Coin Change (Bottom-Up)

Build a one-dimensional table where each amount stores the fewest coins needed to make it.

Algorithm

@steps

Initialize dp[0] = 0 and all other amounts to an unreachable sentinel.
Scan amounts from 1 through 6.
For each coin, read the earlier cell dp[amount - coin] when it exists.
Write the smallest candidate into the current amount.
Print both the final answer and the full DP array. @end @complexity

Time: O(target * coin_count)
Space: O(target) @end

bottom-up dynamic programming `dp[a]` is solved from already-computed smaller amounts, so every table cell has a visible dependency.

C++ DSA Implementation

basic.cpp

#include <iostream>
#include <sstream>
#include <vector>
using namespace std;

string list_string(const vector<int>& values) {
  ostringstream out;
  out << "[";
  for (size_t i = 0; i < values.size(); i++) {
    if (i) out << ", ";
    out << values[i];
  }
  out << "]";
  return out.str();
}

int main() {
  vector<int> coins = {1, 3, 4};
  int target = 6;
  int inf = target + 1;
  vector<int> dp(target + 1, inf);
  dp[0] = 0;
  for (int amount = 1; amount <= target; amount++) {
    for (int coin : coins) {
      if (amount >= coin) {
        int candidate = dp[amount - coin] + 1;
        if (candidate < dp[amount]) dp[amount] = candidate;
      }
    }
  }
  cout << dp[target] << "\n" << list_string(dp) << "\n";
}

@end @output 2 [0, 1, 2, 1, 1, 2, 2] @end