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
Live Demo
#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
Published on 28-Jul-2020 14:40:54
