Count frequencies of all elements in array in O(1) extra space and O(n) time (original) (raw)
Last Updated : 10 May, 2024
Given an unsorted array of **n integers that can contain integers from **1 to **n. Some elements can be repeated multiple times and some other elements can be absent from the array. Count the frequency of all elements that are present and print the missing elements.
**Examples:
**Input: arr[] = {2, 3, 3, 2, 5}
**Output: Below are frequencies of all elements 1 -> 0 2 -> 2 3 -> 2 4 -> 0 5 -> 1
**Explanation: Frequency of elements 1 is 0, 2 is 2, 3 is 2, 4 is 0 and 5 is 1.**Input: arr[] = {4, 4, 4, 4}
**Output: Below are frequencies of all elements 1 -> 0 2 -> 0 3 -> 0 4 -> 4
**Explanation: Frequency of elements 1 is 0, 2 is 0, 3 is 0 and 4 is 4.
**Naive Solution:
Create an extra space of size n, as elements of the array is in the range 1 to n. Use the extra space as HashMap. Traverse the array and update the count of the current element. Finally, print the frequencies of the HashMap along with the indices.
Step-by-step approach:
- Create an extra space of size n (**hm), use it as a HashMap.
- Traverse the array from start to end.
- For every element update **hm[array[i]-1], i.e. **hm[array[i]-1]++
- Run a loop from **0 to **n and print _**hm[array[i]-1]_along with the index **i
Below is the implementation of the above approach:
C++ `
// C++ program to print frequencies of all array // elements in O(n) extra space and O(n) time #include<bits/stdc++.h> using namespace std;
// Function to find counts of all elements present in // arr[0..n-1]. The array elements must be range from // 1 to n void findCounts(int *arr, int n) { //Hashmap int hash[n]={0};
// Traverse all array elements
int i = 0;
while (i<n)
{
//update the frequency of array[i]
hash[arr[i]-1]++;
//increase the index
i++;
}
printf("\nBelow are counts of all elements\n");
for (int i=0; i<n; i++)
printf("%d -> %d\n", i+1, hash[i]);}
// Driver program to test above function int main() { int arr[] = {2, 3, 3, 2, 5}; findCounts(arr, sizeof(arr)/ sizeof(arr[0]));
int arr1[] = {1};
findCounts(arr1, sizeof(arr1)/ sizeof(arr1[0]));
int arr3[] = {4, 4, 4, 4};
findCounts(arr3, sizeof(arr3)/ sizeof(arr3[0]));
int arr2[] = {1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1};
findCounts(arr2, sizeof(arr2)/ sizeof(arr2[0]));
int arr4[] = {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3};
findCounts(arr4, sizeof(arr4)/ sizeof(arr4[0]));
int arr5[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
findCounts(arr5, sizeof(arr5)/ sizeof(arr5[0]));
int arr6[] = {11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
findCounts(arr6, sizeof(arr6)/ sizeof(arr6[0]));
return 0;}
Java
// Java program to print frequencies of all array // elements in O(n) extra space and O(n) time import java.util.*;
class GFG{
// Function to find counts of all elements // present in arr[0..n-1]. The array elements // must be range from 1 to n public static void findCounts(int arr[], int n) {
// Hashmap
int hash[] = new int[n];
Arrays.fill(hash, 0);
// Traverse all array elements
int i = 0;
while (i < n)
{
// Update the frequency of array[i]
hash[arr[i] - 1]++;
// Increase the index
i++;
}
System.out.println("\nBelow are counts " +
"of all elements");
for(i = 0; i < n; i++)
{
System.out.println((i + 1) + " -> " +
hash[i]);
}}
// Driver code public static void main(String []args) { int arr[] = { 2, 3, 3, 2, 5 }; findCounts(arr, arr.length);
int arr1[] = {1};
findCounts(arr1, arr1.length);
int arr3[] = { 4, 4, 4, 4 };
findCounts(arr3, arr3.length);
int arr2[] = { 1, 3, 5, 7, 9,
1, 3, 5, 7, 9, 1 };
findCounts(arr2, arr2.length);
int arr4[] = { 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3 };
findCounts(arr4, arr4.length);
int arr5[] = { 1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11 };
findCounts(arr5, arr5.length);
int arr6[] = { 11, 10, 9, 8, 7,
6, 5, 4, 3, 2, 1 };
findCounts(arr6, arr6.length);} }
// This code is contributed by rag2127
Python
Python3 program to print frequencies
of all array elements in O(n) extra
space and O(n) time
Function to find counts of all
elements present in arr[0..n-1].
