Sorting Big Integers - GeeksforGeeks

Given a array of n positive integers where each integer can have digits upto 10⁶, print the array elements in ascending order.

Input: arr[] = {54, 724523015759812365462, 870112101220845, 8723} 
Output: 54 8723 870112101220845 724523015759812365462
Explanation:
All elements of array are sorted in non-descending(i.e., ascending)
order of their integer value

Input: arr[] = {3643641264874311, 451234654453211101231,
                4510122010112121012121}
Output: 3641264874311 451234654453211101231 4510122010112121012121

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach is to use arbitrary precision data type such as int in python or Biginteger class in Java. But that approach will not be fruitful because internal conversion of string to int and then perform sorting will leads to slow down the calculations of addition and multiplications in binary number system.

Efficient Solution : As size of integer is very large even it can’t be fit in long long data type of C/C++, so we just need to input all numbers as strings and sort them using a comparison function. Following are the key points compare function:-

If lengths of two strings are different, then we need to compare lengths to decide sorting order.
If Lengths are same then we just need to compare both the strings in lexicographically order.

Assumption : There are no leading zeros.

C++

// Below is C++ code to sort the Big integers in 
// ascending order 
#include<bits/stdc++.h> 
using namespace std; 
  
// comp function to perform sorting 
bool comp(const string &left, const string &right) 
{ 
    // if length of both string are equals then sort 
    // them in lexicographically order 
    if (left.size() == right.size()) 
        return left < right; 
  
    // Otherwise sort them according to the length 
    // of string in ascending order 
    else
        return left.size() < right.size(); 
} 
  
// Function to sort arr[] elements according 
// to integer value 
void SortingBigIntegers(string arr[], int n) 
{ 
    // Copy the arr[] elements to sortArr[] 
    vector<string> sortArr(arr, arr + n); 
  
    // Inbuilt sort function using function as comp 
    sort(sortArr.begin(), sortArr.end(), comp); 
  
    // Print the final sorted array 
    for (auto &ele : sortArr) 
        cout << ele << " "; 
} 
  
// Driver code of above implementation 
int main() 
{ 
    string arr[] = {"54", "724523015759812365462", 
                    "870112101220845", "8723"}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
  
    SortingBigIntegers(arr, n); 
  
    return 0; 
} 

Python

# Below is Python code to sort the Big integers 
# in ascending order 
def SortingBigIntegers(arr, n): 
    
  # Direct sorting using lamda operator 
  # and length comparison 
  arr.sort(key = lambda x: (len(x), x)) 
  
# Driver code of above implementation 
arr = ["54", "724523015759812365462", 
        "870112101220845", "8723"] 
n = len(arr) 
  
SortingBigIntegers(arr, n) 
  
# Print the final sorted list using  
# join method 
print " ".join(arr) 

Output: 54 8723 870112101220845 724523015759812365462

Time complexity: O(sum * log(n)) where sum is the total sum of all string length and n is size of array
Auxiliary space: O(n)

Similar Post :
Sort an array of large numbers

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