Hello Everyone,
You are given an array of 0s and 1s in random order. Segregate 0s on left side and 1s on right side of the array. Traverse array only once.
Input array = [0, 1, 0, 1, 0, 0, 1, 1, 1, 0]
Output array = [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]
Method (Count 0s or 1s)
- Count the number of 0s. Let count be C.
- Once we have count, we can put C 0s at the beginning and 1s at the remaining n – C positions in array.
Time Complexity : O(n)
// C++ code to Segregate 0s and 1s in an array
#include <bits/stdc++.h>
using namespace std;
// Function to segregate 0s and 1s
void segregate0and1( int arr[], int n)
{
int count = 0; // Counts the no of zeros in arr
for ( int i = 0; i < n; i++) {
if (arr[i] == 0)
count++;
}
// Loop fills the arr with 0 until count
for ( int i = 0; i < count; i++)
arr[i] = 0;
// Loop fills remaining arr space with 1
for ( int i = count; i < n; i++)
arr[i] = 1;
}
// Function to print segregated array
void print( int arr[], int n)
{
cout << "Array after segregation is " ;
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
}
// Driver function
int main()
{
int arr[] = { 0, 1, 0, 1, 1, 1 };
int n = sizeof (arr) / sizeof (arr[0]);
segregate0and1(arr, n);
print(arr, n);
return 0;
}
Output :
Array after segregation is 0 0 1 1 1 1