**Big Sorting**

### Problem Statement :

Consider an array of numeric strings where each string is a positive number with anywhere from 1 to 10^6 digits. Sort the array's elements in non-decreasing, or ascending order of their integer values and return the sorted array. Example Return the array ['1', '3', '150', '200']. Function Description Complete the bigSorting function in the editor below. bigSorting has the following parameter(s): string unsorted[n]: an unsorted array of integers as strings Returns string[n]: the array sorted in numerical order Input Format The first line contains an integer, n, the number of strings in unsorted. Each of the n subsequent lines contains an integer string, unsorted[ i ]. Constraints 1 <= n <= 2x10^5 Each string is guaranteed to represent a positive integer. There will be no leading zeros. The total number of digits across all strings in is between and (inclusive).

### Solution :

` ````
Solution in C :
In C++ :
#include<iostream>
#include<fstream>
#include<math.h>
#include<algorithm>
#include<string>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<sstream>
#include<set>
using namespace std;
#define forn(i,n) for(int i=0;i<(int)(n); i++)
#define forsn(i,s,n) for(int i=(s);i<(int)(n); i++)
#define esta(x,v) (find((v).begin(),(v).end(),(x)) != (v).end())
#define index(x,v) (find((v).begin(),(v).end(),(x)) - (v).begin())
#define debug(x) cout << #x << " = " << x << endl
#define pb push_back
#define mp make_pair
typedef long long tint;
typedef unsigned long long utint;
typedef long double ldouble;
typedef vector<int> vint;
int toNumber (string s)
{
int Number;
if ( ! (istringstream(s) >> Number) ) Number = 0;
return Number;
}
string toString (int number)
{
ostringstream ostr;
ostr << number;
return ostr.str();
}
int main (){
int n;
cin>>n;
vector< pair<int, string> > v;
forn(i,n){
string s;
cin>>s;
v.pb(mp((int)s.size(), s));
}
sort(v.begin(), v.end());
forn(i, n){
cout<<v[i].second<<endl;
}
}
In Java :
import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
String[] unsorted = new String[n];
for (int unsorted_i = 0; unsorted_i < n; unsorted_i++) {
unsorted[unsorted_i] = in.next();
}
Arrays.sort(unsorted, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
if (o1.length() < o2.length()) {
return -1;
}
if (o1.length() > o2.length()) {
return 1;
}
return o1.compareTo(o2);
}
});
for (String s : unsorted) {
System.out.println(s);
}
}
}
In C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int cmp(const void *p, const void *q)
{
char* s1 = *(char**)p;
char* s2 = *(char**)q;
int s1l=strlen(s1);
int s2l=strlen(s2);
if(s1l>s2l)
return 1;
if(s2l>s1l)
return -1;
int i,flag=0;
for(i=0;i<s1l;i++)
{
if(s1[i]==s2[i])
continue;
else
{
flag=1;
break;
}
}
if(flag)
{
return s1[i]-s2[i];
}
return 0;
}
int main(){
int n,i,tl;
scanf("%d\n",&n);
char* str[n];
char temp[1000001];
for(i=0;i<n;i++)
{
scanf("%s",temp);
tl=strlen(temp);
str[i]=malloc(sizeof(char)*(tl+1));
strcpy(str[i],temp);
}
qsort((void*)str,n,sizeof(str[0]),cmp);
for(i=0;i<n;i++)
printf("%s\n",str[i]);
return 0;
}
In Python3 :
import sys
n = int(input().strip())
unsorted = []
unsorted_i = 0
for unsorted_i in range(n):
unsorted_t = str(input().strip())
unsorted.append(unsorted_t)
unsorted.sort(key = lambda x : int(x))
for u in unsorted:
print(u)
```

## View More Similar Problems

## Polynomial Division

Consider a sequence, c0, c1, . . . , cn-1 , and a polynomial of degree 1 defined as Q(x ) = a * x + b. You must perform q queries on the sequence, where each query is one of the following two types: 1 i x: Replace ci with x. 2 l r: Consider the polynomial and determine whether is divisible by over the field , where . In other words, check if there exists a polynomial with integer coefficie

View Solution →## Costly Intervals

Given an array, your goal is to find, for each element, the largest subarray containing it whose cost is at least k. Specifically, let A = [A1, A2, . . . , An ] be an array of length n, and let be the subarray from index l to index r. Also, Let MAX( l, r ) be the largest number in Al. . . r. Let MIN( l, r ) be the smallest number in Al . . .r . Let OR( l , r ) be the bitwise OR of the

View Solution →## The Strange Function

One of the most important skills a programmer needs to learn early on is the ability to pose a problem in an abstract way. This skill is important not just for researchers but also in applied fields like software engineering and web development. You are able to solve most of a problem, except for one last subproblem, which you have posed in an abstract way as follows: Given an array consisting

View Solution →## Self-Driving Bus

Treeland is a country with n cities and n - 1 roads. There is exactly one path between any two cities. The ruler of Treeland wants to implement a self-driving bus system and asks tree-loving Alex to plan the bus routes. Alex decides that each route must contain a subset of connected cities; a subset of cities is connected if the following two conditions are true: There is a path between ever

View Solution →## Unique Colors

You are given an unrooted tree of n nodes numbered from 1 to n . Each node i has a color, ci. Let d( i , j ) be the number of different colors in the path between node i and node j. For each node i, calculate the value of sum, defined as follows: Your task is to print the value of sumi for each node 1 <= i <= n. Input Format The first line contains a single integer, n, denoti

View Solution →## Fibonacci Numbers Tree

Shashank loves trees and math. He has a rooted tree, T , consisting of N nodes uniquely labeled with integers in the inclusive range [1 , N ]. The node labeled as 1 is the root node of tree , and each node in is associated with some positive integer value (all values are initially ). Let's define Fk as the Kth Fibonacci number. Shashank wants to perform 22 types of operations over his tree, T

View Solution →