# Minimize length of an array by removing similar subarrays from both ends

• Last Updated : 26 Mar, 2021

Given an array arr[] of size N, the task is to minimize the length of the given array by repeatedly removing subarrays from the start and end of the array which consists of the same single element.

Examples:

Input: arr[] = { 3, 1, 2, 1, 1, 2, 1, 3 }
Output: 0
Explanation:

1. Since both the first and last elements are 3, removing them modifies arr[] to {1, 2, 1, 1, 2, 1}.
2. Since both the first and last elements are 1, removing them modifies arr[] to {2, 1, 1, 2}.
3. Since both the first and last elements are 2, removing them modifies arr[] to {1, 1}.
4. Since both the first and last elements are 1, removing them modifies arr[] to {}.

Input: arr[] = {1, 1, 2, 3, 3, 1, 2, 2, 1}
Output: 3
Explanation:

1. Removing { 1, 1 } from the start and { 1 } from the end modifies arr[] to { 2, 3, 3, 1, 2, 2 }.
2. Removing { 2 } from the start and { 2, 2 } from the end modifies arr[] to { 3, 3, 1 }.
3. No more elements can be deleted.

Approach: The idea is to use Two-Pointer technique to solve the problem. Follow the steps below to solve the problem:

1. Initialize two pointers front = 0, back = N – 1 to traverse the array from both ends simultaneously.
2. Traverse the array arr[] till front < back:
• If both the elements are different, then break the loop.
• Otherwise, increment front pointer and decrement back pointer until they point to an element different from the current one.
3. Print the difference between the position of two pointers as the minimized length of the array.

Below is the implementation of the above approach:

## C++

 `// C++ program for``// the above approach` `#include ``using` `namespace` `std;` `// Function to minimize length of``// the array by removing similar``// subarrays from both ends of the array``void` `findMinLength(``int` `arr[], ``int` `N)``{``    ``// Initialize two pointers``    ``int` `front = 0, back = N - 1;` `    ``while` `(front < back) {` `        ``// Stores the current integer``        ``int` `x = arr[front];` `        ``// Check if the elements at``        ``// both ends are same or not``        ``if` `(arr[front] != arr[back])``            ``break``;` `        ``// Move the front pointer``        ``while` `(arr[front] == x``               ``&& front <= back)``            ``front++;` `        ``// Move the rear pointer``        ``while` `(arr[back] == x``               ``&& front <= back)``            ``back--;``    ``}` `    ``// Print the minimized length of the array``    ``cout << back - front + 1 << endl;``}` `// Driver Code``int` `main()``{``    ``// Input``    ``int` `arr[] = { 1, 1, 2, 3, 3, 1, 2, 2, 1 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function call to find the``    ``// minimized length of the array``    ``findMinLength(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;``class` `GFG``{` `  ``// Function to minimize length of``  ``// the array by removing similar``  ``// subarrays from both ends of the array``  ``static` `void` `findMinLength(``int` `arr[], ``int` `N)``  ``{``    ``// Initialize two pointers``    ``int` `front = ``0``, back = N - ``1``;``    ``while` `(front < back) {` `      ``// Stores the current integer``      ``int` `x = arr[front];` `      ``// Check if the elements at``      ``// both ends are same or not``      ``if` `(arr[front] != arr[back])``        ``break``;` `      ``// Move the front pointer``      ``while` `(arr[front] == x``             ``&& front <= back)``        ``front++;` `      ``// Move the rear pointer``      ``while` `(arr[back] == x``             ``&& front <= back)``        ``back--;``    ``}` `    ``// Print the minimized length of the array``    ``System.out.println( back - front + ``1` `);``  ``}`  `  ``// Driver Code``  ``public` `static` `void` `main(String[] args)``  ``{` `    ``// Input``    ``int` `arr[] = { ``1``, ``1``, ``2``, ``3``, ``3``, ``1``, ``2``, ``2``, ``1` `};``    ``int` `N = arr.length;` `    ``// Function call to find the``    ``// minimized length of the array``    ``findMinLength(arr, N);``  ``}``}` `// This code is contributed by sanjoy_62.`

## Python3

 `# Python program for``# the above approach` `# Function to minimize length of``# the array by removing similar``# subarrays from both ends of the array``def` `findMinLength(arr, N):``  ` `    ``# Initialize two pointers``    ``front ``=` `0``    ``back ``=` `N ``-` `1``    ``while` `(front < back):``      ` `        ``# Stores the current integer``        ``x ``=` `arr[front]``        ` `        ``# Check if the elements at``        ``# both ends are same or not``        ``if` `arr[front] !``=` `arr[back]:``            ``break``            ` `            ``# Move the front pointer``        ``while` `(arr[front] ``=``=` `x ``and` `front <``=` `back):``            ``front ``+``=` `1``            ` `            ``# Move the rear pointer``        ``while` `(arr[back] ``=``=` `x ``and` `front <``=` `back):``            ``back ``-``=` `1` `    ``# Print the minimized length of the array``    ``print``(back ``-` `front ``+` `1``)` `# Driver Code``# Input``arr ``=` `[``1``, ``1``, ``2``, ``3``, ``3``, ``1``, ``2``, ``2``, ``1``]``N ``=` `len``(arr)` `# Function call to find the``# minimized length of the array``findMinLength(arr, N)` `# This code is contributed by sudhanshugupta2019a.`

## C#

 `// C# program for the above approach``using` `System;``public` `class` `GFG``{` `  ``// Function to minimize length of``  ``// the array by removing similar``  ``// subarrays from both ends of the array``  ``static` `void` `findMinLength(``int` `[]arr, ``int` `N)``  ``{` `    ``// Initialize two pointers``    ``int` `front = 0, back = N - 1;``    ``while` `(front < back)``    ``{` `      ``// Stores the current integer``      ``int` `x = arr[front];` `      ``// Check if the elements at``      ``// both ends are same or not``      ``if` `(arr[front] != arr[back])``        ``break``;` `      ``// Move the front pointer``      ``while` `(arr[front] == x``             ``&& front <= back)``        ``front++;` `      ``// Move the rear pointer``      ``while` `(arr[back] == x``             ``&& front <= back)``        ``back--;``    ``}` `    ``// Print the minimized length of the array``    ``Console.WriteLine( back - front + 1 );``  ``}` `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{` `    ``// Input``    ``int` `[]arr = { 1, 1, 2, 3, 3, 1, 2, 2, 1 };``    ``int` `N = arr.Length;` `    ``// Function call to find the``    ``// minimized length of the array``    ``findMinLength(arr, N);``  ``}``}` `// This code is contributed by shikhasingrajput`

## Javascript

 ``

Output:

`3`

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

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