Find any one of the multiple repeating elements in read only array
Last Updated :
24 Mar, 2023
Given a read-only array of size ( n+1 ), find one of the multiple repeating elements in the array where the array contains integers only between 1 and n.
A read-only array means that the contents of the array can’t be modified.
Examples:
Input : n = 5
arr[] = {1, 1, 2, 3, 5, 4}
Output : One of the numbers repeated in the array is: 1
Input : n = 10
arr[] = {10, 1, 2, 3, 5, 4, 9, 8, 5, 6, 4}
Output : One of the numbers repeated in the array is: 4 OR 5
Method 1: Since the size of the array is n+1 and elements range from 1 to n then it is confirmed that there will be at least one repeating element. A simple solution is to create a count array and store counts of all elements. As soon as we encounter an element with a count of more than 1, we return it.
Below is the implementation of the above approach:
C++
// C++ program to find one of the repeating
// elements in a read only array
#include <bits/stdc++.h>
using namespace std;
// Function to find one of the repeating
// elements
int findRepeatingNumber(const int arr[], int n)
{
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j])
count++;
}
if (count > 1)
return arr[i];
}
// If no repeating element exists
return -1;
}
// Driver Code
int main()
{
// Read only array
const int arr[] = { 1, 1, 2, 3, 5, 4 };
int n = 5;
cout << "One of the numbers repeated in"
" the array is: "
<< findRepeatingNumber(arr, n) << endl;
}
public class GFG {
// Java program to find one of the repeating
// elements in a read only array
// Function to find one of the repeating
// elements
public static int findRepeatingNumber(int[] arr, int n)
{
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
if (count > 1) {
return arr[i];
}
}
// If no repeating element exists
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Read only array
final int[] arr = { 1, 1, 2, 3, 5, 4 };
int n = 5;
System.out.print("One of the numbers repeated in"
+ " the array is: ");
System.out.print(findRepeatingNumber(arr, n));
System.out.print("\n");
}
}
// This code is contributed by Aarti_Rathi
# Python 3program to find one of the repeating
# elements in a read only array
from math import sqrt
# Function to find one of the repeating
# elements
def findRepeatingNumber(arr, n):
for i in arr:
count = 0;
for j in arr:
if i == j:
count=count+1
if(count>1):
return i
# return -1 if no repeating element exists
return -1
# Driver Code
if __name__ == '__main__':
# read only array, not to be modified
arr = [1, 1, 2, 3, 5, 4]
# array of size 6(n + 1) having
# elements between 1 and 5
n = 5
print("One of the numbers repeated in the array is:",
findRepeatingNumber(arr, n))
# This code is contributed by Arpit Jain
// C# program to find one of the repeating
// elements in a read only array
using System;
public class GFG {
// Function to find one of the repeating
// elements
public static int findRepeatingNumber(int[] arr, int n)
{
for (int i = 0; i < n; i++) {
int count = 0;
for (int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
if (count > 1) {
return arr[i];
}
}
// If no repeating element exists
return -1;
}
// Driver Code
public static void Main(String[] args)
{
// Read only array
int[] arr = { 1, 1, 2, 3, 5, 4 };
int n = 5;
Console.Write("One of the numbers repeated in"
+ " the array is: ");
Console.Write(findRepeatingNumber(arr, n));
Console.Write("\n");
}
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
<script>
// Javascript program to find one of the
// repeating elements in a read only array.
// Function to find one of the repeating
// elements
function findRepeatingNumber(arr, n){
for (let i = 0; i < n; i++) {
let count = 0;
for (let j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count++;
}
}
if (count > 1) {
return arr[i];
}
}
// If no repeating element exists
return -1;
}
const arr = [ 1, 1, 2, 3, 5, 4 ];
let n = 5;
document.write("One of the numbers repeated in the array is: " + findRepeatingNumber(arr, n));
// This code is contributed by lokeshmvs21.
</script>
OutputOne of the numbers repeated in the array is: 1
Time Complexity: O(n2)
Auxiliary Space: O(1)
A space-optimized solution is to break the given range (from 1 to n) into blocks of size equal to sqrt(n). We maintain the count of elements belonging to each block for every block. Now as the size of an array is (n+1) and blocks are of size sqrt(n), then there will be one such block whose size will be more than sqrt(n). For the block whose count is greater than sqrt(n), we can use hashing for the elements of this block to find which element appears more than once.
