, Y = 4
Output: 3
Explanation: The most optimal way is to sell one watermelon. Then, the number of coins increases by 4. Therefore, the total number of coins becomes 12. Therefore, 3 watermelons and 12 coins can be used to buy 3 mangoes. Hence, the maximum number of mangoes that can be bought is 3.
Approach: This problem can be solved using binary search. The idea is to find the maximum number of mangoes in the search space. Follow the steps below to solve the problem:
- Initialize a variable ans as 0 to store the required result.
- Initialize two variables l as 0, r as W to store the boundary regions of the search space for binary search.
- Loop while l?r and perform the following steps:
- Store the middle value in a variable mid as (l+r)/2.
- Check if mid number of mangoes can be bought using the given value of W, C, x, and y.
- If true, then update ans to mid and search in the right part of mid by updating l to mid+1. Otherwise, update the value of r to mid-1.
- Print the value of ans as the result.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to check if mid number// of mangoes can be boughtbool check(int n, int m, int x, int y, int vl){ // Store the coins int temp = m; // If watermelons needed are greater // than given watermelons if (vl > n) return false; // Store remaining watermelons if vl // watermelons are used to buy mangoes int ex = n - vl; // Store the value of coins if these // watermelon get sold ex *= y; // Increment coins by ex temp += ex; // Number of mangoes that can be buyed // if only x coins needed for one mango int cr = temp / x; // If the condition is satisfied, // return true if (cr >= vl) return true; // Otherwise return false return false;}// Function to find the maximum number of mangoes// that can be bought by selling watermelonsint maximizeMangoes(int n, int m, int x, int y){ // Initialize the boundary values int l = 0, r = n; // Store the required result int ans = 0; // Binary Search while (l <= r) { // Store the mid value int mid = l + (r - l) / 2; // Check if it is possible to // buy mid number of mangoes if (check(n, m, x, y, mid)) { ans = mid; l = mid + 1; } // Otherwise, update r to mid -1 else r = mid - 1; } // Return the result return ans;}// Driver Codeint main(){ // Given Input int W = 4, C = 8, x = 4, y = 4; // Function Call cout << maximizeMangoes(W, C, x, y); return 0;} |
Java
// Java program for the above approachclass GFG{// Function to check if mid number// of mangoes can be boughtstatic boolean check(int n, int m, int x, int y, int vl){ // Store the coins int temp = m; // If watermelons needed are greater // than given watermelons if (vl > n) return false; // Store remaining watermelons if vl // watermelons are used to buy mangoes int ex = n - vl; // Store the value of coins if these // watermelon get sold ex *= y; // Increment coins by ex temp += ex; // Number of mangoes that can be buyed // if only x coins needed for one mango int cr = temp / x; // If the condition is satisfied, // return true if (cr >= vl) return true; // Otherwise return false return false;}// Function to find the maximum number // of mangoes that can be bought by // selling watermelonsstatic int maximizeMangoes(int n, int m, int x, int y){ // Initialize the boundary values int l = 0, r = n; // Store the required result int ans = 0; // Binary Search while (l <= r) { // Store the mid value int mid = l + (r - l) / 2; // Check if it is possible to // buy mid number of mangoes if (check(n, m, x, y, mid)) { ans = mid; l = mid + 1; } // Otherwise, update r to mid -1 else r = mid - 1; } // Return the result return ans;}// Driver codepublic static void main(String[] args){ // Given Input int W = 4, C = 8, x = 4, y = 4; // Function Call System.out.println(maximizeMangoes(W, C, x, y));}}// This code is contributed by abhinavjain194 |
Python3
# Python3 program for the above approach# Function to check if mid number# of mangoes can be boughtdef check(n, m, x, y, vl): # Store the coins temp = m # If watermelons needed are greater # than given watermelons if (vl > n): return False # Store remaining watermelons if vl # watermelons are used to buy mangoes ex = n - vl # Store the value of coins if these # watermelon get sold ex *= y # Increment coins by ex temp += ex # Number of mangoes that can be buyed # if only x coins needed for one mango cr = temp // x # If the condition is satisfied, # return true if (cr >= vl): return True # Otherwise return false return False# Function to find the maximum number of mangoes# that can be bought by selling watermelonsdef maximizeMangoes(n, m, x, y): # Initialize the boundary values l = 0 r = n # Store the required result ans = 0 # Binary Search while (l <= r): # Store the mid value mid = l + (r - l) // 2 # Check if it is possible to # buy mid number of mangoes if (check(n, m, x, y, mid)): ans = mid l = mid + 1 # Otherwise, update r to mid -1 else: r = mid - 1 # Return the result return ans# Driver Codeif __name__ == '__main__': # Given Input W = 4 C = 8 x = 4 y = 4 # Function Call print(maximizeMangoes(W, C, x, y)) # This code is contributed by bgangwar59 |
C#
// C# program for the above approachusing System;class GFG{// Function to check if mid number// of mangoes can be boughtstatic bool check(int n, int m, int x, int y, int vl){ // Store the coins int temp = m; // If watermelons needed are greater // than given watermelons if (vl > n) return false; // Store remaining watermelons if vl // watermelons are used to buy mangoes int ex = n - vl; // Store the value of coins if these // watermelon get sold ex *= y; // Increment coins by ex temp += ex; // Number of mangoes that can be buyed // if only x coins needed for one mango int cr = temp / x; // If the condition is satisfied, // return true if (cr >= vl) return true; // Otherwise return false return false;}// Function to find the maximum number of mangoes// that can be bought by selling watermelonsstatic int maximizeMangoes(int n, int m, int x, int y){ // Initialize the boundary values int l = 0, r = n; // Store the required result int ans = 0; // Binary Search while (l <= r) { // Store the mid value int mid = l + (r - l) / 2; // Check if it is possible to // buy mid number of mangoes if (check(n, m, x, y, mid)) { ans = mid; l = mid + 1; } // Otherwise, update r to mid -1 else r = mid - 1; } // Return the result return ans;}// Driver Codepublic static void Main(){ // Given Input int W = 4, C = 8, x = 4, y = 4; // Function Call Console.Write(maximizeMangoes(W, C, x, y));}} // This code is contributed by ukasp |
Javascript
<script>// Javascript program for the above approach// Function to check if mid number// of mangoes can be boughtfunction check(n, m, x, y,vl){ // Store the coins var temp = m; // If watermelons needed are greater // than given watermelons if (vl > n) return false; // Store remaining watermelons if vl // watermelons are used to buy mangoes var ex = n - vl; // Store the value of coins if these // watermelon get sold ex *= y; // Increment coins by ex temp += ex; // Number of mangoes that can be buyed // if only x coins needed for one mango var cr = parseInt(temp / x); // If the condition is satisfied, // return true if (cr >= vl) return true; // Otherwise return false return false;}// Function to find the maximum number of mangoes// that can be bought by selling watermelonsfunction maximizeMangoes(n, m, x, y){ // Initialize the boundary values var l = 0, r = n; // Store the required result var ans = 0; // Binary Search while (l <= r) { // Store the mid value var mid = l + parseInt((r - l) / 2); // Check if it is possible to // buy mid number of mangoes if (check(n, m, x, y, mid)) { ans = mid; l = mid + 1; } // Otherwise, update r to mid -1 else r = mid - 1; } // Return the result return ans;}var W = 4, C = 8, x = 4, y = 4;// Function Calldocument.write( maximizeMangoes(W, C, x, y));//This code is contributed by SoumikMondal</script> |
Output:
3
Time Complexity: O(log(W))
Auxiliary Space: O(1)
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