Question

Create a function l_mode that finds the index of the value that occurs most often in an array. The input to your function is an array of integers A and the number of elements in the array 0 < n <= 10^3. Assuming v is the most frequent element in the array, your function will return the index i of the first (leftmost) occurrence of v in A. You can assume that -500 <= A[i] <= 500, for 0 <= i < n. Ties must be broken by choosing the smallest element.

this is the function name and parameters:

unsigned int l_mode(const int *A, unsigned int n);

It must be written in C++ and can only include

#include <iostream>
#include <cmath>

I only need the function but if you'd like to write a main to test the function out you can use some of the following test cases:

• array of all 0's
• smallest element is chosen if 2 numbers have equal occurrences
• be sure to return the index not the element
• an array where every element is equal (ex. {5, 5, 5, 5, 5})
• array is 2 elements

• {1, 3, -500, -500, 0, 0, 0, -324, 500, 500, 500, 500} = 8

• {50, -500, -500, -500, 0, 1, 0, 3, -500, -500, 500, -500} = 1
• {5, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3} = 6
• {1, 1, 1, 1 ,5, 5, 5, 5} = 0
• {5, 5, 5, 5, 1, 1, 1, 1} = 4
• {-5, -1, -5, -1, -5, -1} = 0 (-5 < -1)

Answer 1

#include <iostream>
#include <climits>
using namespace std;

unsigned int l_mode(const int *A, unsigned int n){
   int value = INT_MAX , count = 0, c, index;
   for(int i = 0;i<n;i++){
      c = 0;
      for(int j = 0;j<n;j++){
         if(A[i] == A[j]){
            c++;
         }
      }
      if(count <= c && A[i]<value){
         value = A[i];
         count = c;
      }
   }  
   for(int i = 0;i<n;i++){
      if(value == A[i]){
         index = i;
         break;
      }
   }
   return index;
}

int main() {
   int A1[] = {1, 3, -500, -500, 0, 0, 0, -324, 500, 500, 500, 500};
   int A2[] = {50, -500, -500, -500, 0, 1, 0, 3, -500, -500, 500, -500};
   int A3[] = {5, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3} ;
   int A4[] = {1, 1, 1, 1 ,5, 5, 5, 5};
   int A5[] = {5, 5, 5, 5, 1, 1, 1, 1} ;
   int A6[] = {-5, -1, -5, -1, -5, -1};
   
   cout<<l_mode(A1,12)<<endl;
   cout<<l_mode(A2,12)<<endl;
   cout<<l_mode(A3,13)<<endl;
   cout<<l_mode(A4,8)<<endl;
   cout<<l_mode(A5,8)<<endl;
   cout<<l_mode(A6,6)<<endl;
   
   return 0;
}

Output:

6 4 Process exited after 0.1834 seconds with returnvalue 0 Press any key to continue . - .

Note: Please comment below if you have any doubts I will update my answer if needs

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