Last duplicate element in a sorted array
We have a sorted array with duplicate elements and we have to find the index of last duplicate element and print index of it and also print the duplicate element. If no such element found print a message.
Examples:
Input : arr[] = {1, 5, 5, 6, 6, 7}
Output :
Last index: 4
Last duplicate item: 6
Input : arr[] = {1, 2, 3, 4, 5}
Output : No duplicate found
We simply iterate through the array in reverse order and compare the current and previous element. If a match is found then we print the index and duplicate element. As this is sorted array it will be the last duplicate. If no such element is found we will print the message for it.
- for i = n-1 to 0
if (arr[i] == arr[i-1])
Print current element and its index.
Return
- If no such element found print a message of no duplicate found.
Implementation:
C++
#include <bits/stdc++.h>
void dupLastIndex( int arr[], int n) {
if (arr == NULL || n <= 0)
return ;
for ( int i = n - 1; i > 0; i--) {
if (arr[i] == arr[i - 1]) {
printf ( "Last index: %d\nLast "
"duplicate item: %d\n" , i, arr[i]);
return ;
}
}
printf ( "no duplicate found" );
}
int main() {
int arr[] = {1, 5, 5, 6, 6, 7, 9};
int n = sizeof (arr) / sizeof ( int );
dupLastIndex(arr, n);
return 0;
}
|
C
#include <stdio.h>
void dupLastIndex( int arr[], int n)
{
if (arr == NULL || n <= 0)
return ;
for ( int i = n - 1; i > 0; i--) {
if (arr[i] == arr[i - 1]) {
printf ( "Last index: %d\nLast "
"duplicate item: %d\n" ,
i, arr[i]);
return ;
}
}
printf ( "no duplicate found" );
}
int main()
{
int arr[] = { 1, 5, 5, 6, 6, 7, 9 };
int n = sizeof (arr) / sizeof ( int );
dupLastIndex(arr, n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static void dupLastIndex( int arr[], int n)
{
if (arr == null || n <= 0 )
return ;
for ( int i = n - 1 ; i > 0 ; i--)
{
if (arr[i] == arr[i - 1 ])
{
System.out.println( "Last index:" + i);
System.out.println( "Last duplicate item: "
+ arr[i]);
return ;
}
}
System.out.print( "no duplicate found" );
}
public static void main (String[] args)
{
int arr[] = { 1 , 5 , 5 , 6 , 6 , 7 , 9 };
int n = arr.length;
dupLastIndex(arr, n);
}
}
|
Python3
def dupLastIndex(arr, n):
if (arr = = None or n < = 0 ):
return
for i in range (n - 1 , 0 , - 1 ):
if (arr[i] = = arr[i - 1 ]):
print ( "Last index:" , i, "\nLast" ,
"duplicate item:" ,arr[i])
return
print ( "no duplicate found" )
arr = [ 1 , 5 , 5 , 6 , 6 , 7 , 9 ]
n = len (arr)
dupLastIndex(arr, n)
|
C#
using System;
class GFG {
static void dupLastIndex( int []arr, int n)
{
if (arr == null || n <= 0)
return ;
for ( int i = n - 1; i > 0; i--)
{
if (arr[i] == arr[i - 1])
{
Console.WriteLine( "Last index:" + i);
Console.WriteLine( "Last duplicate item: "
+ arr[i]);
return ;
}
}
Console.WriteLine( "no duplicate found" );
}
public static void Main ()
{
int []arr = {1, 5, 5, 6, 6, 7, 9};
int n = arr.Length;
dupLastIndex(arr, n);
}
}
|
PHP
<?php
function dupLastIndex( $arr , $n )
{
if ( $arr == null or $n <= 0)
return ;
for ( $i = $n - 1; $i > 0; $i --)
{
if ( $arr [ $i ] == $arr [ $i - 1])
{
echo "Last index:" , $i , "\n" ;
echo "Last duplicate item:" , $arr [ $i ];
return ;
}
}
echo "no duplicate found" ;
}
$arr = array (1, 5, 5, 6, 6, 7, 9);
$n = count ( $arr );
dupLastIndex( $arr , $n );
?>
|
Javascript
<script>
function dupLastIndex(arr, n)
{
if (arr == null || n <= 0)
return ;
for (let i = n - 1; i > 0; i--)
{
if (arr[i] == arr[i - 1])
{
document.write( "Last index:" + i + "<br/>" );
document.write( "Last duplicate item: "
+ arr[i] + "<br/>" );
return ;
}
}
document.write( "no duplicate found" );
}
let arr = [1, 5, 5, 6, 6, 7, 9];
let n = arr.length;
dupLastIndex(arr, n);
</script>
|
Output:
Last index: 4
Last duplicate item: 6
Time Complexity: O(n)
Auxiliary Space: O(1), as no extra space is used.
Last Updated :
03 Aug, 2022
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