Find the non decreasing order array from given array

Given an array A[] of size N / 2, the task is to construct the array B[] of size N such that: 
 

1. B[] is sorted in non-decreasing order.
2. A[i] = B[i] + B[n - i + 1].

Note: Array A[] is given in such a way that the answer is always possible.
Examples: 
 

Input: A[] = {3, 4} 
Output: 0 1 3 3
Input: A[] = {4, 1} 
Output: 0 0 1 4 
 

Approach: Let's present the following greedy approach. The numbers will be restored in pairs (B[0], B[n - 1]), (B[1], B[n - 2]) and so on. Thus, we can have some limits on the values of the current pair (satisfying the criteria about sorted result). 
Initially, l = 0 and r = 10⁹, they are updated with l = a[i] and r = a[n - i + 1]. Let l be minimal possible in the answer. Take a[i] = max(l, b[i] - r) and r = b[i] - l, that way l was chosen in such a way that both l and r are within the restrictions and l is also minimal possible. 
If l was any greater than we would move both l limit up and r limit down leaving less freedom for later choices.
Below is the implementation of the above approach: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to print 
// the contents of the array
void printArr(int b[], int n)
{
    for (int i = 0; i < n; i++)
        cout << b[i] << " ";
}

// Function to build array B[]
void ModifiedArray(int a[], int n)
{
    // Lower and upper limits
    int l = 0, r = INT_MAX;

    // To store the required array
    int b[n] = { 0 };

    // Apply greedy approach
    for (int i = 0; i < n / 2; i++) {
        b[i] = max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }

    // Print the built array b[]
    printArr(b, n);
}

// Driver code
int main()
{
    int a[] = { 5, 6 };
    int n = sizeof(a) / sizeof(a[0]);
    ModifiedArray(a, 2 * n);

    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class solution
{
    // Utility function to print 
    // the contents of the array
    void printArr(int b[], int n)
    {
        for (int i = 0; i < n; i++)
        {
             System.out.print(" " + b[i] + " ");
         }
    }

    // Function to build array B[]
    void ModifiedArray(int a[], int n)
    {
        // Lower and upper limits
        int l = 0, r = Integer.MAX_VALUE;

        // To store the required array
        int[] b = new int[n];
        
    // Apply greedy approach
    for (int i = 0; i < n / 2; i++) {
        b[i] = Math.max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }

    // Print the built array b[]
    printArr(b, n);
}    
// Driver code
public static void main(String args[])
{
   int a[] = { 5, 6 };
   int n = a.length ;
   solution s=new solution();
   s.ModifiedArray(a, 2 * n);

}
}
//This code is contributed by Shivi_Aggarwal

# Python 3 implementation of the approach
import sys

# Utility function to print the 
# contents of the array
def printArr(b, n):
    for i in range(0, n, 1):
        print(b[i], end = " ")

# Function to build array B[]
def ModifiedArray(a, n):
    
    # Lower and upper limits
    l = 0
    r = sys.maxsize

    # To store the required array
    b = [0 for i in range(n)] 

    # Apply greedy approach
    for i in range(0, int(n / 2), 1):
        b[i] = max(l, a[i] - r)
        b[n - i - 1] = a[i] - b[i]
        l = b[i]
        r = b[n - i - 1]

    # Print the built array b[]
    printArr(b, n)

# Driver code
if __name__ == '__main__':
    a = [5, 6]
    n = len(a)
    ModifiedArray(a, 2 * n)

# This code is contributed by
# Shashank_Sharma

// C# implementation of the approach

using System;

public class GFG{

// Utility function to print 
// the contents of the array
static void printArr(int []b, int n)
    {
        for (int i = 0; i < n; i++)
        {
            Console.Write(" " + b[i] + " ");
        }
    }

    // Function to build array B[]
static    void ModifiedArray(int []a, int n)
    {
        // Lower and upper limits
        int l = 0, r = int.MaxValue;

        // To store the required array
        int[] b = new int[n];
        
    // Apply greedy approach
    for (int i = 0; i < n / 2; i++) {
        b[i] = Math.Max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }

    // Print the built array b[]
    printArr(b, n);
} 

        // Driver code
    static public void Main (){
    int []a = { 5, 6 };
    int n = a.Length;
    ModifiedArray(a, 2 * n);
    }
}
// This code is contributed
// by Sach_Code

<?php
// PHP implementation of the approach

// Utility function to print the 
// contents of the array
function printArr($b, $n)
{
    for ($i = 0; $i < $n; $i++)
        echo $b[$i] . " ";
}

// Function to build array B[]
function ModifiedArray($a, $n)
{
    // Lower and upper limits
    $l = 0; $r = PHP_INT_MAX;

    // To store the required array
    $b = array(0);

    // Apply greedy approach
    for ($i = 0; $i < $n / 2; $i++) 
    {
        $b[$i] = max($l, $a[$i] - $r);
        $b[$n - $i - 1] = $a[$i] - $b[$i];
        $l = $b[$i];
        $r = $b[$n - $i - 1];
    }

    // Print the built array b[]
    printArr($b, $n);
}

// Driver code
$a = array( 5, 6 );
$n = sizeof($a);
ModifiedArray($a, 2 * $n);

// This code is contributed
// by Akanksha Rai
?>

<script>

// Javascript program of the above approach

    // Utility function to print 
    // the contents of the array
    function printArr(b, n)
    {
        for (let i = 0; i < n; i++)
        {
             document.write(" " + b[i] + " ");
         }
    }

    // Function to build array B[]
    function ModifiedArray(a, n)
    {
        // Lower and upper limits
        let l = 0, r = Number.MAX_VALUE;

        // To store the required array
        let b = Array(n).fill(0);
        
    // Apply greedy approach
    for (let i = 0; i < n / 2; i++) {
        b[i] = Math.max(l, a[i] - r);
        b[n - i - 1] = a[i] - b[i];
        l = b[i];
        r = b[n - i - 1];
    }

    // Print the built array b[]
    printArr(b, n);
} 

// Driver code

    let a = [ 5, 6 ];
   let n = a.length ;
   ModifiedArray(a, 2 * n);
    
</script>

0 1 5 5

Time Complexity: O(n)

Auxiliary Space: O(n)

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Article Tags :

• Greedy
• Mathematical
• Competitive Programming
• DSA
• programming-puzzle

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Practice Tags :

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