Find the index of the row that has the largest number of duplicate elements (works, but not adequate)

Without the use of containers, this will in fact be pure C code. The only difference is the use of new , not malloc . The most non-trivial part of the task is to find the “largest number of repeating elements in a row”. If it is meant that in line 1, 2, 1, 2, 1 this number is 3 (3 units), the solution of this problem simply requires separation into a separate function. This feature might look like this:

 size_t count_els(size_t size, int row[]) { if (!row || !size) return 0; int *elements = new int[size]; // массив уникальных элементов в row size_t *counter = new size_t[size]; // массив счетчиков количества элементов if (!elements || !counter) { cout << "cannot allocate memory"; return 0; } for (size_t i=0; i<size; ++i) counter[i] = 0; // его нужно обнулить size_t current_index = 0; for (size_t i = 0; i < size; ++i) { int element = row[i]; // find element in elements[0:current_index] bool need_to_add = true; for (size_t j = 0; j < current_index; ++j) { // Проверка counter[j]!=0 должна стоять первой! if (counter[j] != 0 && elements[j] == element) { // нашли need_to_add = false; // добавлять не нужно counter[j]++; break; // т.к. элементы уникальные, дальше искать не нужно } } if (need_to_add) { // элемент row[i] не найден в массиве elements, добавляем elements[current_index] = element; counter[current_index] = 1; ++current_index; } } // Теперь находим максимальный элемент в массиве counter size_t max_counter = 0; for (size_t i = 0; i < current_index; ++i) { if (counter[i] > max_counter) max_counter = counter[i]; } delete[] counter; delete[] elements; return max_counter; } 

In fact, we need to count the number of all unique elements in a string, and find the maximum. For this (IMHO) one cannot do without two additional arrays (the unique elements of the elements and the counter of their number counter ). Their size cannot exceed the length of the string. In principle, nothing prevents you from writing all this code directly in main , but it will look awfully clumsy.

With this function, main (after filling the initial two-dimensional array m size of rows x cols ) may look like this:

 size_t needed_row = 0, longest = 0; for (size_t row = 0; row < rows; ++row) { current = count_els(cols, m[row]) if (current > longest) { longest = current; needed_row = row; } } cout << "needed row index is " << needed_row << endl; 

If your teacher objects to a particular function, argue. If he is not a fool, he will understand that she is needed here. Well, if you're a fool, there's nothing you can do.

Code using standard library containers:

 #include <iostream> #include <vector> #include <utility> using namespace std; template<typename T> size_t maxcounted_element(const vector<T> &v) { // Найти наибольшее число повторяющихся элементов в векторе vector<pair<size_t, T>> t; t.reserve(v.size()); for (const T& el: v) { bool to_push = true; for (size_t i=0; i<t.size(); ++i) { if (t.at(i).second == el) { ++t.at(i).first; to_push = false; break; } } if (to_push) { t.push_back({1, el}); } } size_t maxs = 0; for (auto &e: t) { if (e.first > maxs) maxs = e.first; } return maxs; } template<typename T> size_t find_row(size_t size, const vector<T> m[]) { // найти строку с наибольшим количеством повторяющихся элементов if (!m || !size) return 0; size_t longest = 0; size_t index = 0; for (size_t i=0; i<size; ++i) { size_t current = maxcounted_element(m[i]); if (current > longest) { longest = current; index = i; } } return index; } int main() { vector<double> m[] = { {1, 2, 3}, {1, 1, 1, 4, 5}, {1, 2, 5, 7, 9, 2, 2, 2}, {}, {1.23, 1.23, 3.14} }; size_t row = find_row(5, m); cout << "Row number " << row << " contains " << maxcounted_element(m[row]) << " equals elements" << endl; return 0; } 

The output of the program:

 Row number 2 contains 4 equals elements