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.”
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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]+" ");
}
}
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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);
?>
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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[].
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.
