可以使用给定单位 a 和 b 交叉的最大元素

给定一个包含N 个元素的二进制数组和两个初始值 a 和 b。如果满足以下条件，我们可以交叉第 i 个元素：

1. 如果 a[i] == 0 ，那么我们可以使用 b 或 a 中的 1 个单位来穿过第 i 个元素。
2. 如果a[i] == 1 ，那么如果我们使用 b 中的 1 个单位，则 a 增加 1 个单位。如果从 a 使用 1 个单位，则 a 或 b 都没有增加。

任务是找到可以使用 a 和 b 单位交叉的最大元素数。
注意：当我们在任何一步将 a 增加 1 时，它都不能超过 a 的原始值。
例子：

Input: arr[] = {0, 1, 0, 1, 0}, a = 1, b = 2; 
Output: 5 
Use 1 unit from a to cross 1st element. (a = 0 and b = 2) 
Use 1 unit from b to cross 2nd element. (a = 1 and b = 1) 
Use 1 unit from a to cross 3rd element. (a = 0 and b = 1) 
Use 1 unit from b to cross 4th element. (a = 1 and b = 0) 
Use 1 unit from a to cross 5th element. (a = 0 and b = 0) 
Input: a[] = {1, 0, 0, 1, 0, 1}, a = 1, b = 2 
Use 1 unit from b to cross first element. (a = 1 and b = 1) 
Use 1 unit from b to cross second element. (a = 1 and b = 0) 
Use 1 unit from a to cross third element. (a = 0 and b = 0) 
Output: 3 

方法：迭代数组元素并执行以下步骤：

• 如果我们没有 b 或 a 来传递元素，则中断。
• 否则，如果没有 a，则使用 b，如果 arr[i] == 1，则将 a 增加 1。
• 否则，如果没有 b 就使用 a。
• 否则，如果 arr[i]==1 则使用 b 并将 a 增加 1 直到原始 a 的最大值。
• 否则，只需使用 a 中的 1 个单位。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the number
// of elements crossed
int findElementsCrossed(int arr[], int a, int b, int n)
{
    // Keep a copy of a
    int aa = a;
    int ans = 0;
 
    // Iterate in the binary array
    for (int i = 0; i < n; i++) {
 
        // If no a and b left to use
        if (a == 0 && b == 0)
            break;
 
        // If there is no a
        else if (a == 0) {
 
            // use b and increase a by 1
            // if arr[i] is 1
            if (arr[i] == 1) {
                b -= 1;
                a = min(aa, a + 1);
            }
 
            // simply use b
            else
                b -= 1;
        }
 
        // Use a if theres no b
        else if (b == 0)
            a--;
 
        // Increase a and use b if arr[i] == 1
        else if (arr[i] == 1 && a < aa) {
            b -= 1;
            a = min(aa, a + 1);
        }
 
        // Use a
        else
            a--;
        ans++;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 0, 0, 1, 0, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int a = 1;
    int b = 2;
    cout << findElementsCrossed(arr, a, b, n);
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG
{
 
// Function to find the number
// of elements crossed
static int findElementsCrossed(int arr[],
                        int a, int b, int n)
{
    // Keep a copy of a
    int aa = a;
    int ans = 0;
 
    // Iterate in the binary array
    for (int i = 0; i < n; i++)
    {
 
        // If no a and b left to use
        if (a == 0 && b == 0)
            break;
 
        // If there is no a
        else if (a == 0)
        {
 
            // use b and increase a by 1
            // if arr[i] is 1
            if (arr[i] == 1)
            {
                b -= 1;
                a = Math.min(aa, a + 1);
            }
 
            // simply use b
            else
                b -= 1;
        }
 
        // Use a if theres no b
        else if (b == 0)
            a--;
 
        // Increase a and use b if arr[i] == 1
        else if (arr[i] == 1 && a < aa)
        {
            b -= 1;
            a = Math.min(aa, a + 1);
        }
 
