You are given an array a[] of n values as a[1], a[2]…a[n] which are part of n-equations in n-variables where equations are as:
Eqn1 => X2 + X3 +…..+Xn = a1
Eqn2 => X1 + X3 +…..+Xn = a2
Eqni => X1 + X2 +…+Xi-1 + Xi+1..+Xn = ai
Eqnn => X1 + X2 +…..+Xn-1 = an
As you have n-equations in n-variables find the value of all variables(X1, X2…Xn).
Examples :
Input : a[] = {4, 4, 4, 4, 4} Output : X1 = 1 X2 = 1 X3 = 1 X4 = 1 X5 = 1 Input : a[] = {2, 5, 6, 4, 8} Output : X1 = 4.25 X2 = 1.25 X3 = 0.25 X4 = 2.25 X5 = -1.75
Approach:
Let, X1+X2+X3+….Xn= SUM
Using value SUM we have our equations as
SUM - X1 = a1 -----(1) SUM - X2 = a2 -----(2) SUM - Xi = ai -----(i) SUM - Xn = an -------(n) --------------------------------------------------------------- Now, if we add all these equation we will have an equation as : n*SUM -(X1+X2+...Xn) = a1 + a2 + ...an n*SUM - SUM = a1 + a2 + ...an SUM = (a1+a2+...an)/(n-1) Calculate SUM from above equation. putting value of SUM in (i), (ii).... we have X1 = SUM - a1 X2 = SUM - a2
Solution :
X1 = SUM - a1 X2 = SUM - a2 Xi = SUM - ai Xn = SUM - an
Implementation:
C++
// CPP program to find n-variables #include <bits/stdc++.h> using namespace std;
// function to print n-variable values void findVar( int a[], int n)
{ // calculate value of array SUM
float SUM = 0;
for ( int i = 0; i < n; i++)
SUM += a[i];
// Every variable contributes n-1
// times to sum. So dividing by
// n-1 to get sum of all.
SUM /= (n - 1);
// print the values of n-variables
for ( int i = 0; i < n; i++)
cout << "X" << (i + 1)
<< " = " << SUM - a[i] << endl;
} // driver program int main()
{ int a[] = { 2, 5, 6, 4, 8 };
int n = sizeof (a) / sizeof (a[0]);
findVar(a, n);
return 0;
} |
Java
// Java program to // find n-variables import java.io.*;
class GFG
{ // function to print
// n-variable values
static void findVar( int []a,
int n)
{
// calculate value+
// of array SUM
float SUM = 0 ;
for ( int i = 0 ; i < n; i++)
SUM += a[i];
// Every variable contributes
// n-1 times to sum. So dividing
// by n-1 to get sum of all.
SUM /= (n - 1 );
// print the values
// of n-variables
for ( int i = 0 ; i < n; i++)
System.out.print( "X" + (i + 1 ) +
" = " + (SUM -
a[i]) + "\n" );
}
// Driver Code
public static void main(String args[])
{
int []a = new int []{ 2 , 5 , 6 , 4 , 8 };
int n = a.length;
findVar(a, n);
}
} // This code is contributed by // Manish Shaw(manishshaw1) |
Python3
# Python3 program to # find n-variables # function to print # n-variable values def findVar(a, n):
# calculate value of
# array SUM
SUM = 0 ;
for i in range (n):
SUM + = a[i];
# Every variable contributes
# n-1 times to sum. So
# dividing by n-1 to get sum
# of all.
SUM = SUM / (n - 1 );
# print the values
# of n-variables
for i in range (n):
print ( "X" , (i + 1 ),
" = " , SUM - a[i]);
# Driver Code a = [ 2 , 5 , 6 , 4 , 8 ];
n = len (a);
findVar(a, n); # This code is contributed # by mits |
C#
// C# program to find n-variables using System;
using System.Linq;
using System.Collections.Generic;
class GFG
{ // function to print
// n-variable values
static void findVar( int []a,
int n)
{
// calculate value+
// of array SUM
float SUM = 0;
for ( int i = 0; i < n; i++)
SUM += a[i];
// Every variable contributes
// n-1 times to sum. So dividing
// by n-1 to get sum of all.
SUM /= (n - 1);
// print the values
// of n-variables
for ( int i = 0; i < n; i++)
Console.Write( "X" + (i + 1) +
" = " + (SUM - a[i]) + "\n" );
}
// Driver Code
static void Main()
{
int []a = { 2, 5, 6, 4, 8 };
int n = a.Length;
findVar(a, n);
}
} // This code is contributed by // Manish Shaw(manishshaw1) |
PHP
<?php // PHP program to // find n-variables // function to print // n-variable values function findVar( $a , $n )
{ // calculate value of
// array SUM
$SUM = 0;
for ( $i = 0; $i < $n ; $i ++)
$SUM += $a [ $i ];
// Every variable contributes n-1
// times to sum. So dividing by
// n-1 to get sum of all.
$SUM /= ( $n - 1);
// print the values
// of n-variables
for ( $i = 0; $i < $n ; $i ++)
echo "\nX" , ( $i + 1)
, " = " , $SUM - $a [ $i ];
} // Driver Code
$a = array (2, 5, 6, 4, 8);
$n = count ( $a );
findVar( $a , $n );
// This code is contributed by anuj_67. ?> |
Javascript
<script> // function to print // n-variable values function findVar(a,n)
{ // calculate value+
// of array SUM
let SUM = 0;
for (let i = 0; i < n; i++)
SUM += a[i];
// Every variable contributes
// n-1 times to sum. So dividing
// by n-1 to get sum of all.
SUM /= (n - 1);
// print the values
// of n-variables
for (let i = 0; i < n; i++)
document.write( "X" + (i + 1) + " = " + (SUM - a[i]) + "<br>" );
} let a = [2, 5, 6, 4, 8 ]; let n = a.length; findVar(a, n); // This code is contributed by mohit kumar 29. </script> |
Output
X1 = 4.25 X2 = 1.25 X3 = 0.25 X4 = 2.25 X5 = -1.75
Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(1)