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Construct an array from its pair-sum array
  • Difficulty Level : Medium
  • Last Updated : 07 May, 2021

Given a pair-sum array and size of the original array (n), construct the original array.
A pair-sum array for an array is the array that contains sum of all pairs in ordered form. For example pair-sum array for arr[] = {6, 8, 3, 4} is {14, 9, 10, 11, 12, 7}.
In general, pair-sum array for arr[0..n-1] is {arr[0]+arr[1], arr[0]+arr[2], ……., arr[1]+arr[2], arr[1]+arr[3], ……., arr[2]+arr[3], arr[2]+arr[4], …., arr[n-2]+arr[n-1}.
“Given a pair-sum array, construct the original array.”
We strongly recommend to minimize your browser and try this yourself.
Let the given array be “pair[]” and let there be n elements in original array. If we take a look at few examples, we can observe that arr[0] is half of pair[0] + pair[1] – pair[n-1]. Note that the value of pair[0] + pair[1] – pair[n-1] is (arr[0] + arr[1]) + (arr[0] + arr[2]) – (arr[1] + arr[2]). 
Once we have evaluated arr[0], we can evaluate other elements by subtracting arr[0]. For example arr[1] can be evaluated by subtracting arr[0] from pair[0], arr[2] can be evaluated by subtracting arr[0] from pair[1].
Following is the implementation of the above idea.
 

C++




#include <bits/stdc++.h>
using namespace std;
 
// Fills element in arr[] from its pair sum array pair[].
// n is size of arr[]
void constructArr(int arr[], int pair[], int n)
{
    arr[0] = (pair[0]+pair[1]-pair[n-1]) / 2;
    for (int i=1; i<n; i++)
        arr[i] = pair[i-1]-arr[0];
}
 
// Driver program to test above function
int main()
{
    int pair[] = {15, 13, 11, 10, 12, 10, 9, 8, 7, 5};
    int n = 5;
    int arr[n];
    constructArr(arr, pair, n);
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
    return 0;
}

Java




import java.io.*;
 
class PairSum {
     
    // Fills element in arr[] from its pair sum array pair[].
    // n is size of arr[]
    static void constructArr(int arr[], int pair[], int n)
    {
        arr[0] = (pair[0]+pair[1]-pair[n-1]) / 2;
        for (int i=1; i<n; i++)
            arr[i] = pair[i-1]-arr[0];
    }
     
    // Driver program to test above function
    public static void main(String[] args)
    {
        int pair[] = {15, 13, 11, 10, 12, 10, 9, 8, 7, 5};
        int n = 5;
        int[] arr = new int[n];
        constructArr(arr, pair, n);
        for (int i = 0; i < n; i++)
            System.out.print(arr[i]+" ");       
    }
}
/* This code is contributed by Devesh Agrawal */

Python3




# Fills element in arr[] from its
# pair sum array pair[].
# n is size of arr[]
 
def constructArr(arr,pair,n):
    arr [0] = (pair[0]+pair[1]-pair[n-1])//2
    for i in range(1,n):
        arr[i] = pair[i-1]-arr[0]
 
# Driver code
if __name__=='__main__':
    pair = [15, 13, 11, 10, 12, 10, 9, 8, 7, 5]
    n =5
    arr = [0]*n
    constructArr(arr,pair,n)
    for i in range(n):
        print(arr[i],end =" ")
 
# This code is contributed by
# Shrikant13

C#




// C# program to construct an
// array from its pair-sum array
using System;
 
class PairSum
{
    // Fills element in arr[] from its
    // pair sum array pair[].
    // n is size of arr[]
    static void constructArr(int []arr, int []pair,
                                             int n)
    {
        arr[0] = (pair[0] + pair[1] - pair[n - 1]) / 2;
        for (int i = 1; i < n; i++)
            arr[i] = pair[i - 1] - arr[0];
    }
     
    // Driver program to test above function
    public static void Main()
    {
        int []pair = {15, 13, 11, 10, 12,
                         10, 9, 8, 7, 5};
        int n = 5;
        int []arr = new int[n];
        constructArr(arr, pair, n);
        for (int i = 0; i < n; i++)
            Console.Write(arr[i] + " ");    
    }
}
 
// This code is contributed by nitin mittal

PHP




<?php
// Fills element in arr[] from 
// its pair sum array pair[].
// n is size of arr[]
function constructArr($pair)
{
    $arr = array();
    $n = 5;
    $arr[0] = intval(($pair[0] + $pair[1] -
                      $pair[$n - 1]) / 2);
    for ($i = 1; $i < $n; $i++)
        $arr[$i] = $pair[$i - 1] - $arr[0];
         
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i] . " ";
}
 
// Driver Code
$pair = array(15, 13, 11, 10,
              12, 10, 9, 8, 7, 5);
constructArr($pair);
 
// This code is contributed by Sam007
?>

Javascript




<script>
 
// Fills element in arr[] from
// its pair sum array pair[].
// n is size of arr[]
function constructArr(arr, pair, n)
{
    arr[0] = Math.floor((pair[0]+pair[1]-pair[n-1]) / 2);
    for (let i=1; i<n; i++)
        arr[i] = pair[i-1]-arr[0];
}
 
// Driver program to test above function
 
    let pair = [15, 13, 11, 10, 12, 10, 9, 8, 7, 5];
    let n = 5;
    let arr = new Array(n);
    constructArr(arr, pair, n);
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
 
 
// This code is contributed by Surbhi Tyagi.
 
</script>

Output:  

 8 7 5 3 2 

Time complexity of constructArr() is O(n) where n is number of elements in arr[].
This article is contributed by Abhishek. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 




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