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:NoExplanation:

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:YesExplanation:

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 <bits/stdc++.h> ` `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[0]); ` ` ` ` ` `solve(arr, n); ` ` ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

**Output:**

Yes

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.

## Recommended Posts:

- Reduce the array to a single element with the given operation
- Reduce the array to a single integer with the given operation
- Minimize sum of adjacent difference with removal of one element from array
- Reduce every element of the array to it's half retaining the sum zero
- Reduce the array such that each element appears at most K times
- Reduce the array to atmost one element by the given operations
- Reduce the array such that each element appears at most 2 times
- Count of triplets from the given Array such that sum of any two elements is the third element
- Minimum number of given operations required to reduce the array to 0 element
- Array element moved by k using single moves
- Reduce number to a single digit by subtracting adjacent digits repeatedly
- Maximum removal from array when removal time >= waiting time
- Longest subarray of non-empty cells after removal of at most a single empty cell
- Duplicates Removal in Array using BST
- Count of triplets of numbers 1 to N such that middle element is always largest
- Number of ways to erase exactly one element in the Binary Array to make XOR zero
- Minimize the maximum minimum difference after one removal from array
- Maximize removal of adjacent array elements based on their absolute value
- Maximize length of Subarray of 1's after removal of a pair of consecutive Array elements
- Find Array formed by adding each element of given array with largest element in new array to its left

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.