Find original array from encrypted array (An array of sums of other elements) using C++.

Let us consider we have an array of integers, that array is encrypted array, Suppose the array is A = [10, 14, 12, 13, 11], the original array is B = [5, 1, 3, 2, 4], we can see that each element at index I of A follows this rule: A[i] = sum of all elements at position j in B[j], where I ≠ j. Our task is to find an original array from the encrypted one.

The task is based on arithmetic observation. Suppose the array is of size 4, original array B has four elements B = [a, b, c, d], so A will be like A[b+c+d, a+c+d, a+b+d, a+b+c], if we add all of the elements of B, we will get sum = b+c+d+a+c+d+a+b+d+a+b+c = 3*(a+b+c+d). So the sum of elements of B will be sum/3, now if we see the elements of B will be [sum – A[0], sum – A[1], sum – A[2], sum – A[3]]

Example

 Live Demo

#include<iostream>
using namespace std;
void showOrigianlArray(int arr[], int n) {
   int sum = 0;
   for (int i=0; i<n; i++)
      sum += arr[i];
      sum = sum/(n-1);
   for (int i=0; i<n; i++)
      cout << (sum - arr[i]) << " ";
}
int main() {
   int arr[] = {10, 14, 12, 13, 11};
   int n = sizeof(arr) / sizeof(arr[0]);
   showOrigianlArray(arr, n);
}

Output

5 1 3 2 4

Arnab Chakraborty

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Updated on: 30-Oct-2019

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