dutch national flag problem

Initially all array values are treated as unpartitioned: [unpartitioned values] Array split into four parts as the iteration proceeds: [ values < mid | values = mid | unpartitioned values | values > mid ] procedure three-way-partition(A : array of values, mid : value): i ← 0 j ← 0 k ← size of A - 1 while j <= k: if A[j] < mid: swap A[i] and A[j] i ← i + 1 j ← j + 1 else if A[j] > mid: swap A[j] and A[k] k ← k - 1 else: j ← j + 1

Approach: Count the number of 0s, 1s and 2s in the given array. Then store all the 0s in the beginning followed by all the 1s then all the 2s. Algorithm: 1)Keep three counter c0 to count 0s, c1 to count 1s and c2 to count 2s 2)Traverse through the array and increase the count of c0 is the element is 0,increase the count of c1 if the element is 1 and increase the count of c2 if the element is 2 3)Now again traverse the array and replace first c0 elements with 0, next c1 elements with 1 and next c2 elements with 2. //GfG PseudoCode

dutch national flag algorithm

#include <bits/stdc++.h>

using namespace std;


void dutch(int arr[],int n)
{
  int low=0;
  int mid=0;
  int high=n-1;
  while(mid<=high)
  {
     switch(arr[mid])
     {
     case 0:
        {
            swap(arr[low++],arr[mid++]);
            break;
        }
     case 1:
         {
             mid++;
             break;
         }
     case 2:
        {
            swap(arr[high--],arr[mid]);
            break;
        }
     }
  }
}
int main()
{
    int n;
    cout<<"Enter the size of the array:"<<endl;
    cin>>n;
    int arr[n];
    cout<<"Enter the elements of the array:"<<endl;
    for(int i=0;i<n;i++)
    {
        cin>>arr[i];
    }
    dutch(arr,n);
    cout<<"our sorted array is:"<<endl;
    for(int i=0;i<n;i++)
    {
        cout<<arr[i]<<" ";
    }
    cout<<endl;
    return 0;
}

Posted by: Guest on June-19-2021

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