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by Sumit Chourasia | Oct 05, 2020 | Category :coding | Tags : algorithm data-structure easy greedy leetcode

Lemonade Change - Greedy - Easy - LeetCode

Lemonade Change - Greedy - Easy - LeetCode

At a lemonade stand, each lemonade costs $5. 

Customers are standing in a queue to buy from you, and order one at a time (in the order specified by bills).

Each customer will only buy one lemonade and pay with either a $5, $10, or $20 bill.  You must provide the correct change to each customer, so that the net transaction is that the customer pays $5.

Note that you don't have any change in hand at first.

Return true if and only if you can provide every customer with correct change.

 

Example 1:

Input: [5,5,5,10,20]
Output: true
Explanation: 
From the first 3 customers, we collect three $5 bills in order.
From the fourth customer, we collect a $10 bill and give back a $5.
From the fifth customer, we give a $10 bill and a $5 bill.
Since all customers got correct change, we output true.
Example 2:

Input: [5,5,10]
Output: true
Example 3:

Input: [10,10]
Output: false
Example 4:

Input: [5,5,10,10,20]
Output: false
Explanation: 
From the first two customers in order, we collect two $5 bills.
For the next two customers in order, we collect a $10 bill and give back a $5 bill.
For the last customer, we can't give change of $15 back because we only have two $10 bills.
Since not every customer received correct change, the answer is false.
 

Note:

0 <= bills.length <= 10000
bills[i] will be either 5, 10, or 20.

public class Solution {
    public bool LemonadeChange(int[] bills) {
        int fiveCoinCount = 0;
        int tenCoinCount = 0;
        for(int i=0;i<bills.Length;i++){
            if(bills[i]==5){
                fiveCoinCount++;
            }
            else if(bills[i]==10){
                if(fiveCoinCount>0){
                    fiveCoinCount--;
                    tenCoinCount++;
                }
                else{
                    return false;
                }
            }
            else{
                if(tenCoinCount>0 && fiveCoinCount>0){
                    tenCoinCount--;
                    fiveCoinCount--;
                }
                else if(fiveCoinCount>2){
                    fiveCoinCount-=3;
                }
                else{
                    return false;
                }
            }
        }
        
        return true;
    }
}

Time Complexity: O(n)

Space Complexity: O(1)