# Reduce a given Binary Array to a single element by removal of Triplets

Given an binary array arr[] cof size N, the task is to reduce the array to a single element by the following two operations:

• A triplet of consecutive 0‘s or 1‘s remains unchanged.
• A triplet of consecutive array elements consisting of two 0’s and a single 1 or vice versa can be converted to more frequent element.

Examples:

Input: arr[] = {0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1}
Output: No
Explanation:
Following are the operations performed on the array:
{0, 1, 1} -> 1 modifies the array to {1, 1, 1, 0, 0, 1, 1, 1, 1}
{1, 0, 0} -> 0 modifies the array to {1, 1, 0, 1, 1, 1, 1}
{1, 0, 1} -> 1 modifies the array to {1, 1, 1, 1, 1}
Since, all the remaining elements are 1, they remain unchanged.
Therefore, the array cannot be reduced to a single element.
Input: arr[] = {1, 0, 0, 0, 1, 1, 1}
Output: Yes
Explanation:
Following are the operations performed on the array:
{1, 0, 0} -> 0 {0, 0, 1, 1, 1}
{0, 0, 1} -> 0 {0, 1, 1}
{0, 1, 1} -> 1 {1}

Approach:
Follow the steps below to solve the problem:

• Count the frequency of 0‘s and 1‘s.
• Calculate the absolute difference their respective counts.
• If the difference is 1, only then the array can be reduced to 1. Therefore, print Yes.
• Otherwise, print No.

Below is the implementation of the above approach:

## C++

 `// C++ program to implement ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to check if it is possible to ` `// reduce the array to a single element ` `void` `solve(``int` `arr[], ``int` `n) ` `{ ` `    ``// Stores frequency of 0's ` `    ``int` `countzeroes = 0; ` ` `  `    ``// Stores frequency of 1's ` `    ``int` `countones = 0; ` ` `  `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``if` `(arr[i] == 0) ` `            ``countzeroes++; ` `        ``else` `            ``countones++; ` `    ``} ` ` `  `    ``// Condition for array to be reduced ` `    ``if` `(``abs``(countzeroes - countones) == 1) ` `        ``cout << ``"Yes"``; ` ` `  `    ``// Otherwise ` `    ``else` `        ``cout << ``"No"``; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 1, 0, 0, 1, 1, 1 }; ` ` `  `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``solve(arr, n); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java program to implement ` `// the above approach ` `class` `GFG{ ` ` `  `// Function to check if it is possible to ` `// reduce the array to a single element ` `static` `void` `solve(``int` `arr[], ``int` `n) ` `{ ` `     `  `    ``// Stores frequency of 0's ` `    ``int` `countzeroes = ``0``; ` ` `  `    ``// Stores frequency of 1's ` `    ``int` `countones = ``0``; ` ` `  `    ``for``(``int` `i = ``0``; i < n; i++)  ` `    ``{ ` `        ``if` `(arr[i] == ``0``) ` `            ``countzeroes++; ` `        ``else` `            ``countones++; ` `    ``} ` ` `  `    ``// Condition for array to be reduced ` `    ``if` `(Math.abs(countzeroes - countones) == ``1``) ` `        ``System.out.print(``"Yes"``); ` ` `  `    ``// Otherwise ` `    ``else` `        ``System.out.print(``"No"``); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``0``, ``1``, ``0``, ``0``, ``1``, ``1``, ``1` `}; ` ` `  `    ``int` `n = arr.length; ` ` `  `    ``solve(arr, n); ` `} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

```Yes
```

Time Complexity: O(N)
Auxiliary Space: O(1)

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Improved By : 29AjayKumar