# Maximum elements which can be crossed using given units of a and b

Given a binary array of N elements and two initial values a and b. We can cross the i-th element if:

1. If a[i] == 0, then we can use 1 unit from either of b or a to cross the i-th element.
2. If a[i] == 1, then if we use 1 unit from b, a increases by 1 unit. In case 1 unit is used from a, then there is no increase in either of a or b.

The task is to find the maximum number of elements that can be crossed using a and b units.

Note: When we increase a by 1 at any step, it cannot exceed the original value of a.

Examples:

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Iterate in the array element and perform the following steps:

• Break if we do not have either of b or a to pass the element.
• Else, use b if there is no a left, and increase a by 1 if arr[i] == 1.
• Else, use a if there is no b left.
• Else, use b if arr[i]==1 and increase a by 1 till the maximum of the original a.
• Else, simply use 1 unit from a.

Below is the implementation of the above approach:

## 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 `

## PHP

 ` `

Output:

```3
```

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