Given an array arr[], the task is to find the maximum distance between two unequal elements of the given array.
Examples:
Input: arr[] = {3, 2, 1, 2, 1}
Output: 4
The maximum distance is between the first and the last element.
Input: arr[] = {3, 3, 1, 3, 3}
Output: 2
Naive approach: Traverse the whole array for every single element and find the longest distance of element which is unequal.
Efficient approach: By using the fact that the pair of unequal elements must include either first or last element or both, calculate the longest distance between unequal element by traversing the array either by fixing the first element or fixing the last element.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the maximum distance // between two unequal elements int maxDistance( int arr[], int n)
{ // If first and last elements are unequal
// they are maximum distance apart
if (arr[0] != arr[n - 1])
return (n - 1);
int i = n - 1;
// Fix first element as one of the elements
// and start traversing from the right
while (i > 0) {
// Break for the first unequal element
if (arr[i] != arr[0])
break ;
i--;
}
// To store the distance from the first element
int distFirst = (i == 0) ? -1 : i;
i = 0;
// Fix last element as one of the elements
// and start traversing from the left
while (i < n - 1) {
// Break for the first unequal element
if (arr[i] != arr[n - 1])
break ;
i++;
}
// To store the distance from the last element
int distLast = (i == n - 1) ? -1 : (n - 1 - i);
// Maximum possible distance
int maxDist = max(distFirst, distLast);
return maxDist;
} // Driver code int main()
{ int arr[] = { 4, 4, 1, 2, 1, 4 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << maxDistance(arr, n);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG
{ // Function to return the maximum distance // between two unequal elements static int maxDistance( int arr[], int n)
{ // If first and last elements are unequal
// they are maximum distance apart
if (arr[ 0 ] != arr[n - 1 ])
return (n - 1 );
int i = n - 1 ;
// Fix first element as one of the elements
// and start traversing from the right
while (i > 0 )
{
// Break for the first unequal element
if (arr[i] != arr[ 0 ])
break ;
i--;
}
// To store the distance from the first element
int distFirst = (i == 0 ) ? - 1 : i;
i = 0 ;
// Fix last element as one of the elements
// and start traversing from the left
while (i < n - 1 )
{
// Break for the first unequal element
if (arr[i] != arr[n - 1 ])
break ;
i++;
}
// To store the distance from the last element
int distLast = (i == n - 1 ) ? - 1 : (n - 1 - i);
// Maximum possible distance
int maxDist = Math.max(distFirst, distLast);
return maxDist;
} // Driver code public static void main (String[] args)
{ int arr[] = { 4 , 4 , 1 , 2 , 1 , 4 };
int n = arr.length;
System.out.print(maxDistance(arr, n));
} } // This code is contributed by anuj_67.. |
# Python implementation of the approach # Function to return the maximum distance # between two unequal elements def maxDistance(arr, n):
# If first and last elements are unequal
# they are maximum distance apart
if (arr[ 0 ] ! = arr[n - 1 ]):
return (n - 1 );
i = n - 1 ;
# Fix first element as one of the elements
# and start traversing from the right
while (i > 0 ):
# Break for the first unequal element
if (arr[i] ! = arr[ 0 ]):
break ;
i - = 1 ;
# To store the distance from the first element
distFirst = - 1 if (i = = 0 ) else i;
i = 0 ;
# Fix last element as one of the elements
# and start traversing from the left
while (i < n - 1 ):
# Break for the first unequal element
if (arr[i] ! = arr[n - 1 ]):
break ;
i + = 1 ;
# To store the distance from the last element
distLast = - 1 if (i = = n - 1 ) else (n - 1 - i);
# Maximum possible distance
maxDist = max (distFirst, distLast);
return maxDist;
# Driver code arr = [ 4 , 4 , 1 , 2 , 1 , 4 ];
n = len (arr);
print (maxDistance(arr, n));
# This code has been contributed by 29AjayKumar |
// C# implementation of the approach using System;
class GFG
{ // Function to return the maximum distance // between two unequal elements static int maxDistance( int []arr, int n)
{ // If first and last elements are unequal
// they are maximum distance apart
if (arr[0] != arr[n - 1])
return (n - 1);
int i = n - 1;
// Fix first element as one of the elements
// and start traversing from the right
while (i > 0)
{
// Break for the first unequal element
if (arr[i] != arr[0])
break ;
i--;
}
// To store the distance from the first element
int distFirst = (i == 0) ? -1 : i;
i = 0;
// Fix last element as one of the elements
// and start traversing from the left
while (i < n - 1)
{
// Break for the first unequal element
if (arr[i] != arr[n - 1])
break ;
i++;
}
// To store the distance from the last element
int distLast = (i == n - 1) ? -1 : (n - 1 - i);
// Maximum possible distance
int maxDist = Math.Max(distFirst, distLast);
return maxDist;
} // Driver code static public void Main ()
{ int []arr = { 4, 4, 1, 2, 1, 4 };
int n = arr.Length;
Console.WriteLine(maxDistance(arr, n));
} } // This code is contributed by Tushil.. |
<script> // Javascript implementation of the approach // Function to return the maximum distance // between two unequal elements function maxDistance(arr, n)
{ // If first and last elements are unequal
// they are maximum distance apart
if (arr[0] != arr[n - 1])
return (n - 1);
var i = n - 1;
// Fix first element as one of the elements
// and start traversing from the right
while (i > 0) {
// Break for the first unequal element
if (arr[i] != arr[0])
break ;
i--;
}
// To store the distance from the first element
var distFirst = (i == 0) ? -1 : i;
i = 0;
// Fix last element as one of the elements
// and start traversing from the left
while (i < n - 1) {
// Break for the first unequal element
if (arr[i] != arr[n - 1])
break ;
i++;
}
// To store the distance from the last element
var distLast = (i == n - 1) ? -1 : (n - 1 - i);
// Maximum possible distance
var maxDist = Math.max(distFirst, distLast);
return maxDist;
} // Driver code var arr = [ 4, 4, 1, 2, 1, 4 ];
var n = arr.length;
document.write( maxDistance(arr, n)); </script> |
4
Time Complexity: O(n)
Auxiliary Space: O(1)