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Largest Derangement of a Sequence

Given any sequence  S = {a_1, a_2 dots a_n } , find the largest derangement of  S .

A derangement  D is any permutation of  S , such that no two elements at the same position in  S and  D are equal.

The Largest Derangement is such that  D_i > D_{i+1}, forall i .

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, 23, 21


Since we are interested in generating largest derangement, we start putting larger elements in more significant positions.

Start from left, at any position  i place the next largest element among the values of the sequence which have not yet been placed in positions before  i .

To scan all positions takes N iteration. In each iteration we are required to find a maximum numbers, so a trivial implementation would be  O(N^2) complexity,

However if we use a data structure like max-heap to find the maximum element, then the complexity reduces to  O(N * log{N})

Below is C++ implementation.

// CPP program to find the largest derangement
#include <bits/stdc++.h>
using namespace std;
  
void printLargest(int seq[], int N)
{
    int res[N]; // Stores result
  
    // Insert all elements into a priority queue
    std::priority_queue<int> 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 poped 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(" Largest Derangement ");
    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;
}

Output:

Sequence: 
92 3 52 13 2 31 1 
Largest Derangement 
52 92 31 3 13 1 2 

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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