problem

classical bag problem


322. Coin Change

You are given an integer array 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 = [1,2,5], amount = 11
Output: 3
Explanation: 11 = 5 + 5 + 1

Example 2:

Input: coins = [2], amount = 3
Output: -1

Example 3:

Input: coins = [1], amount = 0
Output: 0

wrong approach

59/188， Greedy


class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        if(coins.size() == 1 ){
            if(amount == 0)return 0;
            if(amount < coins[0])return -1;
            if(amount%coins[0])return -1;
        }
        sort(coins.begin(), coins.end());
        int mins = 100000;
        for(int i=0; i<coins.size(); i++){
            if(amount%coins[i] == 0){// can complete divide
                int tmp_min = amount/coins[i]; 
                if(tmp_min < mins) mins = tmp_min;
            }else{// can't complete divide
                int tmp_min = 0, mid = amount;
                for(int t=i; t>=0; t--){
                    tmp_min += mid/coins[t];
                    mid = mid%coins[t];
                    if(mid==0)break;
                }
                if(mid==0 && tmp_min < mins) mins = tmp_min;
            }
        }
        if(mins == 100000) return -1;
        else return mins;
    }
};

wrong approach 2

TLE, recursive


class Solution {
public:
    vector<int> coins_t;
    set<int> coinset;
    int rec(int num){
        int mincoins = num, numcoins;
        if(coinset.find(num) != coinset.end()) return 1;
        for(auto coin : coins_t){
            if(num - coin > 0)
                numcoins =  1 + rec(num - coin);
            if(numcoins < mincoins) mincoins = numcoins;
        }
        return mincoins;
    }
    int coinChange(vector<int>& coins, int amount) {
        if(coins.size() == 1 ){
            if(amount == 0)return 0;
            if(amount < coins[0])return -1;
            if(amount%coins[0])return -1;
        }
        coins_t = coins;
        coinset = set<int>(coins.begin(), coins.end());
        return rec(amount);
    }
};

appraoch 3

memorize dp


class Solution {
public:
    vector<int> dp;
    int rec(vector<int>& coins, int num){
        if(num < 0)return -1;
        if(num == 0)return 0;
        if(dp[num])return dp[num];
        
        int mins = INT_MAX;
        for(auto coin : coins){
            int res = rec(coins, num-coin);
            if(res >=0 && res < mins) mins = res + 1;
        }
        dp[num] = mins==INT_MAX ? -1 : mins;
        return dp[num];
    }
    int coinChange(vector<int>& coins, int amount) {
        if(coins.size() == 0) return -1;
        dp = vector<int>(amount+1, 0);
        return rec(coins, amount);
    }
};

approach

bag problem , dp


class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        vector<int> dp(amount+1, amount+1);
        dp[0] = 0;
        for(auto coin : coins)
            for(int i=coin; i<=amount; i++)
                dp[i] = min(dp[i], dp[i-coin]+1);
        return dp[amount] != amount+1 ? dp[amount] : -1;
    }
};