计算可消耗糖果的最大数量
给定两个数组A【】和B【】，分别由表示每种糖果数量和最大消耗限额的 N 个整数和表示添加的未知糖果数量的整数 M 组成，任务是找出一个人在眼罩中可以消耗的最大糖果数量。
示例:
输入: A[] = {4，5，2，3}，B[] = {8，13，6，4}，M = 5 输出: 4 说明:直接消耗第 4类型的全部 3 颗糖果，再消耗一颗可以是任意类型的糖果。因此，一个人总共只能安全食用 4 颗糖果。
输入: A[] = {2，4，1，9，6}，B[] = {8，7，3，12，7}，M = 0 输出: 2 说明:一个人可以直接消费所有的糖果，因为所有类型的糖果都在安全范围内。
方法:给定的问题可以基于以下观察来解决:
- If a [i]+m ≤ b [i] for each type of candy, it is safe to consume all available candy.
- Otherwise, all 0 ≤ I < N.
只能消耗最少分钟(A[i] + M，B[i])
按照以下步骤解决问题:
- 初始化两个变量，比如和和总计，以存储可安全消费的最大糖果数和糖果总数
- 初始化一个变量，比如 allSafe = true ，检查所有类型的糖果是否可以安全食用。
- 遍历范围【0，N–1】如果 A[i] + M > B[i] ，则设置 allSafe = false 并更新 ans = min(ans，B[i]) 。否则，更新 ans = min(ans，A[i])。
- 如果 allSafe 为真，则打印数组 A[] 的总和。
- 否则，在和中打印结果。
下面是上述方法的实现:
C++
// C++ implementation
// of the above approach
#include <bits/stdc++.h>
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[] = { 2, 4, 1, 9, 6 };
int B[] = { 8, 7, 3, 12, 7 };
int M = 0;
int N = sizeof(A) / sizeof(A[0]);
// Function call to find
// maximum consumable candies
cout << maximumCandy(A, B, N, M);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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.
java 描述语言
<script>
// javascript program of the above approach
// Function to find the count of
// maximum consumable candies
function maximumCandy(candies, safety,
N, M)
{
// Store the count of total candies
let total = 0;
// Stores the count of maximum
// consumable candies
let ans = Number.MAX_VALUE;
// Checks if it is safe
// to counsume all candies
let all_safe = true;
// Traverse the array arr[]
for (let 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
let A = [ 4, 5, 2, 3 ];
let B = [ 8, 13, 6, 4 ];
let M = 5;
let N = A.length;
// Function call to find
// maximum consumable candies
document.write(maximumCandy(A, B, N, M));
</script>
Output:
4
时间复杂度:O(N) T5辅助空间:** O(1)
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