Find d to maximize the number of zeros in array c[] created as c[i] = d*a[i] + b[i] in C++

Concept

With respect of two given arrays of M integers, assume an array C, where the i-th integer will be d*a[i] + b[i] where d is indicated as any arbitrary real number. Our task is to display or print d such that array C has largest number of zeros and also prints the number of zeros.

Input

a[] = {15, 40, 45}
b[] = {4, 5, 6}

Output

Value of d is: -0.133333
The number of zeros in array C is: 1
If we choose d as -0.133333 then we get one zero in the array C which is the maximum possible.

Methods

We follow the below mentioned steps to solve the above problem −

• We rewrite the equation as d = -b[i]/a[i]
• Implement hash-table to count the largest number of occurrence of any real number to obtain the value of d.
• Now,we conclude that number of zeros will be the largest count + (number of pairs a[i] and b[i] where both are 0).

Example

 Live Demo

// C++ program to implement the above
// approach
#include <bits/stdc++.h>
using namespace std;
// Shows function to find the value of d
// and find the number of zeros in the array
void findDandZeros1(int a[], int b[], int m){
   // Shows hash table
   unordered_map<long double, int> mpp1;
   int count1 = 0;
   // Performs iteration for i-th element
   for (int i = 0; i < m; i++) {
      // Now if both are not 0
      if (b[i] != 0 && a[i] != 0) {
         long double val1 = (long double)(-1.0 * b[i]) /
         (long double)(a[i]);
         mpp1[val1] += 1;
      }
      // Now if both are 0
      else if (b[i] == 0 && a[i] == 0)
         count1 += 1;
      }
      // Used to find max occurring d
      int maxi1 = 0;
      for (auto it : mpp1) {
         maxi1 = max(it.second, maxi1);
   }
   // Used to print the d which occurs max times
   for (auto it : mpp1) {
      if (it.second == maxi1) {
         cout << "Value of d is: "
         << it.first << endl;
         break;
      }
   }
   // Used to print the number of zeros
   cout << "The number of zeros in array C is: "
   << maxi1 + count1;
}
// Driver code
int main(){
   int a[] = { 15, 40, 45 };
   int b[] = { 4, 5, 6 };
   int m = sizeof(a) / sizeof(a[0]);
   findDandZeros1(a, b, m);
   return 0;
}

Output

Value of d is: -0.133333
The number of zeros in array C is: 1