The array elements must be range
from 1 to n
def findCounts(arr, n):
# Hashmap
hash = [0 for i in range(n)]
# Traverse all array elements
i = 0
while (i < n):
# Update the frequency of array[i]
hash[arr[i] - 1] += 1
# Increase the index
i += 1
print("Below are counts of all elements")
for i in range(n):
print(i + 1, "->", hash[i], end = " ")
print()Driver code
arr = [ 2, 3, 3, 2, 5 ] findCounts(arr, len(arr))
arr1 = [1] findCounts(arr1, len(arr1))
arr3 = [ 4, 4, 4, 4 ] findCounts(arr3, len(arr3))
arr2 = [ 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1 ] findCounts(arr2, len(arr2))
arr4 = [ 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 ] findCounts(arr4, len(arr4))
arr5 = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 ] findCounts(arr5, len(arr5))
arr6 = [ 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 ] findCounts(arr6, len(arr6))
This code is contributed by avanitrachhadiya2155
C#
// C# program to print frequencies of all array // elements in O(n) extra space and O(n) time using System;
class GFG {
// Function to find counts of all elements // present in arr[0..n-1]. The array elements // must be range from 1 to n public static void findCounts(int[] arr, int n) {
// Hashmap
int[] hash = new int[n];
// Traverse all array elements
int i = 0;
while (i < n)
{
// Update the frequency of array[i]
hash[arr[i] - 1]++;
// Increase the index
i++;
}
Console.WriteLine("\nBelow are counts "
+ "of all elements");
for (i = 0; i < n; i++)
{
Console.WriteLine((i + 1) + " -> " + hash[i]);
}}
// Driver code static public void Main() { int[] arr = new int[] { 2, 3, 3, 2, 5 }; findCounts(arr, arr.Length); int[] arr1 = new int[] { 1 }; findCounts(arr1, arr1.Length); int[] arr3 = new int[] { 4, 4, 4, 4 }; findCounts(arr3, arr3.Length); int[] arr2 = new int[] { 1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1 }; findCounts(arr2, arr2.Length); int[] arr4 = new int[] { 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3 }; findCounts(arr4, arr4.Length); int[] arr5 = new int[] { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 }; findCounts(arr5, arr5.Length); int[] arr6 = new int[] { 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 }; findCounts(arr6, arr6.Length); } }
// This code is contributed by Dharanendra L V
JavaScript
`
Output
Below are counts of all elements 1 -> 0 2 -> 2 3 -> 2 4 -> 0 5 -> 1
Below are counts of all elements 1 -> 1
Below are counts of all elements 1 -> 0 2 -> 0 3 -> 0 4 -> 4
Below are counts of all elements 1 -> 3 2 -> 0 3 -> 2 4 -> 0 5 -> 2 6 -> 0 7 -> 2 8 -> 0 9 -> 2 10 -> 0 11 -> 0
Below are counts of all elements 1 -> 0 2 -> 0 3 -> 11 4 -> 0 5 -> 0 6 -> 0 7 -> 0 8 -> 0 9 -> 0 10 -> 0 11 -> 0
Below are counts of all elements 1 -> 1 2 -> 1 3 -> 1 4 -> 1 5 -> 1 6 -> 1 7 -> 1 8 -> 1 9 -> 1 10 -> 1 11 -> 1
Below are counts of all elements 1 -> 1 2 -> 1 3 -> 1 4 -> 1 5 -> 1 6 -> 1 7 -> 1 8 -> 1 9 -> 1 10 -> 1 11 -> 1
