Recursion and Dynamic Programming

Coin Change (Bottom-Up)

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

Algorithm

Steps

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

Complexity

  • Time: O(target * coin_count)
  • Space: O(target)
bottom-up dynamic programming `dp[a]` is solved from already-computed smaller amounts, so every table cell has a visible dependency.

Visual walkthrough

Dart DSA Implementation

basic.dart 
String listString(List<int> values) => "[${values.join(", ")}]";

void main() {
  final coins = [1, 3, 4];
  const target = 6;
  const inf = target + 1;
  final dp = List<int>.filled(target + 1, inf);
  dp[0] = 0;
  for (var amount = 1; amount <= target; amount++) {
    for (final coin in coins) {
      if (amount >= coin) {
        final candidate = dp[amount - coin] + 1;
        if (candidate < dp[amount]) dp[amount] = candidate;
      }
    }
  }
  print(dp[target]);
  print(listString(dp));
}

The pinned coins are [1, 3, 4] and target is 6. The diagrams show the one-dimensional DP table becoming reachable from left to right.

Step 1 - Initialize reachable amount 0

dp[0] = 0; every other amount starts as the sentinel 7.

Initial DP table for target 6.a0a1a2a3a4a5a60777777

Step 2 - Early amounts become reachable

With coins 1, 3, and 4, amounts 1 through 4 fill as [1, 2, 1, 1].

Table after filling amounts 1 through 4.a0a1a2a3a4a5a60121177base11+134todotodo

Step 3 - Final answer at amount 6

dp[5] = 2 and dp[6] = 2, so the target needs two coins.

Final DP table: [0, 1, 2, 1, 1, 2, 2].a0a1a2a3a4a5a6012112211+1341+43+3

Output

2
[0, 1, 2, 1, 1, 2, 2]