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.
1- for i = n-1 to 0
if (arr[i] == arr[i-1])
Print current element and its index.
Return
2- If no such element found print a message
of no duplicate found.
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 |
Output:
Last index: 4 Last duplicate item: 6
Time Complexity: O(n), where n represents the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Last duplicate element in a sorted array for more details!
