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# Largest Derangement of a Sequence

Given any sequence, find the largest derangement of .
A derangement is any permutation of, such that no two elements at the same position in and are equal.
The Largest Derangement is such that.

Examples:

Input : seq[] = {5, 4, 3, 2, 1}
Output : 4 5 2 1 3

Input : seq[] = {56, 21, 42, 67, 23, 74}
Output : 74, 67, 56, 42, 21, 23

Since we are interested in generating the largest derangement, we start putting larger elements in more significant positions.
Start from left, at any position place the next largest element among the values of the sequence which have not yet been placed in positions before.
To scan all positions takes N iteration. In each iteration we are required to find a maximum number, so a trivial implementation would be complexity,
However, if we use a data structure like max-heap to find the maximum element, then the complexity reduces to

Below is the implementation.

## C++

 // C++ program to find the largest derangement #include using namespace std;   void printLargest(int seq[], int N) {     int res[N]; // Stores result       // Insert all elements into a priority queue     std::priority_queue pq;     for (int i = 0; i < N; i++)         pq.push(seq[i]);          // Fill Up res[] from left to right     for (int i = 0; i < N; i++) {         int d = pq.top();         pq.pop();         if (d != seq[i] || i == N - 1) {             res[i] = d;         } else {               // New Element popped equals the element             // in original sequence. Get the next             // largest element             res[i] = pq.top();             pq.pop();             pq.push(d);         }     }       // If given sequence is in descending order then     // we need to swap last two elements again     if (res[N - 1] == seq[N - 1]) {         res[N - 1] = res[N - 2];         res[N - 2] = seq[N - 1];     }       printf("\nLargest Derangement \n");     for (int i = 0; i < N; i++)         printf("%d ", res[i]); }   // Driver code int main() {     int seq[] = { 92, 3, 52, 13, 2, 31, 1 };     int n = sizeof(seq)/sizeof(seq[0]);     printLargest(seq, n);     return 0; }

## Java

 // Java program to find the largest derangement import java.io.*; import java.util.Collections; import java.util.PriorityQueue;   class GFG{       public static void printLargest(int a[],int n) {      PriorityQueue pq = new PriorityQueue<>(          Collections.reverseOrder());               // Insert all elements into a priority queue       for(int i = 0; i < n; i++)     {         pq.add(a[i]);     }           // Stores result       int res[] = new int[n];               // Fill Up res[] from left to right     for(int i = 0; i < n; i++)     {         int p = pq.peek();         pq.remove();                   if (p != a[i] || i == n - 1)         {             res[i] = p;         }         else         {                           // New Element popped equals the element             // in original sequence. Get the next             // largest element             res[i] = pq.peek();             pq.remove();             pq.add(p);         }     }             // If given sequence is in descending       // order then we need to swap last two       // elements again       if (res[n - 1] == a[n - 1])     {         res[n - 1] = res[n - 2];         res[n - 2] = a[n - 1];     }           System.out.println("Largest Derangement");     for(int i = 0; i < n; i++)     {         System.out.print(res[i] + " ");     } }   // Driver code public static void main(String[] args) {       int n = 7;     int seq[] = { 92, 3, 52, 13, 2, 31, 1 };             printLargest(seq, n); } }   // This code is contributed by aditya7409

## Python3

 # Python3 program to find the largest derangement def printLargest(seq, N) :       res = [0]*N # Stores result         # Insert all elements into a priority queue     pq = []     for i in range(N) :         pq.append(seq[i])           # Fill Up res[] from left to right     for i in range(N) :            pq.sort()         pq.reverse()         d = pq[0]         del pq[0]         if (d != seq[i] or i == N - 1) :             res[i] = d                else :                        # New Element popped equals the element             # in original sequence. Get the next             # largest element             res[i] = pq[0]             del pq[0]             pq.append(d)         # If given sequence is in descending order then     # we need to swap last two elements again     if (res[N - 1] == seq[N - 1]) :            res[N - 1] = res[N - 2]         res[N - 2] = seq[N - 1]                print("Largest Derangement")     for i in range(N) :         print(res[i], end = " ")   # Driver code seq = [ 92, 3, 52, 13, 2, 31, 1 ] n = len(seq) printLargest(seq, n)   # This code is contributed by divyesh072019.

## C#

 // C# program to find the largest derangement using System; using System.Collections.Generic; class GFG {           static void printLargest(int[] seq, int N)     {         int[] res = new int[N]; // Stores result                // Insert all elements into a priority queue         List pq = new List();         for (int i = 0; i < N; i++)             pq.Add(seq[i]);                   // Fill Up res[] from left to right         for (int i = 0; i < N; i++)         {             pq.Sort();             pq.Reverse();             int d = pq[0];             pq.RemoveAt(0);             if (d != seq[i] || i == N - 1)             {                 res[i] = d;             }           else             {                        // New Element popped equals the element                 // in original sequence. Get the next                 // largest element                 res[i] = pq[0];                 pq.RemoveAt(0);                 pq.Add(d);             }         }                // If given sequence is in descending order then         // we need to swap last two elements again         if (res[N - 1] == seq[N - 1])         {             res[N - 1] = res[N - 2];             res[N - 2] = seq[N - 1];         }             Console.WriteLine("Largest Derangement");         for (int i = 0; i < N; i++)             Console.Write(res[i] + " ");     }     // Driver code   static void Main()   {     int[] seq = { 92, 3, 52, 13, 2, 31, 1 };     int n = seq.Length;     printLargest(seq, n);   } }   // This code is contributed by divyeshrabadiya07

## Javascript



Output

Largest Derangement
52 92 31 3 13 1 2

Time Complexity: O(n log n)
Auxiliary Space: O(N), because, we use an N size array to store results.

Note:

The method can be easily modified to obtain the smallest derangement as well.
Instead of a Max Heap, we should use a Min Heap to consecutively get minimum elements

This article is contributed by Sayan Mahapatra. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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