Print k different sorted permutations of a given array in C Program.

CServer Side ProgrammingProgramming

Given an array a[] containing N integers, the challenge is to print k different permutations of indices such that the values at those indices form a non-decreasing sequence. Print -1 if it is not possible.

Example

Input: arr[] = {2,5,6,2,2,2,2}, k = 4
Output:
   0 3 4 5 6 1 2
   3 0 4 5 6 1 2
   0 3 4 5 6 1 2
   3 0 4 5 6 1 2

Sort the given array and keep track of the original indices of each element. That gives one required permutation. Now if any 2 continuous elements are equal then they can be swapped to get another permutation. Similarly, the third permutation can be generated.

Algorithm

START
Step 1 -> Declare Function void indice(int n, pair<int, int> array[])
   Loop For int i=0 and i<n and i++
      Print array[i].second
   End
Step 2 -> Declare Function void permutation(int n, int a[], int k)
   Use STL pair<int, int> arr[n]
   Loop for int i=0 and i<n and i++
      Set arr[i].first = a[i]
      Set arr[i].second = i
   End
   Call sort(arr, arr + n)
   Declare int count to 1
   Loop For int i=1 and i<n and i++
      IF (arr[i].first == arr[i - 1].first)
         Increment count by 1
      End
   End
   IF count < k
      Return -1
   End
   Loop For int i = 0 and i < k – 1 and i++
      Call indice(n, arr)
      Loop For int j = 1 and j < n and j++
         IF arr[j].first == arr[j - 1].first
            Call swap(arr[j], arr[j - 1])
            Break
         End
      End
   End
   Call indice(n, arr)
Step 3 -> In main()
   Declare array a[]={2,5,6,2,2,2,2}
   Declare int n= sizeof(a)/sizeof(a[0])
   Declare int k=4
   Call permutation(n,a,k)
STOP

Example

#include <bits/stdc++.h>
using namespace std;
void indice(int n, pair<int, int> array[]){
   for (int i = 0; i < n; i++)
      cout << array[i].second << " ";
   cout << endl;
}
void permutation(int n, int a[], int k){
   pair<int, int> arr[n];
   for (int i = 0; i < n; i++){
      arr[i].first = a[i];
      arr[i].second = i;
   }
   sort(arr, arr + n);
   int count = 1;
   for (int i = 1; i < n; i++)
      if (arr[i].first == arr[i - 1].first)
         count++;
   if (count < k){
      cout << "-1";
      return;
   }
   for (int i = 0; i < k - 1; i++){
      indice(n, arr);
      for (int j = 1; j < n; j++){
         if (arr[j].first == arr[j - 1].first){
            swap(arr[j], arr[j - 1]);
            break;
         }
      }
   }
   indice(n, arr);
}
int main(){
   int a[] ={2,5,6,2,2,2,2};
   int n = sizeof(a) / sizeof(a[0]);
   int k = 4;
   permutation(n, a, k);
   return 0;
}

Output

if we run above program then it will generate following output

0 3 4 5 6 1 2
3 0 4 5 6 1 2
0 3 4 5 6 1 2
3 0 4 5 6 1 2
raja
Published on 22-Aug-2019 06:49:37
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