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📜  计算消耗性糖果的最大数量

📅  最后修改于: 2021-05-14 05:07:13             🧑  作者: Mango

给定两个数组A []B [],它们分别N个代表每种类型的糖果的数量和最大消耗限制的整数以及一个代表所添加的未知糖果的数量的整数M组成,任务是找到一个人可以在眼罩中食用的糖果。

例子:

方法:可以根据以下观察结果解决给定问题:

请按照以下步骤解决问题:

  1. 初始化两个变量,例如anstotal ,以存储可以安全食用的最大糖果数和糖果总数
  2. 初始化一个变量,例如allSafe = true ,以检查所有类型的糖果是否可以安全食用。
  3. 遍历范围[0,N – 1],并且如果A [i] + M> B [i] ,则将allSafe = false设置并更新ans = min(ans,B [i]) 。否则,更新ans = min(ans,A [i])。
  4. 如果allSafe为true,则打印数组A []的总和。
  5. 否则,将结果打印在ans中

下面是上述方法的实现:

C++
// C++ implememtation
// of the above approach
#include 
using namespace std;
 
// Function to find the count of
// maximum consumable candies
int maximumCandy(int candies[],
                 int safety[],
                 int N, int M)
{
 
    // Store the count of total candies
    int total = 0;
 
    // Stores the count of maximum
    // consumable candies
    int ans = INT_MAX;
 
    // Checks if it is safe
    // to counsume all candies
    bool all_safe = true;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++) {
 
        // If A[i] + M is greater than B[i]
        if (candies[i] + M > safety[i]) {
 
            // Mark all_safe as false
            all_safe = false;
 
            // Update ans
            ans = min(ans, safety[i]);
        }
        else {
 
            // Update ans
            ans = min(ans, candies[i] + M);
        }
 
        // Increment total by A[i]
        total += candies[i];
    }
 
    // If all_safe is true
    if (all_safe)
        return total;
 
    // Otherwise,
    else
        return ans;
}
 
// Driver Code
int main()
{
    int A[] = { 4, 5, 2, 3 };
    int B[] = { 8, 13, 6, 4 };
    int M = 5;
 
    int N = sizeof(A) / sizeof(A[0]);
 
    // Function call to find
    // maximum consumable candies
    cout << maximumCandy(A, B, N, M);
 
    return 0;
}


Java
// Java implememtation
// of the above approach
public class GFG
{
 
  // Function to find the count of
  // maximum consumable candies
  static int maximumCandy(int []candies,
                          int []safety,
                          int N, int M)
  {
 
    // Store the count of total candies
    int total = 0;
 
    // Stores the count of maximum
    // consumable candies
    int ans = Integer.MAX_VALUE;
 
    // Checks if it is safe
    // to counsume all candies
    boolean all_safe = true;
 
    // Traverse the array arr[]
    for (int i = 0; i < N; i++)
    {
 
      // If A[i] + M is greater than B[i]
      if (candies[i] + M > safety[i])
      {
 
        // Mark all_safe as false
        all_safe = false;
 
        // Update ans
        ans = Math.min(ans, safety[i]);
      }
      else
      {
 
        // Update ans
        ans = Math.min(ans, candies[i] + M);
      }
 
      // Increment total by A[i]
      total += candies[i];
    }
 
    // If all_safe is true
    if (all_safe)
      return total;
 
    // Otherwise,
    else
      return ans;
  }
 
  // Driver Code
  public static void main (String[] args)
  {
    int A[] = { 4, 5, 2, 3 };
    int B[] = { 8, 13, 6, 4 };
    int M = 5;
 
    int N = A.length;
 
    // Function call to find
    // maximum consumable candies
    System.out.println(maximumCandy(A, B, N, M));
 
  }
 
}
 
// This code is contributed by AnkThon


Python3
# Python3 implememtation
# of the above approach
 
# Function to find the count of
# maximum consumable candies
def maximumCandy(candies, safety, N, M):
 
    # Store the count of total candies
    total = 0
 
    # Stores the count of maximum
    # consumable candies
    ans = 10**8
 
    # Checks if it is safe
    # to counsume all candies
    all_safe = True
 
    # Traverse the array arr
    for i in range(N):
 
        # If A[i] + M is greater than B[i]
        if (candies[i] + M > safety[i]):
 
            # Mark all_safe as false
            all_safe = False
 
            # Update ans
            ans = min(ans, safety[i])
        else:
 
            # Update ans
            ans = min(ans, candies[i] + M)
 
        # Increment total by A[i]
        total += candies[i]
 
    # If all_safe is true
    if (all_safe):
        return total
 
    # Otherwise,
    else:
        return ans
 
# Driver Code
if __name__ == '__main__':
    A = [4, 5, 2, 3]
    B = [ 8, 13, 6, 4]
    M = 5
 
    N = len(A)
 
    # Function call to find
    # maximum consumable candies
    print (maximumCandy(A, B, N, M))
 
    # This code is contributed by mohit kumar 29.


C#
// C# implememtation
// of the above approach
using System;
class GFG {
     
    // Function to find the count of
    // maximum consumable candies
    static int maximumCandy(int[] candies, int[] safety, int N, int M)
    {
      
        // Store the count of total candies
        int total = 0;
      
        // Stores the count of maximum
        // consumable candies
        int ans = Int32.MaxValue;
      
        // Checks if it is safe
        // to counsume all candies
        bool all_safe = true;
      
        // Traverse the array arr[]
        for (int i = 0; i < N; i++) {
      
            // If A[i] + M is greater than B[i]
            if (candies[i] + M > safety[i]) {
      
                // Mark all_safe as false
                all_safe = false;
      
                // Update ans
                ans = Math.Min(ans, safety[i]);
            }
            else {
      
                // Update ans
                ans = Math.Min(ans, candies[i] + M);
            }
      
            // Increment total by A[i]
            total += candies[i];
        }
      
        // If all_safe is true
        if (all_safe)
            return total;
      
        // Otherwise,
        else
            return ans;
    }
 
  // Driver code
  static void Main()
  {
    int[] A = { 4, 5, 2, 3 };
    int[] B = { 8, 13, 6, 4 };
    int M = 5;
  
    int N = A.Length;
  
    // Function call to find
    // maximum consumable candies
    Console.WriteLine(maximumCandy(A, B, N, M));
  }
}
 
// This code is contributed by divyeshrabadiya07.


输出:
4

时间复杂度: O(N)
辅助空间: O(1)