add solution: coin-change

Author: dusunax

coin-change/dusunax.py

@@ -0,0 +1,49 @@
+'''
+# 322. Coin Change
+
+use a queue for BFS & iterate through the coins and check the amount is down to 0.
+use a set to the visited check.
+
+## Time and Space Complexity
+
+```
+TC: O(n * Amount)
+SC: O(Amount)
+```
+
+#### TC is O(n * Amount):
+- sorting the coins = O(n log n)
+- reversing the coins = O(n)
+- iterating through the queue = O(Amount)
+- iterating through the coins and check the remaining amount is down to 0 = O(n)
+
+#### SC is O(Amount):
+- using a queue to store (the remaining amount, the count of coins) tuple = O(Amount)
+- using a set to store the visited check = O(Amount)
+'''
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        if amount == 0:
+            return 0
+        if len(coins) == 1 and coins[0] == amount:
+            return 1
+
+        coins.sort() # TC: O(n log n)
+        coins.reverse() # TC: O(n)
+        
+        queue = deque([(amount, 0)]) # SC: O(Amount)
+        visited = set() # SC: O(Amount)
+
+        while queue: # TC: O(Amount)
+            remain, count = queue.popleft()
+
+            for coin in coins: # TC: O(n)
+                next_remain = remain - coin
+
+                if next_remain == 0:
+                    return count + 1
+                if next_remain > 0 and next_remain not in visited:
+                    queue.append((next_remain, count + 1))
+                    visited.add(next_remain)
+    
+        return -1