Explanation:
The method described above works because of the following two reasons:
- There would always be a block that has a count greater than sqrt(n) because of one extra element. Even when one extra element has been added it will occupy a position in one of the blocks only, making that block to be selected.
- The selected block definitely has a repeating element. Consider that ith block is selected. The size of the block is greater than sqrt(n) (Hence, it is selected) Maximum distinct elements in this block = sqrt(n). Thus, size can be greater than sqrt(n) only if there is a repeating element in range ( i*sqrt(n), (i+1)*sqrt(n) ].
Note: The last block formed may or may not have a range equal to sqrt(n). Thus, checking if this block has a repeating element will be different than other blocks. However, this difficulty can be overcome from the implementation point of view by initializing the selected block with the last block. This is safe because at least one block has to get selected.
Below is the step-by-step algorithm to solve this problem:
- Divide the array into blocks of size sqrt(n).
- Make a count array that stores the count of elements for each block.
- Pick up the block which has a count of more than sqrt(n), setting the last block
as default. - For the elements belonging to the selected block, use the method of hashing(explained in the next step) to find the repeating element in that block.
- We can create a hash array of key-value pairs, where the key is the element in the block and the value is the count of a number of times the given key is appearing. This can be easily implemented using unordered_map in C++ STL.
Below is the implementation of the above idea:
C++
// C++ program to find one of the repeating
// elements in a read only array
#include <bits/stdc++.h>
using namespace std;
// Function to find one of the repeating
// elements
int findRepeatingNumber(const int arr[], int n)
{
// Size of blocks except the
// last block is sq
int sq = sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
int range = (n / sq) + 1;
// Count array maintains the count for
// all blocks
int count[range] = {0};
// Traversing the read only array and
// updating count
for (int i = 0; i <= n; i++)
{
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[(arr[i] - 1) / sq]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
int selected_block = range - 1;
for (int i = 0; i < range - 1; i++)
{
if (count[i] > sq)
{
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
unordered_map<int, int> m;
for (int i = 0; i <= n; i++)
{
// checks if the element belongs to the
// selected_block
if ( ((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq)))
{
m[arr[i]]++;
// repeating element found
if (m[arr[i]] > 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver Program
int main()
{
// read only array, not to be modified
const int arr[] = { 1, 1, 2, 3, 5, 4 };
// array of size 6(n + 1) having
// elements between 1 and 5
int n = 5;
cout << "One of the numbers repeated in"
" the array is: "
<< findRepeatingNumber(arr, n) << endl;
}
// Java program to find one of the repeating
// elements in a read only array
import java.io.*;
import java.util.*;
class GFG
{
// Function to find one of the repeating
// elements
static int findRepeatingNumber(int[] arr, int n)
{
// Size of blocks except the
// last block is sq
int sq = (int) Math.sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
int range = (n / sq) + 1;
// Count array maintains the count for
// all blocks
int[] count = new int[range];
// Traversing the read only array and
// updating count
for (int i = 0; i <= n; i++)
{
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[(arr[i] - 1) / sq]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
int selected_block = range - 1;
for (int i = 0; i < range - 1; i++)
{
if (count[i] > sq)
{
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
HashMap<Integer, Integer> m = new HashMap<>();
for (int i = 0; i <= n; i++)
{
// checks if the element belongs to the
// selected_block
if ( ((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq)))
{
m.put(arr[i], 1);
// repeating element found
if (m.get(arr[i]) == 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver code
public static void main(String args[])
{
// read only array, not to be modified
int[] arr = { 1, 1, 2, 3, 5, 4 };
// array of size 6(n + 1) having
// elements between 1 and 5
int n = 5;
System.out.println("One of the numbers repeated in the array is: " +
findRepeatingNumber(arr, n));
}
}
// This code is contributed by rachana soma
# Python 3program to find one of the repeating
# elements in a read only array
from math import sqrt