**Time Complexity: O(n). As a single traversal of array takes O(n) time.
**Auxiliary Space Complexity: O(n). To store all the elements in a HashMap O(n) space is needed.
**Efficient Approach: By making elements negative.
The idea is to traverse the given array, use elements as an index and store their counts at the index. Consider the basic approach, a Hashmap of size n is needed and the array is also of size n. So the array can be used as a hashmap, all the elements of the array are from 1 to n, i.e. all are positive elements. So the frequency can be stored as negative. This might lead to a problem. Let _i-th element be a then the count should be stored at **array[a-1], but when the frequency will be stored the element will be lost. To deal with this problem, first, replace the ith element by **array[a-1] and then put **-1 at **array[a-1]. So our idea is to replace the element by frequency and store the element in the current index and if the element at array[a-1] is already negative, then it is already replaced by a frequency so decrement **array[a-1].
Step-by-step approach:
- Traverse the array from start to end.
- For each element check if the element is less than or equal to zero or not. If negative or zero skip the element as it is frequency.
- If an element (_e = array[i] - 1 ) is positive, then check if _array[e] is positive or not.
- If positive then that means it is the first occurrence of e in the array and replace _array[i] with array[e], i.e__array[i] = array[e] and assign _array[e] = -1.
- If _array[e] is negative, then it is not the first occurrence, the update _array[e] as _array[e]-- and update _array[i] as _array[i] = 0.
- Again, traverse the array and print i+1 as value and array[i] as frequency.
Below is the implementation of the above approach:
C++ `
// C++ program to print frequencies of all array // elements in O(1) extra space and O(n) time #include<bits/stdc++.h> using namespace std;
// Function to find counts of all elements present in // arr[0..n-1]. The array elements must be range from // 1 to n void findCounts(int *arr, int n) { // Traverse all array elements int i = 0; while (i<n) { // If this element is already processed, // then nothing to do if (arr[i] <= 0) { i++; continue; }
// Find index corresponding to this element
// For example, index for 5 is 4
int elementIndex = arr[i]-1;
// If the elementIndex has an element that is not
// processed yet, then first store that element
// to arr[i] so that we don't lose anything.
if (arr[elementIndex] > 0)
{
arr[i] = arr[elementIndex];
// After storing arr[elementIndex], change it
// to store initial count of 'arr[i]'
arr[elementIndex] = -1;
}
else
{
// If this is NOT first occurrence of arr[i],
// then decrement its count.
arr[elementIndex]--;
// And initialize arr[i] as 0 means the element
// 'i+1' is not seen so far
arr[i] = 0;
i++;
}
}
printf("\nBelow are counts of all elements\n");
for (int i=0; i<n; i++)
printf("%d -> %d\n", i+1, abs(arr[i]));}
// Driver program to test above function int main() { int arr[] = {2, 3, 3, 2, 5}; findCounts(arr, sizeof(arr)/ sizeof(arr[0]));
int arr1[] = {1};
findCounts(arr1, sizeof(arr1)/ sizeof(arr1[0]));
int arr3[] = {4, 4, 4, 4};
findCounts(arr3, sizeof(arr3)/ sizeof(arr3[0]));
int arr2[] = {1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1};
findCounts(arr2, sizeof(arr2)/ sizeof(arr2[0]));
int arr4[] = {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3};
findCounts(arr4, sizeof(arr4)/ sizeof(arr4[0]));
int arr5[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
findCounts(arr5, sizeof(arr5)/ sizeof(arr5[0]));
int arr6[] = {11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
findCounts(arr6, sizeof(arr6)/ sizeof(arr6[0]));
return 0;}
Java
// Java program to print frequencies of all array // elements in O(1) extra space and O(n) time
class CountFrequencies { // Function to find counts of all elements present in // arr[0..n-1]. The array elements must be range from // 1 to n void findCounts(int arr[], int n) { // Traverse all array elements int i = 0; while (i < n) { // If this element is already processed, // then nothing to do if (arr[i] <= 0) { i++; continue; }
// Find index corresponding to this element
// For example, index for 5 is 4
int elementIndex = arr[i] - 1;
// If the elementIndex has an element that is not
// processed yet, then first store that element
// to arr[i] so that we don't lose anything.
