Lemonade Change - Greedy - Easy - LeetCode
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Lemonade Change - Greedy - Easy - LeetCode

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  • 1Customers pay for lemonade using $5, $10, or $20 bills, requiring correct change to complete transactions.
  • 2The solution tracks the count of $5 and $10 bills to determine if change can be provided.
  • 3If change cannot be given to any customer, the function returns false; otherwise, it returns true.

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"Customers pay for lemonade using $5, $10, or $20 bills, requiring correct change to complete transactions."

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)

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sumitc91

Published on 5 October 2020 · 1 min read · 299 words

Part of AskGif Blog · coding

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