# Smallest array that can be obtained by replacing adjacent pairs with their products

Given an array **arr[] **of size **N, **the task is to print the least possible size the given array can be reduced to by performing the following operations:

- Remove any two adjacent elements, say
**arr[i]**and**arr[i+1]**and insert a single element**arr[i] * arr[i+1]**at that position in the array. - If all array elements become equal, then print the size of the array.

**Examples:**

Input:arr[] = {1, 7, 7, 1, 7, 1}Output:1Explanation:

Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {7, 7, 1, 7, 1}

Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {49, 1, 7, 1}

Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {49, 7, 1}

Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {343, 1}

Pick arr[0] and arr[1] and replace with arr[0]*arr[1], arr[] = {343}

Input: arr[]= {2, 2, 2, 2}Output:4

**Approach:** The approach is based on the idea that if all array elements are the same, then the given operations can’t be performed on the given array. Otherwise, in every case, the array size can be reduced to 1. Below are the steps:

- Iterate over the given array
**arr[]**. - If all elements of the array are same, then print
**N**as the required answer. - Otherwise, always pick adjacent elements which give the maximum product to reduce the size of
**arr[]**to 1. Therefore, the minimum possible size will be**1**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to minimize the size of` `// array by performing given operations` `int` `minLength(` `int` `arr[], ` `int` `N)` `{` ` ` `for` `(` `int` `i = 1; i < N; i++) {` ` ` `// If all array elements` ` ` `// are not same` ` ` `if` `(arr[0] != arr[i]) {` ` ` `return` `1;` ` ` `}` ` ` `}` ` ` `// If all array elements` ` ` `// are same` ` ` `return` `N;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `arr[] = { 2, 1, 3, 1 };` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `// Function Call` ` ` `cout << minLength(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the above approach` `import` `java.io.*;` `class` `GFG{` ` ` `// Function to minimize the size of` `// array by performing given operations` `static` `int` `minLength(` `int` `arr[], ` `int` `N)` `{` ` ` `for` `(` `int` `i = ` `1` `; i < N; i++)` ` ` `{` ` ` ` ` `// If all array elements` ` ` `// are not same` ` ` `if` `(arr[` `0` `] != arr[i])` ` ` `{` ` ` `return` `1` `;` ` ` `}` ` ` `}` ` ` ` ` `// If all array elements` ` ` `// are same` ` ` `return` `N;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `[] arr = { ` `2` `, ` `1` `, ` `3` `, ` `1` `};` ` ` `int` `N = arr.length;` ` ` `// Function call` ` ` `System.out.print(minLength(arr, N));` `}` `}` `// This code is contributed by akhilsaini` |

## Python3

`# Python3 implementation of the above approach` `# Function to minimize the size of` `# array by performing given operations` `def` `minLength(arr, N):` ` ` ` ` `for` `i ` `in` `range` `(` `1` `, N):` ` ` ` ` `# If all array elements` ` ` `# are not same` ` ` `if` `(arr[` `0` `] !` `=` `arr[i]):` ` ` `return` `1` ` ` ` ` `# If all array elements` ` ` `# are same` ` ` `return` `N` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` ` ` `arr ` `=` `[ ` `2` `, ` `1` `, ` `3` `, ` `1` `]` ` ` `N ` `=` `len` `(arr)` ` ` ` ` `# Function call` ` ` `print` `(minLength(arr, N))` `# This code is contributed by akhilsaini` |

## C#

`// C# implementation of the above approach` `using` `System;` `class` `GFG{` ` ` `// Function to minimize the size of` `// array by performing given operations` `static` `int` `minLength(` `int` `[] arr, ` `int` `N)` `{` ` ` `for` `(` `int` `i = 1; i < N; i++)` ` ` `{` ` ` ` ` `// If all array elements` ` ` `// are not same` ` ` `if` `(arr[0] != arr[i])` ` ` `{` ` ` `return` `1;` ` ` `}` ` ` `}` ` ` `// If all array elements` ` ` `// are same` ` ` `return` `N;` `}` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `int` `[] arr = { 2, 1, 3, 1 };` ` ` `int` `N = arr.Length;` ` ` `// Function call` ` ` `Console.Write(minLength(arr, N));` `}` `}` `// This code is contributed by akhilsaini` |

## Javascript

`<script>` `// Javascript program to implement` `// the above approach` ` ` `// Function to minimize the size of` `// array by performing given operations` `function` `minLength(arr, N)` `{` ` ` `for` `(let i = 1; i < N; i++)` ` ` `{` ` ` ` ` `// If all array elements` ` ` `// are not same` ` ` `if` `(arr[0] != arr[i])` ` ` `{` ` ` `return` `1;` ` ` `}` ` ` `}` ` ` ` ` `// If all array elements` ` ` `// are same` ` ` `return` `N;` `}` `// Driver Code` ` ` `let arr = [ 2, 1, 3, 1 ];` ` ` `let N = arr.length;` ` ` `// Function call` ` ` `document.write(minLength(arr, N));` `</script>` |

**Output:**

1

**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. To complete your preparation from learning a language to DS Algo and many more, please refer **Complete Interview Preparation Course****.**

In case you wish to attend **live classes **with experts, please refer **DSA Live Classes for Working Professionals **and **Competitive Programming Live for Students**.