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; ` `} ` |

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## 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 ` |

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## Python3

`# Python3 program to implement ` `# the above approach ` ` ` `# Function to check if it is possible to ` `# reduce the array to a single element ` `def` `solve(arr, n): ` ` ` ` ` `# Stores frequency of 0's ` ` ` `countzeroes ` `=` `0` `; ` ` ` ` ` `# Stores frequency of 1's ` ` ` `countones ` `=` `0` `; ` ` ` ` ` `for` `i ` `in` `range` `(n): ` ` ` `if` `(arr[i] ` `=` `=` `0` `): ` ` ` `countzeroes ` `+` `=` `1` `; ` ` ` `else` `: ` ` ` `countones ` `+` `=` `1` `; ` ` ` ` ` `# Condition for array to be reduced ` ` ` `if` `(` `abs` `(countzeroes ` `-` `countones) ` `=` `=` `1` `): ` ` ` `print` `(` `"Yes"` `); ` ` ` ` ` `# Otherwise ` ` ` `else` `: ` ` ` `print` `(` `"No"` `); ` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `arr ` `=` `[ ` `0` `, ` `1` `, ` `0` `, ` `0` `, ` `1` `, ` `1` `, ` `1` `]; ` ` ` ` ` `n ` `=` `len` `(arr); ` ` ` ` ` `solve(arr, n); ` ` ` `# This code is contributed by Amit Katiyar ` |

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## C#

`// C# program to implement ` `// the above approach ` `using` `System; ` ` ` `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) ` ` ` `Console.Write(` `"Yes"` `); ` ` ` ` ` `// Otherwise ` ` ` `else` ` ` `Console.Write(` `"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 ` |

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**Output:**

Yes

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

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