if (arr[elementIndex] > 0)
{
arr[i] = arr[elementIndex];
// After storing arr[elementIndex], change it
// to store initial count of 'arr[i]'
arr[elementIndex] = -1;
}
else
{
// If this is NOT first occurrence of arr[i],
// then decrement its count.
arr[elementIndex]--;
// And initialize arr[i] as 0 means the element
// 'i+1' is not seen so far
arr[i] = 0;
i++;
}
}
System.out.println("Below are counts of all elements");
for (int j = 0; j < n; j++)
System.out.println(j+1 + "->" + Math.abs(arr[j]));
}
// Driver program to test above functions
public static void main(String[] args)
{
CountFrequencies count = new CountFrequencies();
int arr[] = {2, 3, 3, 2, 5};
count.findCounts(arr, arr.length);
int arr1[] = {1};
count.findCounts(arr1, arr1.length);
int arr3[] = {4, 4, 4, 4};
count.findCounts(arr3, arr3.length);
int arr2[] = {1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1};
count.findCounts(arr2, arr2.length);
int arr4[] = {3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3};
count.findCounts(arr4, arr4.length);
int arr5[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11};
count.findCounts(arr5, arr5.length);
int arr6[] = {11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1};
count.findCounts(arr6, arr6.length);
}}
// This code has been contributed by Mayank Jaiswal(mayank_24)
Python3
Python3 program to print frequencies of all array
elements in O(1) extra space and O(n) time
Function to find counts of all elements present in
arr[0..n-1]. The array elements must be range from
1 to n
def findCounts(arr, n):
# Traverse all array elements
i = 0
while i<n:
# If this element is already processed,
# then nothing to do
if arr[i] <= 0:
i += 1
continue
# Find index corresponding to this element
# For example, index for 5 is 4
elementIndex = arr[i] - 1
# If the elementIndex has an element that is not
# processed yet, then first store that element
# to arr[i] so that we don't lose anything.
if arr[elementIndex] > 0:
arr[i] = arr[elementIndex]
# After storing arr[elementIndex], change it
# to store initial count of 'arr[i]'
arr[elementIndex] = -1
else:
# If this is NOT first occurrence of arr[i],
# then decrement its count.
arr[elementIndex] -= 1
# And initialize arr[i] as 0 means the element
# 'i+1' is not seen so far
arr[i] = 0
i += 1
print ("Below are counts of all elements")
for i in range(0,n):
print ("%d -> %d"%(i+1, abs(arr[i])))
print ("")Driver program to test above function
arr = [2, 3, 3, 2, 5] findCounts(arr, len(arr))
arr1 = [1] findCounts(arr1, len(arr1))
arr3 = [4, 4, 4, 4] findCounts(arr3, len(arr3))
arr2 = [1, 3, 5, 7, 9, 1, 3, 5, 7, 9, 1] findCounts(arr2, len(arr2))
arr4 = [3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3] findCounts(arr4, len(arr4))
arr5 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11] findCounts(arr5, len(arr5))
arr6 = [11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1] findCounts(arr6, len(arr6))
This code is contributed
by shreyanshi_19
C#
// C# program to print frequencies of // all array elements in O(1) extra // space and O(n) time using System;
class GFG { // Function to find counts of all // elements present in arr[0..n-1]. // The array elements must be range // from 1 to n void findCounts(int[] arr, int n) { // Traverse all array elements int i = 0; while (i < n) { // If this element is already // processed, then nothing to do if (arr[i] <= 0) { i++; continue; }
// Find index corresponding to
// this element. For example,
// index for 5 is 4
int elementIndex = arr[i] - 1;
// If the elementIndex has an element
// that is not processed yet, then
// first store that element to arr[i]
// so that we don't loose anything.
