Last duplicate element in a sorted array (original) (raw)

Last Updated : 1 Jun, 2026

Given a sorted array **arr[] that may contain duplicate elements. The task is to find the index of the last occurrence of any duplicate element and return the **index along with the **value of that element. If no duplicate element is found, return **[-1, -1].

**Examples:

**Input: arr[] = [1, 5, 5, 6, 6, 7]
**Output: [4, 6]
**Explanation: Last duplicate element is 6 having index 4.

**Input: arr[] = [1, 2, 3, 4, 5]
**Output: [-1, -1]
**Explanation: No duplicate elements are present in the array.

Try It Yourself

Table of Content

• [Naive Approach] Using Linear Scan from Left to Right - O(n) Time O(1) Space
• [Expected Approach] Using Linear Scan from Right to Left - O(n) Time O(1) Space

[Naive Approach] Using Linear Scan from Left to Right - O(n) Time O(1) Space

The idea is to traverse the sorted array from left to right and compare every element with its next element. Since duplicates appear consecutively in a sorted array, whenever two adjacent elements are equal, store their index and value.

C++ `

#include #include using namespace std;

vector dupLastIndex(vector &arr) { vector res = {-1, -1}; int n = arr.size();

// Traverse the array
for (int i = 0; i < n - 1; i++)
{

    // If duplicate found
    if (arr[i] == arr[i + 1])
    {
        res = {i + 1, arr[i]};
    }
}

return res;

}

// Driver Code int main() { vector arr = {1, 5, 5, 6, 6, 7};

vector<int> res = dupLastIndex(arr);

cout << "[" << res[0] << ", " << res[1] << "]";

return 0;

}

Java

import java.util.Arrays;

public class GfG { public static int[] dupLastIndex(int[] arr) { int[] res = { -1, -1 }; int n = arr.length;

    // Traverse the array
    for (int i = 0; i < n - 1; i++) {

        // If duplicate found
        if (arr[i] == arr[i + 1]) {
            res[0] = i + 1;
            res[1] = arr[i];
        }
    }

    return res;
}

public static void main(String[] args)
{
    int[] arr = { 1, 5, 5, 6, 6, 7 };

    int[] res = dupLastIndex(arr);

    System.out.println(Arrays.toString(res));
}

}

Python

def dupLastIndex(arr): res = [-1, -1] n = len(arr)

# Traverse the array
for i in range(n - 1):

    # If duplicate found
    if arr[i] == arr[i + 1]:
        res = [i + 1, arr[i]]

return res

Driver Code

if name == "main": arr = [1, 5, 5, 6, 6, 7]

res = dupLastIndex(arr)

print(res)

C#

using System;

public class GfG { public static int[] dupLastIndex(int[] arr) { int[] res = { -1, -1 }; int n = arr.Length;

    // Traverse the array
    for (int i = 0; i < n - 1; i++) {
        // If duplicate found
        if (arr[i] == arr[i + 1]) {
            res[0] = i + 1;
            res[1] = arr[i];
        }
    }

    return res;
}

public static void Main()
{
    int[] arr = { 1, 5, 5, 6, 6, 7 };

    int[] res = dupLastIndex(arr);

    Console.WriteLine("[" + res[0] + ", " + res[1]
                      + "]");
}

}

` JavaScript ``

function dupLastIndex(arr) { let res = [ -1, -1 ]; let n = arr.length;

// Traverse the array
for (let i = 0; i < n - 1; i++) {

    // If duplicate found
    if (arr[i] === arr[i + 1]) {
        res = [ i + 1, arr[i] ];
    }
}

return res;

}

// Driver Code let arr = [ 1, 5, 5, 6, 6, 7 ];

let res = dupLastIndex(arr);

console.log([${res[0]}, ${res[1]}]);

``

**Time Complexity: O(n)
**Auxiliary Space: O(1)

[Expected Approach] Using Linear Scan from Right to Left - O(n) Time O(1) Space

The idea is to traverse the sorted array from right to left and compare every element with its previous element. Since duplicate elements occur consecutively, the first duplicate pair found during this traversal will be the last duplicate.

**Let us understand with example:
Input: arr[] = [1, 5, 5, 6, 6, 7]

• Start traversing the array from the end.
• At index 5, arr[5] = 7 and arr[4] = 6, so no duplicate is found.
• At index 4, arr[4] = 6 and arr[3] = 6, so a duplicate is found.
• Return {4, 6} as the index and value of the last duplicate element.

Output: **[4, 6]

C++ `

#include #include using namespace std;

vector dupLastIndex(vector &arr) { int n = arr.size();

// Traverse from end
for (int i = n - 1; i > 0; i--)
{

    // If duplicate found
    if (arr[i] == arr[i - 1])
    {
        vector<int> res = {i, arr[i]};
        return res;
    }
}

return {-1, -1};

}

// Driver Code int main() { vector arr = {1, 5, 5, 6, 6, 7};

vector<int> res = dupLastIndex(arr);

cout << "[" << res[0] << ", " << res[1] << "]";

return 0;

}

Java

import java.util.Arrays;

public class GfG {

public static int[] dupLastIndex(int[] arr)
{
    int n = arr.length;

    // Traverse from end
    for (int i = n - 1; i > 0; i--) {

        // If duplicate found
        if (arr[i] == arr[i - 1]) {
            return new int[] { i, arr[i] };
        }
    }

    return new int[] { -1, -1 };
}

public static void main(String[] args)
{
    int[] arr = { 1, 5, 5, 6, 6, 7 };

    int[] res = dupLastIndex(arr);

    System.out.println(Arrays.toString(res));
}

}

Python

def dupLastIndex(arr): n = len(arr)

# Traverse from end
for i in range(n - 1, 0, -1):

    # If duplicate found
    if arr[i] == arr[i - 1]:
        return [i, arr[i]]

return [-1, -1]

Driver Code

if name == "main": arr = [1, 5, 5, 6, 6, 7]

res = dupLastIndex(arr)

print(res)

C#

using System; using System.Collections.Generic;

public class GfG { public static List dupLastIndex(int[] arr) { int n = arr.Length;

    // Traverse from end
    for (int i = n - 1; i > 0; i--) {

        // If duplicate found
        if (arr[i] == arr[i - 1]) {
            List<int> res = new List<int>{ i, arr[i] };
            return res;
        }
    }

    return new List<int>{ -1, -1 };
}

// Driver Code
public static void Main()
{
    int[] arr = { 1, 5, 5, 6, 6, 7 };

    List<int> res = dupLastIndex(arr);

    Console.WriteLine("[" + res[0] + ", " + res[1]
                      + "]");
}

}

` JavaScript ``

function dupLastIndex(arr) { let n = arr.length;

// Traverse from end
for (let i = n - 1; i > 0; i--) {

    // If duplicate found
    if (arr[i] === arr[i - 1]) {
        return [i, arr[i]];
    }
}

return [-1, -1];

}

// Driver Code let arr = [1, 5, 5, 6, 6, 7];

let res = dupLastIndex(arr);

console.log([${res[0]}, ${res[1]}]);

``

**Time Complexity: O(n)
**Auxiliary Space: O(1)