'Sort the array as per the rules of card game using a generic method
cards = ['Jack', 8, 2, 6, 'King', 5, 3, 'Queen', "Jack", "Queen", "King"] <!- Required Output = [2,3,5,6,8,'Jack','Queen','King'] Question: Sort the array as per the rules of card game using a generic method.
Solution 1:[1]
One approach is to use an array with all the cards in the right order as a reference, and rank each card being sorted by its index in the reference array.
let cards = ['Jack', 8, 2, 6, 'King', 5, 3, 'Queen',"Jack","Queen","King"];
// change this to match the rules of card game
let theRightOrder = ["Ace", 2, 3, 4, 5, 6, 7, 8, 9, 10, "Jack", "Queen", "King"];
cards.sort((a, b) => theRightOrder.indexOf(a) - theRightOrder.indexOf(b));
console.log(cards);
Solution 2:[2]
The required output to this problem has to be displayed in an ascending order. Other items greater than 10 (Jack, Queen, King) are limited, so we can attach a weight to each of them.
let cards = ['Jack', 8, 2, 6, 'King', 5, 3,"Queen"];
// change this to match the rules of card game
const otherItems = {
'Jack': 11,
'Queen': 12
'King': 13
}
cards.sort((a, b) => {
if(isNaN(a)) a = otherItems[a];
if(isNaN(b)) b = otherItems[b];
return a-b;
});
console.log(cards);
Instead of looking for index in an array, items with weight greater than 10 can be looked up
on otherItems
Solution 3:[3]
let cards = ['Jack', 8, 2, 6, 'King', 5, 3, 'Queen', "Jack",
"Queen","King"]
const otherItems = {
'Ace':1,
'Jack': 11,
'Queen': 12,
'King': 13
}
const solutionTwo= (someArray)=>{
someArray.sort((a,b)=>{
a= isNaN(a)? otherItems[a]:a;
b= isNaN(b)? otherItems[b]:b;
return a-b;
});
console.log("Solution Two array", someArray);
}
solutionTwo(cards);
Solution 4:[4]
This is unrelated language but if anyone wants to solve this using python, here is the code i came up with during an interview that asked for the same solution with 3 different test cases.
def test_case(cards):
int_list = sorted([x for x in cards if type(x) == int])
str_list = sorted([x for x in cards if type(x) == str])
for x in str_list:
if x == "King":
str_list.remove(x)
str_list.append(x)
sorted_cards = int_list + str_list
print(f"TEST CASE = {sorted_cards}")
test_case(['Jack', 8, 2, 6, 'King', 5, 3, 'Queen'])
test_case(['Jack', 8, 2, 6, 'King', 5, 3, 'Queen', 'Jack', 'King', 'Queen', 'Queen', 'King', 'Jack'])
test_case(['Jack', 8, 2, 6, 5, 3])
Solution 5:[5]
This solution avoids mutating the original array and instead provides a shallow-copy of the cards in the sorted-order. It allows being called with parameters to indicate whether:
a. Ace
is low (default) or high
b. Duplicates need to be removed (default) or retained
c. The sorted array needs to be ascending (default) or descending
const handleDupes = (fl, ar) => (fl ? [...new Set(ar)] : ar);
const sortCards = (arr, aceHigh = false, removeDupes = false, descending = false) => {
const sorted = [...Array(9)].map(
(x, i) => (i + 2)
).concat(
"Jack, Queen, King".split(', ')
);
if (aceHigh) sorted.push("Ace");
else sorted.splice(0, 1, "Ace");
return handleDupes(
removeDupes,
[...arr]
.sort(
(a, b) => (
sorted.indexOf(a) > sorted.indexOf(b)
? descending ? -1 : 1
: descending ? 1 : -1
)
)
);
};
const unsorted = ['Jack', 8, 2, 6, 'King', 5, 3, 'Queen', "Jack", "Queen", "King"];
console.log(sortCards(unsorted, true, true));
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | James |
Solution 2 | Devmaleeq |
Solution 3 | Eze Onyekachukwu |
Solution 4 | |
Solution 5 |