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.

Python DSA Implementation

basic.py

def list_string(values):
    return "[" + ", ".join(str(v) for v in values) + "]"

coins = [1, 3, 4]
target = 6
inf = target + 1
dp = [inf] * (target + 1)
dp[0] = 0

for amount in range(1, target + 1):
    for coin in coins:
        if amount >= coin:
            candidate = dp[amount - coin] + 1
            if candidate < dp[amount]:
                dp[amount] = candidate

print(dp[target])
print(list_string(dp))

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