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;
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];
}
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 {
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 ];
}
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]+ " " );
}
}
|
Python3
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 ]
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 = " " )
|
C#
using System;
class PairSum
{
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];
}
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] + " " );
}
}
|
PHP
<?php
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 ] . " " ;
}
$pair = array (15, 13, 11, 10,
12, 10, 9, 8, 7, 5);
constructArr( $pair );
?>
|
Javascript
<script>
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];
}
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] + " " );
</script>
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Output:
8 7 5 3 2
Time complexity of constructArr() is O(n) where n is number of elements in arr[].
Auxiliary Space: O(1)
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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