# Function to find one of the repeating
# elements
def findRepeatingNumber(arr, n):
# Size of blocks except the
# last block is sq
sq = sqrt(n)
# Number of blocks to incorporate 1 to
# n values blocks are numbered from 0
# to range-1 (both included)
range__= int((n / sq) + 1)
# Count array maintains the count for
# all blocks
count = [0 for i in range(range__)]
# Traversing the read only array and
# updating count
for i in range(0, n + 1, 1):
# arr[i] belongs to block number
# (arr[i]-1)/sq i is considered
# to start from 0
count[int((arr[i] - 1) / sq)] += 1
# The selected_block is set to last
# block by default. Rest of the blocks
# are checked
selected_block = range__ - 1
for i in range(0, range__ - 1, 1):
if (count[i] > sq):
selected_block = i
break
# after finding block with size > sq
# method of hashing is used to find
# the element repeating in this block
m = {i:0 for i in range(n)}
for i in range(0, n + 1, 1):
# checks if the element belongs
# to the selected_block
if (((selected_block * sq) < arr[i]) and
(arr[i] <= ((selected_block + 1) * sq))):
m[arr[i]] += 1
# repeating element found
if (m[arr[i]] > 1):
return arr[i]
# return -1 if no repeating element exists
return -1
# Driver Code
if __name__ == '__main__':
# read only array, not to be modified
arr = [1, 1, 2, 3, 5, 4]
# array of size 6(n + 1) having
# elements between 1 and 5
n = 5
print("One of the numbers repeated in the array is:",
findRepeatingNumber(arr, n))
# This code is contributed by
# Sahil_shelangia
// C# program to find one of the repeating
// elements in a read only array
using System;
using System.Collections.Generic;
class GFG
{
// Function to find one of the repeating
// elements
static int findRepeatingNumber(int[] arr, int n)
{
// Size of blocks except the
// last block is sq
int sq = (int) Math.Sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
int range = (n / sq) + 1;
// Count array maintains the count for
// all blocks
int[] count = new int[range];
// Traversing the read only array and
// updating count
for (int i = 0; i <= n; i++)
{
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[(arr[i] - 1) / sq]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
int selected_block = range - 1;
for (int i = 0; i < range - 1; i++)
{
if (count[i] > sq)
{
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
Dictionary<int,int> m = new Dictionary<int,int>();
for (int i = 0; i <= n; i++)
{
// checks if the element belongs to the
// selected_block
if ( ((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq)))
{
m.Add(arr[i], 1);
// repeating element found
if (m[arr[i]] == 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver code
public static void Main(String []args)
{
// read only array, not to be modified
int[] arr = { 1, 1, 2, 3, 5, 4 };
// array of size 6(n + 1) having
// elements between 1 and 5
int n = 5;
Console.WriteLine("One of the numbers repeated in the array is: " +
findRepeatingNumber(arr, n));
}
}
// This code contributed by Rajput-Ji
<script>
// JavaScript program to find one of the repeating
// elements in a read only array
// Function to find one of the repeating
// elements
function findRepeatingNumber(arr, n) {
// Size of blocks except the
// last block is sq
let sq = Math.sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
let range = Math.floor(n / sq) + 1;
// Count array maintains the count for
// all blocks
let count = new Array(range).fill(0);
// Traversing the read only array and
// updating count
for (let i = 0; i <= n; i++) {
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[(Math.floor((arr[i] - 1) / sq))]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
let selected_block = range - 1;
for (let i = 0; i < range - 1; i++) {
if (count[i] > sq) {
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
let m = new Map();
for (let i = 0; i <= n; i++) {
// checks if the element belongs to the
// selected_block
if (((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq))) {
m[arr[i]]++;
if (m.has(arr[i])) {
m.set(arr[i], m.get(arr[i]) + 1)
} else {
m.set(arr[i], 1)
}
// repeating element found
if (m.get(arr[i]) > 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver Program
// read only array, not to be modified
const arr = [1, 1, 2, 3, 5, 4];
// array of size 6(n + 1) having
// elements between 1 and 5
let n = 5;
document.write("One of the numbers repeated in"
+ " the array is: "
+ findRepeatingNumber(arr, n) + "<br>");
</script>
OutputOne of the numbers repeated in the array is: 1
Time Complexity: O(N)
Auxiliary Space: sqrt(N)
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