Solution 1 - Greedy Change

Prefer one ten and one five for a twenty, preserving five-dollar bills for future customers.

TimeComplexity: O(n)

SpaceComplexity: O(1)

class Solution:
 def lemonadeChange(self,b):
  f=t=0
  for x in b:
   if x==5:f+=1
   elif x==10:f-=1;t+=1
   elif t and f:t-=1;f-=1
   else:f-=3
   if f<0:return False
  return True
if __name__=='__main__':print(Solution().lemonadeChange([5,5,5,10,20]))

class Solution{lemonadeChange(b){let f=0,t=0;for(const x of b){if(x===5)f++;else if(x===10){f--;t++}else if(t&&f){t--;f--}else f-=3;if(f<0)return false}return true}}console.log(new Solution().lemonadeChange([5,5,5,10,20]));

class Solution{lemonadeChange(b:number[]):boolean{let f=0,t=0;for(const x of b){if(x===5)f++;else if(x===10){f--;t++}else if(t&&f){t--;f--}else f-=3;if(f<0)return false}return true}}console.log(new Solution().lemonadeChange([5,5,10,10,20]));

package main
import"fmt"
func change(b []int)bool{f,t:=0,0;for _,x:=range b{if x==5{f++}else if x==10{f--;t++}else if t>0&&f>0{t--;f--}else{f-=3};if f<0{return false}};return true}
func main(){fmt.Println(change([]int{5,5,5,10,20}))}

fn change(b:&[i32])->bool{let(mut f,mut t)=(0,0);for &x in b{if x==5{f+=1}else if x==10{f-=1;t+=1}else if t>0&&f>0{t-=1;f-=1}else{f-=3}if f<0{return false}}true}fn main(){println!("{}",change(&[5,5,5,10,20]));}

0860 - Lemonade Change

Solution 1 - Greedy Change