Arrange numbers to form a valid sequence

Given an array arr[] with N distinct numbers and another array arr1[] with N-1 operators (either < or >), the task is to organize the numbers to form a valid sequence which obeys relational operator rules with respect to provided operators.

Examples:

Input: arr[] = {3, 12, 7, 8, 5}; arr1= {‘<‘, ‘>’, ‘>’, ‘<‘}
Output: {3, 12, 8, 5, 7}
Explanation:
3 < 12 > 8 > 5 < 7
There can be more such combinations. The task is to return one of the combinations.

Input: arr[] = {8, 2, 7, 1, 5, 9}; arr1[] = {‘>’, ‘>’, ‘<‘, ‘>’, ‘<‘}
Output:{9, 8, 1, 7, 2, 5}
Explanation:
9 > 8 > 1 < 7 > 2 < 5

Naive Approach:
A naive approach is to try all different possible arrangement of numbers and check if the sequence is valid.



Time Complexity: O(2N).

Efficient Approach: The idea is to first sort the given array of numbers in ascending order. Then solve the problem using two pointers technique: one pointing at the front and other pointing at the end.

  1. Take one resultant array of size same as given array.
  2. If the current operator is ‘<‘ then include the element which the top pointer is pointing to in resultant array and increment it by 1.
  3. If the current operator is ‘>’ then include the element which the last pointer is pointing to in resultant array and decrement it by 1.

Below is the implementation of the above approach.

C++

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// C++ implemenattion of the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to organize the given numbers
// to form a valid sequence.
vector<int> orgazineInOrder(vector<int> vec,
                            vector<int> op, int n)
{
    vector<int> result(n);
    // Sorting the array
    sort(vec.begin(), vec.end());
  
    int i = 0, j = n - 1, k = 0;
    while (i <= j && k <= n - 2) {
        // Two pointer technique
        // to organize the numbers
        if (op[k] == '<') {
            result[k] = vec[i++];
        }
        else {
            result[k] = vec[j--];
        }
        k++;
    }
    result[n - 1] = vec[i];
  
    return result;
}
  
// Driver code
int main()
{
    vector<int> vec({ 8, 2, 7, 1, 5, 9 });
  
    vector<int> op({ '>', '>', '<',
                     '>', '<' });
  
    vector<int> result
        = orgazineInOrder(vec,
                          op, vec.size());
  
    for (int i = 0; i < result.size(); i++) {
        cout << result[i] << " ";
    }
    return 0;
}

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Java

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// Java implemenattion of the above approach
import java.util.*;
  
class GFG
{
  
// Function to organize the given numbers
// to form a valid sequence.
static int[] orgazineInOrder(int []vec,int[] op, int n)
{
    int []result = new int[n];
      
    // Sorting the array
    Arrays.sort(vec);
  
    int i = 0, j = n - 1, k = 0;
    while (i <= j && k <= n - 2
    {
        // Two pointer technique
        // to organize the numbers
        if (op[k] == '<'
        {
            result[k] = vec[i++];
        }
        else
        {
            result[k] = vec[j--];
        }
        k++;
    }
    result[n - 1] = vec[i];
  
    return result;
}
  
// Driver code
public static void main(String[] args)
{
    int []vec ={ 8, 2, 7, 1, 5, 9 };
  
    int[] op ={ '>', '>', '<',
                    '>', '<' };
  
    int []result = orgazineInOrder(vec,
                        op, vec.length);
  
    for (int i = 0; i < result.length; i++) 
    {
        System.out.print(result[i]+ " ");
    }
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python3 implemenattion of the above approach 
  
# Function to organize the given numbers 
# to form a valid sequence. 
def orgazineInOrder(vec, op, n) : 
  
    result = [0] * n;
      
    # Sorting the array
    vec.sort();
    i = 0;
    j = n - 1;
    k = 0
      
    while (i <= j and k <= n - 2) :
          
        # Two pointer technique 
        # to organize the numbers 
        if (op[k] == '<') :
            result[k] = vec[i]; 
            i += 1;
          
        else :
            result[k] = vec[j]; 
            j -= 1;
          
        k += 1
  
    result[n - 1] = vec[i]; 
  
    return result; 
  
# Driver code 
if __name__ == "__main__"
  
    vec = [ 8, 2, 7, 1, 5, 9 ];
    op = [ '>', '>', '<', '>', '<' ]; 
  
    result = orgazineInOrder(vec, op, len(vec)); 
  
    for i in range(len(result)) :
        print(result[i], end = " "); 
  
# This code is contributed by AnkitRai01

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C#

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// C# implementation of the approach 
using System;
  
class GFG 
      
    // Function to organize the given numbers 
    // to form a valid sequence. 
    static int[] orgazineInOrder(int []vec,int[] op, int n) 
    
        int []result = new int[n]; 
          
        // Sorting the array 
        Array.Sort(vec); 
      
        int i = 0, j = n - 1, k = 0; 
        while (i <= j && k <= n - 2) 
        
            // Two pointer technique 
            // to organize the numbers 
            if (op[k] == '<'
            
                result[k] = vec[i++]; 
            
            else
            
                result[k] = vec[j--]; 
            
            k++; 
        
        result[n - 1] = vec[i]; 
      
        return result; 
    
      
    // Driver code 
    public static void Main() 
    
        int []vec ={ 8, 2, 7, 1, 5, 9 }; 
      
        int[] op ={ '>', '>', '<'
                        '>', '<' }; 
      
        int []result = orgazineInOrder(vec, 
                            op, vec.Length); 
      
        for (int i = 0; i < result.Length; i++) 
        
            Console.Write(result[i] + " "); 
        
    
  
// This code is contributed by AnkitRai01 

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Output:

9 8 1 7 2 5

Time Complexity: O(NlogN)

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Improved By : 29AjayKumar, AnkitRai01