if (arr[elementIndex] > 0)
{
arr[i] = arr[elementIndex];
// After storing arr[elementIndex],
// change it to store initial count
// of 'arr[i]'
arr[elementIndex] = -1;
}
else
{
// If this is NOT first occurrence
// of arr[i], then decrement its count.
arr[elementIndex]--;
// And initialize arr[i] as 0 means
// the element 'i+1' is not seen so far
arr[i] = 0;
i++;
}
}
Console.Write("\nBelow are counts of " +
"all elements" + "\n");
for (int j = 0; j < n; j++)
Console.Write(j + 1 + "->" +
Math.Abs(arr[j]) + "\n");}
// Driver Code public static void Main() { GFG count = new GFG(); int[] arr = {2, 3, 3, 2, 5}; count.findCounts(arr, arr.Length);
int[] arr1 = {1};
count.findCounts(arr1, arr1.Length);
int[] arr3 = {4, 4, 4, 4};
count.findCounts(arr3, arr3.Length);
int[] arr2 = {1, 3, 5, 7, 9, 1,
3, 5, 7, 9, 1};
count.findCounts(arr2, arr2.Length);
int[] arr4 = {3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3};
count.findCounts(arr4, arr4.Length);
int[] arr5 = {1, 2, 3, 4, 5, 6,
7, 8, 9, 10, 11};
count.findCounts(arr5, arr5.Length);
int[] arr6 = {11, 10, 9, 8, 7, 6,
5, 4, 3, 2, 1};
count.findCounts(arr6, arr6.Length);} }
// This code is contributed by ChitraNayal
JavaScript
`
Output
Below are counts of all elements 1 -> 0 2 -> 2 3 -> 2 4 -> 0 5 -> 1
Below are counts of all elements 1 -> 1
Below are counts of all elements 1 -> 0 2 -> 0 3 -> 0 4 -> 4
Below are counts of all elements 1 -> 3 2 -> 0 3 -> 2 4 -> 0 5 -> 2 6 -> 0 7 -> 2 8 -> 0 9 -> 2 10 -> 0 11 -> 0
Below are counts of all elements 1 -> 0 2 -> 0 3 -> 11 4 -> 0 5 -> 0 6 -> 0 7 -> 0 8 -> 0 9 -> 0 10 -> 0 11 -> 0
Below are counts of all elements 1 -> 1 2 -> 1 3 -> 1 4 -> 1 5 -> 1 6 -> 1 7 -> 1 8 -> 1 9 -> 1 10 -> 1 11 -> 1
Below are counts of all elements 1 -> 1 2 -> 1 3 -> 1 4 -> 1 5 -> 1 6 -> 1 7 -> 1 8 -> 1 9 -> 1 10 -> 1 11 -> 1
**Time Complexity: O(n). As a single traversal of the array takes O(n) time.
**Auxiliary Space: O(1). Since no extra space is needed.
**Efficient Approach: By adding 'n' to keep track of counts.
All the array elements are from **1 to **n. Reduce every element by 1. The array elements now are in range **0 to n-1 so it can be said that for every index i, **floor(array[i]/n) = 0**.
So as previously said that the idea is to traverse the given array, use elements as an index and store their counts at the index. Consider the basic approach, a Hashmap of size n is needed and the array is also of size n. So the array can be used as a hashmap but the difference is that instead of replacing elements add n to the **array[i] th index.
So after updating the **array[i]%n will give the ith element and **array[i]/n will give the frequency of **i+1.
Step-by-step approach:
- Traverse the array from start to end and reduce all the elements by 1.
- Again traverse the array from start to end.
- For each element _array[i]%n update _array[array[i]%n], i.e _array[array[i]%n]+n
- Traverse the array from start to end and print _i + 1 as value and _array[i]/n as frequency.
