Replace duplicates with greater than previous duplicate value
Last Updated :
31 Dec, 2022
Given an array of elements and change the array in such a way that all the elements on the array are distinct. If you are replacing a value, then the replacing value should be greater than the previous value and, after modification, the sum of the elements should be as small as possible.
Examples:
Input : arr = {1, 2, 2, 5, 8, 8, 8}
Output : 1 2 3 5 8 9 10
8 is replaced with 9 (A non-existing element greater than 8). Next, duplicate occurrence of 8 is replaced with 10.
Input : arr = {1, 2, 5, 7, 8, 8, 7}
Output : 1 2 5 7 8 9 10
Input : arr = {9, 9, 9, 9, 9}
Output : 9 10 11 12 13
Source: Paytm Interview Experience 30.
This problem is a variation of this article. In the above article, we had to replace the duplicates with the greatest value so far. But in this problem, we have to replace elements with duplicates greater than the previous duplicate value. The idea is to find the next greater element to be replaced to minimize the total sum. To find the next greater element, we have to iterate from the repeated element till the INT_MAX until the next greater element is found
Below is the implementation of the above approach.
C++
#include <bits/stdc++.h>
using namespace std;
void replaceElements(int arr[], int n)
{
unordered_set<int> s;
for (int i = 0; i < n; i++) {
if (s.find(arr[i]) == s.end())
s.insert(arr[i]);
else {
for (int j = arr[i] + 1; j < INT_MAX; j++) {
if (s.find(j) == s.end()) {
arr[i] = j;
s.insert(j);
break;
}
}
}
}
}
int main()
{
int arr[] = { 1, 2, 5, 7, 8, 8, 7 };
int n = sizeof(arr) / sizeof(arr[0]);
replaceElements(arr, n);
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
cout << "\n";
}
|
Java
import java.util.HashSet;
import java.util.Set;
public class ReplaceDuplicateWithGreaterThanPreviousDuplicate {
private static void replaceElements(int[] arr, int n)
{
Set<Integer> st = new HashSet<>();
for (int i = 0; i < n; i++) {
if (!st.contains(arr[i])) {
st.add(arr[i]);
}
else {
for (int j = arr[i] + 1; j < Integer.MAX_VALUE; j++) {
if (!st.contains(j)) {
arr[i] = j;
st.add(j);
break;
}
}
}
}
}
public static void main(String[] args)
{
int[] arr = new int[] { 1, 2, 5, 7, 8, 8, 7 };
int n = arr.length;
replaceElements(arr, n);
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
}
}
|
Python3
import sys
def replaceElements(arr, n):
s = []
for i in range (n):
if arr[i] not in s:
s.append(arr[i])
else :
for j in range(arr[i] + 1, sys.maxsize) :
if j not in s:
arr[i] = j
s.append(j)
break
if __name__ == "__main__":
arr = [ 1, 2, 5, 7, 8, 8, 7 ]
n = len(arr)
replaceElements(arr, n)
for i in range(n):
print (arr[i], end = " ")
print ()
|
C#
using System;
using System.Collections.Generic;
public class ReplaceDuplicateWithGreaterThanPreviousDuplicate
{
private static void replaceElements(int[] arr, int n)
{
HashSet<int> st = new HashSet<int>();
for (int i = 0; i < n; i++)
{
if (!st.Contains(arr[i]))
{
st.Add(arr[i]);
}
else
{
for (int j = arr[i] + 1; j < int.MaxValue; j++)
{
if (!st.Contains(j))
{
arr[i] = j;
st.Add(j);
break;
}
}
}
}
}
public static void Main(String[] args)
{
int[] arr = new int[] { 1, 2, 5, 7, 8, 8, 7 };
int n = arr.Length;
replaceElements(arr, n);
for (int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
}
|
Javascript
<script>
function replaceElements(arr,n)
{
let st = new Set();
for (let i = 0; i < n; i++) {
if (!st.has(arr[i])) {
st.add(arr[i]);
}
else {
for (let j = arr[i] + 1; j < Number.MAX_VALUE; j++) {
if (!st.has(j)) {
arr[i] = j;
st.add(j);
break;
}
}
}
}
}
let arr=[1, 2, 5, 7, 8, 8, 7];
let n = arr.length;
replaceElements(arr, n);
for (let i = 0; i < n; i++) {
document.write(arr[i] + " ");
}
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
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