Step by Step solution about How to resolve the algorithm Count the coins step by step in the C++ programming language

The provided code implements the coin change problem using a stack-based depth-first search (DFS) approach. The problem is to find the number of ways to make a change for a given amount of money using a given set of coins of different denominations.

The code uses a stack to store the possible combinations of coins and their corresponding sums. Each element of the stack is a DataFrame containing three fields: sum, coins, and avail_coins. sum is the remaining amount of money to be changed, coins is the list of coins used to make up the change so far, and avail_coins is the list of available coins that can be used to make up the change.

The initial state of the stack is a DataFrame with sum set to the total amount of money to be changed, coins set to an empty list, and avail_coins set to a list of the available coins.

The main loop of the code pops the top element from the stack and checks if the sum is negative. If the sum is negative, the combination is invalid and is discarded. If the sum is zero, the combination is valid and the number of ways to make the change is incremented by 1. If the sum is positive and the list of available coins is not empty, two new combinations are created from the current combination. In the first combination, the sum is reduced by the value of the first coin in the list of available coins and that coin is added to the list of coins. In the second combination, the first coin in the list of available coins is removed. Both combinations are then pushed onto the stack.

The code continues to pop elements from the stack and generate new combinations until the stack is empty. The final count of valid combinations is stored in the ways variable and printed to the console.

In the example provided, the code finds that there are 8 ways to make change for 100 cents using coins of denominations 25, 10, 5, and 1. The 8 combinations are:

• 25, 25, 25, 25
• 25, 25, 25, 10, 10
• 25, 25, 10, 10, 5, 5
• 25, 25, 10, 5, 5, 5, 1
• 25, 10, 10, 10, 10, 5
• 25, 10, 10, 5, 5, 5, 1, 1
• 25, 10, 5, 5, 5, 5, 5, 1, 1
• 10, 10, 10, 10, 10, 10, 10

#include <iostream>
#include <stack>
#include <vector>

struct DataFrame {
  int sum;
  std::vector<int> coins;
  std::vector<int> avail_coins;
};

int main() {
  std::stack<DataFrame> s;
  s.push({ 100, {}, { 25, 10, 5, 1 } });
  int ways = 0;
  while (!s.empty()) {
    DataFrame top = s.top();
    s.pop();
    if (top.sum < 0) continue;
    if (top.sum == 0) {
      ++ways;
      continue;
    }
    if (top.avail_coins.empty()) continue;
    DataFrame d = top;
    d.sum -= top.avail_coins[0];
    d.coins.push_back(top.avail_coins[0]);
    s.push(d);
    d = top;
    d.avail_coins.erase(std::begin(d.avail_coins));
    s.push(d);
  }
  std::cout << ways << std::endl;
  return 0;
}