Find n-variables from n sum equations with one missing in C++

In this problem, we are given an array sum[] consisting of a sum of (n-1) variables given as,

Sum[1] = x2 + x3 + x4 + … xn Sum[2] = x1 + x3 + x4 + … xn . . Sum[i] = x2 + x3 + x4 + … x(i-1) + x(i+1) + … + xn . . Sum[n] = x1 + x2 + x3 + … x(n-1) Our task is to find the value of x1, x2,... xn.

Let’s take an example to understand the problem,

Input

sum[] = {6, 6, 6, 6, 6, 6, 6}

Output

x1 = 1, x2 = 1, x3 = 1, x4 = 1, x5 = 1, x6 = 1, x7 = 1

Explanation

arr[1] = 1 + 1 + 1 + 1 + 1 + 1 = 6

Solution Approach

Let the sum of all variables be sumX,

sumX = x1 + x2 + x3 + … + xn

So, the values of sum array are −

sum[1] = x2 + x3 + x4 + … xn = -x1 + x1 + x2 + x3 + x4 + … xn = sumX - x1

Similarly,

sum[2] = sumX - x2 sum[3] = sumX - x3 . sum[i] = sumX - xi . sum[n] = sumX - xn

Adding all the sum arrays we get,

Sum[1] + sum[2] + … sum[n] = sumX - x1 + sumX - x2 + … + sumX - xn arrSum = n*sumX - (x1 + x2 + x3 … xn) arrSum = n*SumX - (x1 + x2 + x3 … xn) arrSum = sumX*(n-1) sumX = arrSum/ (n-1)

Using this value of sumX, we can find the values of x1, x2…

So,

x1 = sumX - sum[1] x2 = sumX - sum[2] .. xi = sumX - sum[i] .. xn = sumX - sum[n]

Program to illustrate the working of our solution,

Example

 Live Demo

#include <iostream> using namespace std; void calcSumVariables(int sum[], int n) {    float SUMX = 0;    for (int i = 0; i < n; i++) {       SUMX += sum[i];    }    SUMX /= (n - 1);    for (int i = 0; i < n; i++)       cout<<"\nx"<<(i + 1)<<" = "<<(SUMX - sum[i]); } int main(){    int sum[] = {3, 8, 6, 7, 4, 5, 9 };    int N = sizeof(sum) / sizeof(sum[0]);    cout<<"The value of variables that form the sum are ";    calcSumVariables(sum, N);    return 0; }

Output

The value of variables that form the sum are

x1 = 4 x2 = -1 x3 = 1 x4 = 0 x5 = 3 x6 = 2 x7 = -2