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. @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.

SQL DSA Implementation

basic.sql 
CREATE TEMP TABLE coins(value INTEGER);
INSERT INTO coins VALUES (1), (3), (4);
CREATE TEMP TABLE dp(amount INTEGER PRIMARY KEY, value INTEGER);
INSERT INTO dp VALUES (0, 0), (1, 7), (2, 7), (3, 7), (4, 7), (5, 7), (6, 7);
UPDATE dp SET value = (SELECT MIN(prev.value + 1) FROM coins JOIN dp AS prev ON prev.amount = 1 - coins.value WHERE 1 >= coins.value) WHERE amount = 1;
UPDATE dp SET value = (SELECT MIN(prev.value + 1) FROM coins JOIN dp AS prev ON prev.amount = 2 - coins.value WHERE 2 >= coins.value) WHERE amount = 2;
UPDATE dp SET value = (SELECT MIN(prev.value + 1) FROM coins JOIN dp AS prev ON prev.amount = 3 - coins.value WHERE 3 >= coins.value) WHERE amount = 3;
UPDATE dp SET value = (SELECT MIN(prev.value + 1) FROM coins JOIN dp AS prev ON prev.amount = 4 - coins.value WHERE 4 >= coins.value) WHERE amount = 4;
UPDATE dp SET value = (SELECT MIN(prev.value + 1) FROM coins JOIN dp AS prev ON prev.amount = 5 - coins.value WHERE 5 >= coins.value) WHERE amount = 5;
UPDATE dp SET value = (SELECT MIN(prev.value + 1) FROM coins JOIN dp AS prev ON prev.amount = 6 - coins.value WHERE 6 >= coins.value) WHERE amount = 6;
SELECT value FROM dp WHERE amount = 6;
SELECT '[' || group_concat(value, ', ') || ']' FROM dp ORDER BY amount;

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