Discussion on Big Sorting Challenge

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948 Discussions

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2 weeks ago+ 0 comments

MERGE SORT IS still the fastest algorithm since o(n) complexity

def merge(left, right):
    merged = []
    i = j = 0
    while i < len(left) and j < len(right):
        if compare_numbers(left[i], right[j]) <= 0:
            merged.append(left[i])
            i += 1
        else:
            merged.append(right[j])
            j += 1
    merged.extend(left[i:])
    merged.extend(right[j:])
    return merged
def compare_numbers(a, b):
    # First, compare by length.
    if len(a) != len(b):
        return len(a) - len(b)
    if a < b:
        return -1
    elif a > b:
        return 1
    else:
        return 0
        
def merge_sort(arr):
    if len(arr) <= 1:
        return arr
    mid = len(arr) // 2
    left = merge_sort(arr[:mid])
    right = merge_sort(arr[mid:])
    return merge(left, right)

this is insertion sort (0)^2 ` def compare(res, current): if len(res) == 0: res = [current] return res for idx, pip in enumerate(res): if pip >= current: rep = idx if rep<0: rep = 0 res.insert(rep, current) return res res.append(current) print("final", res) return res

def bigSorting(unsorted): # Write your code here res =[] for datas in unsorted: data = int(datas) res = compare(res,data) fix = [] for bub in res: fix.append(str(bub)) print(bub) return fix `

2 weeks ago+ 0 comments

Here is my c++ solution, you can watch the explanation here : https://youtu.be/GAvltofMdYc

vector<string> bigSorting(vector<string> u) {
    sort(u.begin(), u.end(), [](string l, string r){
        if(l.size() == r.size()) return l < r;
        return l.size() < r.size();
    });
    return u;
}

1 month ago+ 1 comment

Python3, tried this way

from functools import cmp_to_key

def compare_big(num1, num2):
    if len(num1) > len(num2):
        return 1
    elif len(num1) < len(num2):
        return -1
    else:
        for d1, d2 in zip(num1, num2):
            if d1 > d2:
                return 1
            elif d1 < d2:
                return -1
        return 0

def bigSorting(unsorted):
    return sorted(unsorted, key=cmp_to_key(compare_big))
    # Write your code herefrom functools import cmp_to_key

def compare_big(num1, num2):
    if len(num1) > len(num2):
        return 1
    elif len(num1) < len(num2):
        return -1
    else:
        for d1, d2 in zip(num1, num2):
            if d1 > d2:
                return 1
            elif d1 < d2:
                return -1
        return 0

def bigSorting(unsorted):
    return sorted(unsorted, key=cmp_to_key(compare_big))
    # Write your code here

1 month ago+ 0 comments

But you can always do

return sorted(unsorted, key = lambda x: (len(x), x))

2 months ago+ 0 comments

function bigSorting(unsorted) {
  return unsorted.sort((a, b) => {
    if (a.length !== b.length) {
      return a.length - b.length;
    }

    for (let i = 0; i < a.length; i++) {
      if (a[i] !== b[i]) return a[i] - b[i];
    }
  });
}

2 months ago+ 0 comments

def bigSorting(unsorted):
    return sorted(unsorted, key=lambda x: (len(x), x))

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