Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Count frequencies of all elements in array in O(1) extra space and O(n) time in C++
We are given with an array of elements ranging from values 1 to n. Some elements are repeated, and some are missing. The goal is to find the frequencies of all elements in O(n) time and O(1) extra space.
Input
Arr[]= { 1,2,2,3,4,4,4,5 }Output
1→ 1, 2 → 2, 3→ 1, 4→ 3, 5→ 5
Explanation − The highest element is 5, the output shows the number of times each element appears in the array.
Input
Arr[]= { 1,4,4,5,5,5,5 }Output
1→ 1, 2 →0, 3→ 0, 4→ 2, 5→ 4
Explanation − The highest element is 5, the output shows the number of times each element appears in the array.
Approach used in the below program is as follows
Below program works for arrays with numbers between 1 to 10.
Function printfrequency(int arr[],int n) takes an array and its size n as input and returns the count of numbers between 1 to 10 present in the array.
We make arr[i]=arr[i]-1 so that each index stores the frequency of number i, 1 stored at index 0 and so on.
Using for loop at each frequency arr[arr[i]%10] add 10 to the original value.
For x times 10 will be added if number i occurs x times in array.
Now using for loop print frequencies using arr[i]/10 for all elements i+1 at index i.
Example
#include<bits/stdc++.h>
using namespace std;
void printfrequency(int arr[],int n){
int i=0;
//1 becomes 0, 2 becomes 1 .....10 becomes 9 so arr[i] will have count of i
for ( i =0; i<n; i++)
arr[i] = arr[i]-1;
//as numbers are between 1-10 add 10 to all (num%10 is num itself)
for (i=0; i<n; i++)
arr[arr[i]%10] = arr[arr[i]%10] + 10;
for (i =0; i<10; i++)
cout << i + 1 << " -> " << arr[i]/10 << endl;
}
int main(){
int arr[] = {2, 3, 3, 2, 5, 6, 7, 7, 7, 8, 8, 9, 9};
int n = sizeof(arr)/sizeof(arr[0]);
printfrequency(arr,n);
return 0;
}Output
1 -> 0 2 -> 2 3 -> 2 4 -> 0 5 -> 1 6 -> 1 7 -> 3 8 -> 2 9 -> 5 10 -> 0
966 Views
