Arrange numbers to form a valid sequence

0 0 4 minutes read

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(2^N).

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.

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

// C++ implementation 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;
}

Java

// Java implementation 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

Python3

# Python3 implementation 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

C#

// 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 

Javascript

<script>
 
// Javascript implementation of the above approach
 
// Function to organize the given numbers
// to form a valid sequence.
function orgazineInOrder(vec, op, n)
{
    let result = [n];
     
    // Sorting the array
    vec.sort();
 
    let 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
let vec = [ 8, 2, 7, 1, 5, 9 ];
let op = [ '>', '>', '<', '>', '<' ];
 
let result = orgazineInOrder(vec,
                             op, vec.length);
 
for(let i = 0; i < result.length; i++) 
{
   document.write(result[i] + " ");
}
 
// This code is contributed by sravan kumar 
 
</script>

Output:

9 8 1 7 2 5

Time Complexity: O(NlogN)

Auxiliary Space: O(N)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!