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 valuesvoid 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 programint 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-variablesimport 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 valuesdef 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 Codea = [2, 5, 6, 4, 8];n = len(a);findVar(a, n); # This code is contributed # by mits |
C#
// C# program to find n-variablesusing 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 valuesfunction 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 valuesfunction 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)
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