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++
// To print last duplicate element and its// index in a sorted array#include <bits/stdc++.h>void dupLastIndex(int arr[], int n) { // if array is null or size is less // than equal to 0 return if (arr == NULL || n <= 0) return; // compare elements and return last // duplicate and its index 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; } } // If we reach here, then no duplicate // found. 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
// C code To print last duplicate element and its// index in a sorted array#include <stdio.h>void dupLastIndex(int arr[], int n){ // if array is null or size is less // than equal to 0 return if (arr == NULL || n <= 0) return; // compare elements and return last // duplicate and its index 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; } } // If we reach here, then no duplicate // found. 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;}// This code is contributed by muditj148. |
Java
// Java code to print last duplicate element // and its index in a sorted arrayimport java.io.*;class GFG{ static void dupLastIndex(int arr[], int n) { // if array is null or size is less // than equal to 0 return if (arr == null || n <= 0) return; // compare elements and return last // duplicate and its index 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; } } // If we reach here, then no duplicate // found. System.out.print("no duplicate found"); } // Driver code public static void main (String[] args) { int arr[] = {1, 5, 5, 6, 6, 7, 9}; int n = arr.length; dupLastIndex(arr, n); }}// This code is contributed by vt_m |
Python3
# Python3 code to print last duplicate # element and its index in a sorted arraydef dupLastIndex(arr, n): # if array is null or size is less # than equal to 0 return if (arr == None or n <= 0): return # compare elements and return last # duplicate and its index for i in range(n - 1, 0, -1): if (arr[i] == arr[i - 1]): print("Last index:", i, "\nLast", "duplicate item:",arr[i]) return # If we reach here, then no duplicate # found. print("no duplicate found") arr = [1, 5, 5, 6, 6, 7, 9]n = len(arr) dupLastIndex(arr, n)# This code is contributed by # Smitha Dinesh Semwal |
C#
// C# code to print last duplicate element // and its index in a sorted arrayusing System;class GFG { static void dupLastIndex(int []arr, int n) { // if array is null or size is less // than equal to 0 return if (arr == null || n <= 0) return; // compare elements and return last // duplicate and its index 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; } } // If we reach here, then no duplicate // found. Console.WriteLine("no duplicate found"); } // Driver code public static void Main () { int []arr = {1, 5, 5, 6, 6, 7, 9}; int n = arr.Length; dupLastIndex(arr, n); }}// This code is contributed by vt_m. |
PHP
<?php// PHP program to print last // duplicate element and its// index in a sorted arrayfunction dupLastIndex($arr, $n){ // if array is null or size is less // than equal to 0 return if ($arr == null or $n <= 0) return; // compare elements and return last // duplicate and its index 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; }} // If we reach here, then // no duplicate found. echo "no duplicate found";}// Driver Code$arr = array(1, 5, 5, 6, 6, 7, 9);$n = count($arr);dupLastIndex($arr, $n);// This code is contributed by anuj_67.?> |
Javascript
<script>// JavaScript program to print last duplicate element // and its index in a sorted array function dupLastIndex(arr, n) { // if array is null or size is less // than equal to 0 return if (arr == null || n <= 0) return; // compare elements and return last // duplicate and its index 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; } } // If we reach here, then no duplicate // found. document.write("no duplicate found"); } // Driver code 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.
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