Restore a shuffled Queue as per given Conditions

Given N people standing in a queue and two arrays A[] and B[]. The array A[] represent the name of the person and array B[] represents how many people are taller than a particular person standing in front of that. Now the queue is shuffled. The task is to print the original sequence of the queue following the above property.
Examples:

Input: N = 4, A[] = {‘a’, ‘b’, ‘c’, ‘d’}, B[] = {0, 2, 0, 0} 
Output: 
a 1 
c 3 
d 4 
b 2 
Explanation: 
Looking at the output queue and their generated heights, it can be easily understood that: 
1) a is the first one in the queue and so we have the person with 0th index in front of him. So a is associated with 0 in the input. 
2) c has only a in front of him/her but a is shorter than c. Therefore c is associated with 0 in the input. 
3) d has c and a in front of him/her but they are both shorter than d . Therefore d is associated with 0 in the input. 
4) b has d, c and a in front of b. But only c and d are taller than b. So, b is associated with 2 in the input.

Input: N = 4, A[] = { ‘a’, ‘b’, ‘c’, ‘d’}, B[] = { 0, 1, 3, 3} 
Output: -1 
Explanation: 
The given order is the original order of the queue.

Approach:

  • Firstly make a pair of the person’s name and their associated integers and sort the pairs.
  • Create an array answer[] to store the possible heights of the person.
  • Iterate over all the pair and if the number of persons standing in front of is taller and is greater than their current standing position, then return -1.
  • Otherwise, store the difference between the current standing position and the height of the person taller than him, in the answer array.
  • For every person iterate over the pair and if the value of the answer array for our current person is greater than the person with whom we are comparing, increment in the answer array for current pair.
  • Finally, print the possible pairs from the given sequence according to values stored in answer[] array.

Below is the implementation of the above approach:



C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to generate the Queue
void OriginalQueue(char A[], int B[],
                   int N)
{
    // Making a pair
    pair<int, string> a[N + 1];
  
    // Answer array
    int ans[N + 1];
    bool possible = true;
  
    // Store the values in the pair
    for (int i = 0; i < N; i++) {
        a[i].second = A[i];
        a[i].first = B[i];
    }
  
    // Sort the pair
    sort(a, a + N);
  
    // Traverse the pairs
    for (int i = 0; i < N; i++) {
  
        int len = i - a[i].first;
  
        // If it is not possible to
        // generate the Queue
        if (len < 0) {
  
            cout << "-1";
            possible = false;
        }
  
        if (!possible)
            break;
        else {
            ans[i] = len;
  
            for (int j = 0; j < i; j++) {
  
                // Increment the answer
                if (ans[j] >= ans[i])
                    ans[j]++;
            }
        }
  
        // Finally printing the answer
        if (i == N - 1 && possible) {
            for (int i = 0; i < N; i++) {
                cout << a[i].second << " "
                     << ans[i] + 1 << endl;
            }
        }
    }
}
  
// Driver Code
int main()
{
    int N = 4;
  
    // Given name of person as char
    char A[N] = { 'a', 'b', 'c', 'd' };
  
    // Associated integers
    int B[N] = { 0, 2, 0, 0 };
  
    // Function Call
    OriginalQueue(A, B, N);
    return 0;
}

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// Java program for the above approach 
import java.util.*;
import java.io.*;
  
class GFG{
      
// Function to generate the Queue
static void OriginalQueue(char A[], int B[],
                                    int N)
{
    // Making a pair
    int[][] a = new int[N][2];
  
    // Answer array
    int[] ans = new int[N];
    boolean possible = true;
  
    // Store the values in the pair
    for(int i = 0; i < N; i++)
    {
        a[i][0] = B[i];
        a[i][1] = (int)A[i];
    }
  
    // Sort the pair
    Arrays.sort(a, (o1, o2) -> o1[0] - o2[0]);
      
    // Traverse the pairs
    for(int i = 0; i < N; i++)
    {
        int len = i - a[i][0];
  
        // If it is not possible to
        // generate the Queue
        if (len < 0)
        {
            System.out.print("-1");
            possible = false;
        }
  
        if (!possible)
            break;
        else 
        {
            ans[i] = len;
  
            for(int j = 0; j < i; j++)
            {
  
                // Increment the answer
                if (ans[j] >= ans[i])
                    ans[j]++;
            }
        }
  
        // Finally printing the answer
        if (i == N - 1 && possible)
        {
            for(int k = 0; k < N; k++)
            {
                System.out.println((char)a[k][1] +
                                 " "+ (ans[k] + 1));
            }
        }
    }
}
  
// Driver Code
public static void main (String[] args)
{
    int N = 4;
      
    // Given name of person as char
    char A[] = { 'a', 'b', 'c', 'd' };
      
    // Associated integers
    int B[] = { 0, 2, 0, 0 };
      
    // Function Call
    OriginalQueue(A, B, N);
}
}
  
// This code is contributed by offbeat

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Output: 

a 1
c 3
d 4
b 2

Time Complexity: O(N2
Auxiliary Space: O(N)
 

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