        // Use a
        else
            a--;
        ans++;
    }
 
    return ans;
}
 
// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 0, 0, 1, 0, 1 };
    int n = arr.length;
    int a = 1;
    int b = 2;
    System.out.println(findElementsCrossed(arr, a, b, n));
 
}
}
 
// This code is contributed by
// Surendra_Gangwar

Python3

# Python3 program to implement
# the above approach
 
# Function to find the number
# of elements crossed
def findElementsCrossed(arr, a, b, n):
 
    # Keep a copy of a
    aa = a
    ans = 0
 
    # Iterate in the binary array
    for i in range(n):
 
        # If no a and b left to use
        if (a == 0 and b == 0):
            break
 
        # If there is no a
        elif (a == 0):
 
            # use b and increase a by 1
            # if arr[i] is 1
            if (arr[i] == 1):
                b -= 1
                a = min(aa, a + 1)
             
            # simply use b
            else:
                b -= 1
         
        # Use a if theres no b
        elif (b == 0):
            a -= 1
 
        # Increase a and use b if arr[i] == 1
        elif (arr[i] == 1 and a < aa):
            b -= 1
            a = min(aa, a + 1)
         
        # Use a
        else:
            a -= 1
        ans += 1
     
    return ans
 
# Driver code
arr = [1, 0, 0, 1, 0, 1]
n = len(arr)
a = 1
b = 2
print(findElementsCrossed(arr, a, b, n))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the above approach
using System;
 
class GFG
{
 
// Function to find the number
// of elements crossed
static int findElementsCrossed(int []arr,
                        int a, int b, int n)
{
    // Keep a copy of a
    int aa = a;
    int ans = 0;
 
    // Iterate in the binary array
    for (int i = 0; i < n; i++)
    {
 
        // If no a and b left to use
        if (a == 0 && b == 0)
            break;
 
        // If there is no a
        else if (a == 0)
        {
 
            // use b and increase a by 1
            // if arr[i] is 1
            if (arr[i] == 1)
            {
                b -= 1;
                a = Math.Min(aa, a + 1);
            }
 
            // simply use b
            else
                b -= 1;
        }
 
        // Use a if theres no b
        else if (b == 0)
            a--;
 
        // Increase a and use b if arr[i] == 1
        else if (arr[i] == 1 && a < aa)
        {
            b -= 1;
            a = Math.Min(aa, a + 1);
        }
 
        // Use a
        else
            a--;
        ans++;
    }
 
    return ans;
}
 
// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 0, 0, 1, 0, 1 };
    int n = arr.Length;
    int a = 1;
    int b = 2;
    Console.WriteLine(findElementsCrossed(arr, a, b, n));
 
}
}
 
// This code contributed by Rajput-Ji

Javascript

// javascript program to implement
// the above approach
 
    // Function to find the number
    // of elements crossed
    function findElementsCrossed(arr , a , b , n) {
        // Keep a copy of a
        var aa = a;
        var ans = 0;
 
        // Iterate in the binary array
        for (i = 0; i < n; i++) {
 
            // If no a and b left to use
            if (a == 0 && b == 0)
                break;
 
            // If there is no a
            else if (a == 0) {
 
                // use b and increase a by 1
                // if arr[i] is 1
                if (arr[i] == 1) {
                    b -= 1;
                    a = Math.min(aa, a + 1);
                }
 
                // simply use b
                else
                    b -= 1;
            }
 
            // Use a if theres no b
            else if (b == 0)
                a--;
 
            // Increase a and use b if arr[i] == 1
            else if (arr[i] == 1 && a < aa) {
                b -= 1;
                a = Math.min(aa, a + 1);
            }
 
            // Use a
            else
                a--;
            ans++;
        }
 
        return ans;
    }
 
    // Driver code
     
        var arr = [ 1, 0, 0, 1, 0, 1 ];
        var n = arr.length;
        var a = 1;
        var b = 2;
        document.write(findElementsCrossed(arr, a, b, n));
 
 
// This code contributed by umadevi9616

输出：

3

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。