Below is the implementation of the above approach:
C++ `
// C++ program to print frequencies of all array // elements in O(1) extra space and O(n) time #include<bits/stdc++.h> using namespace std;
// Function to find counts of all elements present in // arr[0..n-1]. The array elements must be range from // 1 to n void printfrequency(int arr[],int n) { // Subtract 1 from every element so that the elements // become in range from 0 to n-1 for (int j =0; j<n; j++) arr[j] = arr[j]-1;
// Use every element arr[i] as index and add 'n' to
// element present at arr[i]%n to keep track of count of
// occurrences of arr[i]
for (int i=0; i<n; i++)
arr[arr[i]%n] = arr[arr[i]%n] + n;
// To print counts, simply print the number of times n
// was added at index corresponding to every element
for (int i =0; i<n; i++)
cout << i + 1 << " -> " << arr[i]/n << endl;}
// Driver program to test above function int main() { int arr[] = {2, 3, 3, 2, 5}; int n = sizeof(arr)/sizeof(arr[0]); printfrequency(arr,n); return 0; }
Java
class CountFrequency { // Function to find counts of all elements present in // arr[0..n-1]. The array elements must be range from // 1 to n void printfrequency(int arr[], int n) { // Subtract 1 from every element so that the elements // become in range from 0 to n-1 for (int j = 0; j < n; j++) arr[j] = arr[j] - 1;
// Use every element arr[i] as index and add 'n' to
// element present at arr[i]%n to keep track of count of
// occurrences of arr[i]
for (int i = 0; i < n; i++)
arr[arr[i] % n] = arr[arr[i] % n] + n;
// To print counts, simply print the number of times n
// was added at index corresponding to every element
for (int i = 0; i < n; i++)
System.out.println(i + 1 + "->" + arr[i] / n);
}
// Driver program to test above functions
public static void main(String[] args)
{
CountFrequency count = new CountFrequency();
int arr[] = {2, 3, 3, 2, 5};
int n = arr.length;
count.printfrequency(arr, n);
}}
// This code has been contributed by Mayank Jaiswal
Python3
Python 3 program to print frequencies
of all array elements in O(1) extra
space and O(n) time
Function to find counts of all elements
present in arr[0..n-1]. The array
elements must be range from 1 to n
def printfrequency(arr, n):
# Subtract 1 from every element so that
# the elements become in range from 0 to n-1
for j in range(n):
arr[j] = arr[j] - 1
# Use every element arr[i] as index
# and add 'n' to element present at
# arr[i]%n to keep track of count of
# occurrences of arr[i]
for i in range(n):
arr[arr[i] % n] = arr[arr[i] % n] + n
# To print counts, simply print the
# number of times n was added at index
# corresponding to every element
for i in range(n):
print(i + 1, "->", arr[i] // n)Driver code
arr = [2, 3, 3, 2, 5] n = len(arr) printfrequency(arr, n)
This code is contributed
by Shrikant13
C#
using System;
internal class CountFrequency { // Function to find counts of all elements present in // arr[0..n-1]. The array elements must be range from // 1 to n internal virtual void printfrequency(int[] arr, int n) { // Subtract 1 from every element so that the elements // become in range from 0 to n-1 for (int j = 0; j < n; j++) { arr[j] = arr[j] - 1; }
// Use every element arr[i] as index and add 'n' to
// element present at arr[i]%n to keep track of count of
// occurrences of arr[i]
for (int i = 0; i < n; i++)
{
arr[arr[i] % n] = arr[arr[i] % n] + n;
}
// To print counts, simply print the number of times n
// was added at index corresponding to every element
for (int i = 0; i < n; i++)
{
Console.WriteLine(i + 1 + "->" + arr[i] / n);
}
}
// Driver program to test above functions
public static void Main(string[] args)
{
CountFrequency count = new CountFrequency();
int[] arr = new int[] {2, 3, 3, 2, 5};
int n = arr.Length;
count.printfrequency(arr, n);
}}
// This code is contributed by Shrikant13
JavaScript
PHP
`
Output
1 -> 0 2 -> 2 3 -> 2 4 -> 0 5 -> 1
**Time Complexity: O(n). Only two traversals of the array are needed and a single traversal of the array takes O(n) time.
**Auxiliary Space: O(1). Since no extra space is needed.