Fining the mode in a vector of numbers: Logic errors

Hi,

Could you please help point out the logic error in the following codes, & give a hint for fixing the error? This is an exercise for control loop and vectors, so I'm not allowed to use a function with an argument as a vector.

Thanks a lot.

Code:

// Exercise 16 - Chapter 4
// Control statements, if statements & vectors

#include "std_lib_facilities.h"

int main()
{
	int x = 0;
	unsigned int i = 0;
	unsigned int j = 0;
	vector<int> numbers;
	cout << "Enter a series of positive integers: ";
	while (cin>>x) {
		numbers.push_back(x);
	}

	cout << "You have entered: ";
	for (i=0; i<numbers.size(); ++i)
		cout << numbers[i] << ' ';
	
	sort(numbers.begin(), numbers.end());
	cout << "\nIn ascending sorted order: ";
	for (i=0; i<numbers.size(); ++i)
		cout << numbers[i] << ' ';
	
	int prev_count = 0;
	int curr_count = 0;
	int mode = 0;
	for (i=0; i<numbers.size(); ++i) { // for each element of the vector "numbers"
		for (j=0; j<numbers.size(); ++j) { // compare it to all its other elements
			if (numbers[j]==numbers[i]) // if 2 elements equal, increment the count (curr_count)
				++curr_count;
		}
		if (curr_count>prev_count) { // if the current count greater than previous count
			mode = numbers[i]; // the mode is set to equal that element of the vector
			prev_count = curr_count; // set previous count == current count & repeat above steps
			curr_count = 0; // set current count to zero after finishing the check on each of the elements
		}
	}
	
	cout << "The mode is " << mode << '\n';
}

You are not resetting curr_count to zero every time through the loop. You only set it to zero when that if statement evaluates to true.

Originally Posted by bithub

You are not resetting curr_count to zero every time through the loop. You only set it to zero when that if statement evaluates to true.

Thanks so much for your help! With the solution given, things look so much, much clearer. Before it, it took me days to work on this error, and I couldn't locate the logic error. Thanks a lot, buddy.