Arrange given numbers to form the biggest number | Set 1
Given an array of numbers, arrange them in a way that yields the largest value. For example, if the given numbers are {54, 546, 548, 60}, the arrangement 6054854654 gives the largest value. And if the given numbers are {1, 34, 3, 98, 9, 76, 45, 4}, then the arrangement 998764543431 gives the largest value.
A simple solution that comes to our mind is to sort all numbers in descending order, but simply sorting doesn’t work. For example, 548 is greater than 60, but in output 60 comes before 548. As a second example, 98 is greater than 9, but 9 comes before 98 in output.
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So how do we go about it? The idea is to use any comparison based sorting algorithm.
In the used sorting algorithm, instead of using the default comparison, write a comparison function myCompare() and use it to sort numbers.
Given two numbers X and Y, how should myCompare() decide which number to put first – we compare two numbers XY (Y appended at the end of X) and YX (X appended at the end of Y). If XY is larger, then X should come before Y in output, else Y should come before. For example, let X and Y be 542 and 60. To compare X and Y, we compare 54260 and 60542. Since 60542 is greater than 54260, we put Y first.
Following is the implementation of the above approach.
To keep the code simple, numbers are considered as strings, the vector is used instead of a normal array.
Below is the implementation of the above approach:
C++
// Given an array of numbers,// program to arrange the numbers// to form the largest number#include <algorithm>#include <iostream>#include <string>#include <vector>using namespace std;// A comparison function which// is used by sort() in// printLargest()int myCompare(string X, string Y){ // first append Y at the end of X string XY = X.append(Y); // then append X at the end of Y string YX = Y.append(X); // Now see which of the two // formed numbers is greater return XY.compare(YX) > 0 ? 1 : 0;}// The main function that prints// the arrangement with the// largest value. The function// accepts a vector of stringsvoid printLargest(vector<string> arr){ // Sort the numbers using // library sort function. The // function uses our comparison // function myCompare() to // compare two strings. See // algorithm/sort/ // for details sort(arr.begin(), arr.end(), myCompare); for (int i = 0; i < arr.size(); i++) cout << arr[i];}// Driver codeint main(){ vector<string> arr; // output should be 6054854654 arr.push_back("54"); arr.push_back("546"); arr.push_back("548"); arr.push_back("60"); printLargest(arr); return 0;} |
Java
// Given an array of numbers, program to// arrange the numbers to form the// largest numberimport java.util.*;class GFG { // The main function that prints the // arrangement with the largest value. // The function accepts a vector of strings static void printLargest(Vector<String> arr) { Collections.sort(arr, new Comparator<String>() { // A comparison function which is used by // sort() in printLargest() @Override public int compare(String X, String Y) { // first append Y at the end of X String XY = X + Y; // then append X at the end of Y String YX = Y + X; // Now see which of the two // formed numbers // is greater return XY.compareTo(YX) > 0 ? -1 : 1; } }); Iterator it = arr.iterator(); while (it.hasNext()) System.out.print(it.next()); } // Driver code public static void main(String[] args) { Vector<String> arr; arr = new Vector<>(); // output should be 6054854654 arr.add("54"); arr.add("546"); arr.add("548"); arr.add("60"); printLargest(arr); }}// This code is contributed by Shubham Juneja |
Python3
# Python3 Program to get the maximum# possible integer from given array# of integers...# custom comparator to sort according# to the ab, ba as mentioned in descriptiondef comparator(a, b): ab = str(a) + str(b) ba = str(b) + str(a) return ((int(ba) & gt int(ab)) - (int(ba) & lt int(ab)))def myCompare(mycmp): # Convert a cmp= function into a key= function class K(object): def __init__(self, obj, *args): self.obj = obj def __lt__(self, other): return mycmp(self.obj, other.obj) & lt 0 def __gt__(self, other): return mycmp(self.obj, other.obj) & gt 0 def __eq__(self, other): return mycmp(self.obj, other.obj) == 0 def __le__(self, other): return mycmp(self.obj, other.obj) & lt = 0 def __ge__(self, other): return mycmp(self.obj, other.obj) & gt = 0 def __ne__(self, other): return mycmp(self.obj, other.obj) != 0 return K# driver codeif __name__ == "__main__": a = [54, 546, 548, 60, ] sorted_array = sorted(a, key=myCompare(comparator)) number = "".join([str(i) for i in sorted_array]) print(number)# This code is Contributed by SaurabhTewary |
C#
// C# program for above approachusing System.Collections.Generic;using System;namespace LargestNumberClass {class LargestNumberClass { // Given a list of non-negative // integers, // arrange them such that they // form the largest number. // Note: The result may be very // large, so you need to // return a string instead // of an integer. public static void LargestNumberMethod(List<int> inputList) { string output = string.Empty; List<string> newList = inputList.ConvertAll<string>( delegate(int i) { return i.ToString(); }); newList.Sort(MyCompare); for (int i = 0; i < inputList.Count; i++) { output = output + newList[i]; } if (output[0] == '0' && output.Length > 1) { Console.Write("0"); } Console.Write(output); } internal static int MyCompare(string X, string Y) { // first append Y at the end of X string XY = X + Y; // then append X at the end of Y string YX = Y + X; // Now see which of the two // formed numbers is greater return XY.CompareTo(YX) > 0 ? -1 : 1; }}class Program { // Driver code static void Main(string[] args) { List<int> inputList = new List<int>() { 54, 546, 548, 60 }; LargestNumberClass.LargestNumberMethod(inputList); }}// This code is contributed by Deepak Kumar Singh} |
PHP
<?php// Given an array of numbers, program// to arrange the numbers to form the// largest number// A comparison function which is used// by sort() in printLargest()function myCompare($X, $Y){ // first append Y at the end of X $XY = $Y.$X; // then append X at the end of Y $YX = $X.$Y; // Now see which of the two formed // numbers is greater return strcmp($XY, $YX) > 0 ? 1: 0;}// The main function that prints the// arrangement with the largest value.// The function accepts a vector of stringsfunction printLargest($arr){ // Sort the numbers using library sort // function. The function uses our // comparison function myCompare() to // compare two strings. // algorithm/sort/ // for details usort($arr, "myCompare"); for ($i = 0; $i < count($arr) ; $i++ ) echo $arr[$i];}// Driver Code$arr = array("54", "546", "548", "60");printLargest($arr);// This code is contributed by// rathbhupendra?> |
6054854654
Time Complexity: O(nlogn) ,sorting is considered to have running time complexity of O(nlogn) and the for loop runs in O(n) time.
Auxiliary Space: O(1).
Another approach:(using itertools)
Using the inbuilt library of the Python, itertools library can be used to perform this task.
Python3
# Python3 implementation this is to use itertools.# permutations as coded below:from itertools import permutationsdef largest(l): lst = [] for i in permutations(l, len(l)): # provides all permutations of the list values, # store them in list to find max lst.append("".join(map(str,i))) return max(lst)print(largest([54, 546, 548, 60])) #Output 6054854654# This code is contributed by Raman Monga |
6054854654
Time Complexity: O(nlogn)
Auxiliary Space: O(1).
This article is compiled by Ravi Chandra Enaganti. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
