Minimum Coin Change Problem Dynamic Programming

Minimum Coin Change Problem in Dynamic Programming is the most asked frequent questions asked in Software Coding Interview. Today we will be looking at the solution of this problem.

Coin Change – LeetCode

You are given an integer array of coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

You may assume that you have an infinite number of each kind of coin.

Example 1:

Input: coins = [1,3,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [1], amount = 3
Output: -1

Solution in Python

from  collections import deque

class Solution:
	def coinChange(self, coins: List[int], amount: int) -> int:
		qset = set()
		qeu = deque()
		qeu.append((0, amount)) # (num coins,  amount)
		while q:
			(num_coins, rem)  = qeu.popleft()
			if rem == 0:
				return num_coins
			if rem < 0:
			for coin in coins:
				x = rem - coin
				if  x not in qset:
					qeu.append((num_coins + 1, x))
		return -1

Solution in C++

class Solution {
	int coinChange(vector<int>& coins, int amount)
		int getCoinCount = coins.size();
		//cache to store minimum number of coins needed for each sub amount less than the target
		vector<int> cache(amount + 1, INT_MAX);
		//If amount is 0, no coin is needed
		cache[0] = 0;

		//Calculate the minimum number of coins needed for each amount less than target amount.
		//Use the minimum cains needed for lower amounts in calculating coins needed for higher amount.
		for (int currAmount = 1; currAmount <= amount; currAmount++) {
			for (int coinIdx = 0; coinIdx < getCoinCount; coinIdx++) {
				if (coins[coinIdx] <= currAmount) {
					int sub_result = cache[currAmount - coins[coinIdx]];

					if (sub_result != INT_MAX && sub_result + 1 < cache[currAmount])
						cache[currAmount] = sub_result + 1;
		return cache[amount] != INT_MAX ? cache[amount] : -1;

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