Given a array of **n** positive integers where each integer can have digits upto 10^{6}, 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

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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