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:
- Since both the first and last elements are 3, removing them modifies arr[] to {1, 2, 1, 1, 2, 1}.
- Since both the first and last elements are 1, removing them modifies arr[] to {2, 1, 1, 2}.
- Since both the first and last elements are 2, removing them modifies arr[] to {1, 1}.
- 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:
- Removing { 1, 1 } from the start and { 1 } from the end modifies arr[] to { 2, 3, 3, 1, 2, 2 }.
- Removing { 2 } from the start and { 2, 2 } from the end modifies arr[] to { 3, 3, 1 }.
- 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:
- Initialize two pointers front = 0, back = N – 1 to traverse the array from both ends simultaneously.
-
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.
- 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 <bits/stdc++.h> 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[0]);
// 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
<script> // JavaScript program for // the above approach // Function to minimize length of // the array by removing similar // subarrays from both ends of the array function findMinLength(arr, N)
{ // Initialize two pointers
let front = 0, back = N - 1;
while (front < back)
{
// Stores the current integer
let 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
document.write(back - front + 1);
document.write( "<br>" );
} // Driver Code // Input
let arr = [ 1, 1, 2, 3, 3, 1, 2, 2, 1 ];
let N = arr.length;
// Function call to find the
// minimized length of the array
findMinLength(arr, N);
// This code is contributed by Surbhi Tyagi. </script> |
Output:
3
Time Complexity: O(N)
Auxiliary Space: O(1)