Replace repeating elements with greater that greatest values

Given an integer array, if an integer is repeating then replace it with a number greater than that number that has not been inserted yet in the array.

Examples:

Input : arr = {1, 3, 4, 5, 3}
Output : 1 3 4 5 6
Explanation: Here 3 is repeating so it is replaced with 6

Input : arr = {1, 3, 4, 4, 5, 3} 
Output : 1 3 4 6 5 7

We need to replace repeating numbers with a number that does not appear in the array and is one greater than the largest present in the array.

Source: Paytm Interview Experience (Backend Developer).

Calculate the maximum element in the array and replace the repeated elements with the maxx+1 and update the maxx element as visited. In the implementation, basic concept of hashing is used. 

Implementation:

C++

// CPP program to replace repeating elements
// with greater than the greatest.
#include <bits/stdc++.h>
using namespace std;

void replaceElements(int arr[], int n)
{
    // Maximum element in an array
    int maxx = *max_element(arr, arr+n);

    unordered_set<int> s;

    for (int i = 0; i < n; i++) {

        // check whether the element is
        // repeated or not
        if (s.find(arr[i]) == s.end())
            s.insert(arr[i]);

        else {

            // update the repeated element with the
            // maxx element
            arr[i] = maxx + 1;

            maxx++; // update the max

            // mark the maximum element as visited
            s.insert(maxx);
        }
    }
}

// Driver code
int main()
{
    int arr[] = { 1, 3, 4, 5, 3 };
    int n = sizeof(arr)/sizeof(arr[0]);
    replaceElements(arr, n);
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

Java

// Java  program to replace repeating elements
// with greater than the greatest.
import java.util.*;
class Solution
{
    //returns the maximum element
    static int max_element(int arr[])
    {
        int max=arr[0];
        for(int i=1;i<arr.length;i++)
        {
            if(max<arr[i])
            max=arr[i];
        }
        return max;
    }

static void replaceElements(int arr[], int n)
{
    // Maximum element in an array
    int maxx = max_element(arr);

    Vector<Integer> s=new Vector<Integer>();

    for (int i = 0; i < n; i++) {

        // check whether the element is
        // repeated or not
        if (!s.contains(arr[i]))
            s.add(arr[i]);

        else {

            // update the repeated element with the
            // maxx element
            arr[i] = maxx + 1;

            maxx++; // update the max

            // mark the maximum element as visited
            s.add(maxx);
        }
    }
}

// Driver code
public static void main(String args[])
{
    int arr[] = { 1, 3, 4, 5, 3 };
    int n = arr.length;
    replaceElements(arr, n);
    for (int i = 0; i < n; i++)
        System.out.print( arr[i] + " ");

}
}
//contributed by Arnab Kundu

Python3

# Python 3 program to replace repeating
# elements with greater than the greatest.

def replaceElements( arr, n):

    # Maximum element in an array
    maxx = max(arr)

    s = []

    for i in range (n) :

        # check whether the element is
        # repeated or not
        if arr[i] not in s:
            s.append(arr[i])

        else:

            # update the repeated element
            # with the maxx element
            arr[i] = maxx + 1

            maxx += 1 # update the max

            # mark the maximum element
            # as visited
            s.append(maxx)

# Driver code
if __name__ =="__main__":

    arr = [ 1, 3, 4, 5, 3 ]
    n = len(arr)
    replaceElements(arr, n)
    for i in range( n):
        print (arr[i], end = " ")

# This code is contributed by ita_c

C#

// C# program to replace repeating elements
// with greater than the greatest.
using System;
using System.Collections.Generic;

class GFG
{
    //returns the maximum element
    static int max_element(int []arr)
    {
        int max = arr[0];
        for(int i = 1; i < arr.Length; i++)
        {
            if(max < arr[i])
            max = arr[i];
        }
        return max;
    }

static void replaceElements(int []arr, int n)
{
    // Maximum element in an array
    int maxx = max_element(arr);

    List<int> s = new List<int>();

    for (int i = 0; i < n; i++)
    {

        // check whether the element is
        // repeated or not
        if (!s.Contains(arr[i]))
            s.Add(arr[i]);

        else
        {

            // update the repeated element with the
            // maxx element
            arr[i] = maxx + 1;

            maxx++; // update the max

            // mark the maximum element as visited
            s.Add(maxx);
        }
    }
}

// Driver code
public static void Main()
{
    int []arr = { 1, 3, 4, 5, 3 };
    int n = arr.Length;
    replaceElements(arr, n);
    for (int i = 0; i < n; i++)
        Console.Write( arr[i] + " ");
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript  program to replace repeating elements
// with greater than the greatest.

    //returns the maximum element
    function max_element(arr)
    {
        let max=arr[0];
        for(let i=1;i<arr.length;i++)
        {
            if(max<arr[i])
                max=arr[i];
        }
        return max;
    }

    function replaceElements(arr,n)
    {
        // Maximum element in an array
    let maxx = max_element(arr);

    let s=[];

    for (let i = 0; i < n; i++) {

        // check whether the element is
        // repeated or not
        if (!s.includes(arr[i]))
            s.push(arr[i]);

        else {

            // update the repeated element with the
            // maxx element
            arr[i] = maxx + 1;

            maxx++; // update the max

            // mark the maximum element as visited
            s.push(maxx);
        }
    }
    }

    // Driver code
    let arr=[1, 3, 4, 5, 3];
    let n = arr.length;
    replaceElements(arr, n);
    for (let i = 0; i < n; i++)
        document.write( arr[i] + " ");


// This code is contributed by rag2127

</script>

Output

1 3 4 5 6 

Time complexity: O(n) where n is the number of elements in the given array
Auxiliary space: O(n) for storing elements in the HashSet.