Given an array arr[] of N non-negative integers, the task is to sort these integers according to the product of their digits.
Examples:
Input: arr[] = {12, 10, 102, 31, 15}
Output: 10 102 12 31 15
10 -> 1 * 0 = 0
102 -> 1 * 0 * 2 = 0
12 -> 1 * 2 = 2
31 -> 3 * 1 = 3
15 -> 1 * 5 = 5
Input: arr[] = {12, 10}
Output: 10 12
Approach: The idea is to store each element with its product of digits in a vector pair and then sort all the elements of the vector according to the digit products stored. Finally, print the elements in order.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the product// of the digits of nint productOfDigit(int n){ int product = 1; while (n > 0) { product *= n % 10; n = n / 10; } return product;}// Function to sort the array according to// the product of the digits of elementsvoid sortArr(int arr[], int n){ // Vector to store the digit product // with respective elements vector<pair<int, int> > vp; // Inserting digit product with elements // in the vector pair for (int i = 0; i < n; i++) { vp.push_back(make_pair(productOfDigit(arr[i]), arr[i])); } // Sort the vector, this will sort the pair // according to the product of the digits sort(vp.begin(), vp.end()); // Print the sorted vector content for (int i = 0; i < vp.size(); i++) cout << vp[i].second << " ";}// Driver codeint main(){ int arr[] = { 12, 10, 102, 31, 15 }; int n = sizeof(arr) / sizeof(arr[0]); sortArr(arr, n); return 0;} |
Java
// Java implementation of the// above approachimport java.util.*;class GFG { // Function to return the product // of the digits of n static int productOfDigit(int n) { int product = 1; while (n > 0) { product *= n % 10; n /= 10; } return product; } // Function to sort the array according to // the product of the digits of elements static void sortArr(int[] arr, int n) { // Vector to store the digit product // with respective elements ArrayList<ArrayList<Integer> > vp = new ArrayList<ArrayList<Integer> >(); // Loop to create 2D array using 1D array for (int i = 0; i < n; i++) { vp.add(new ArrayList<Integer>()); } // Inserting digit product with elements // in the vector pair for (int i = 0; i < n; i++) { ArrayList<Integer> l1 = vp.get(i); l1.add(productOfDigit(arr[i])); l1.add(arr[i]); vp.set(i, l1); } // Sort the vector, this will sort the pair // according to the product of the digits Collections.sort( vp, new Comparator<ArrayList<Integer> >() { public int compare(ArrayList<Integer> o1, ArrayList<Integer> o2) { if (o1.get(0) != o2.get(0)) return o1.get(0).compareTo( o2.get(0)); return o1.get(1).compareTo(o2.get(1)); } }); // Print the sorted vector content for (int i = 0; i < n; i++) System.out.print(vp.get(i).get(1) + " "); } // Driver code public static void main(String[] args) { int[] arr = { 12, 10, 102, 31, 15 }; int n = arr.length; sortArr(arr, n); }}// This code is contributed by phasing17 |
Python3
# Python3 implementation of the approach # Function to return the product # of the digits of n def productOfDigit(n) : product = 1; while (n > 0) : product *= (n % 10); n = n // 10; return product; # Function to sort the array according to # the product of the digits of elements def sortArr(arr, n) : # Vector to store the digit product # with respective elements vp = []; # Inserting digit product with elements # in the vector pair for i in range(n) : vp.append((productOfDigit(arr[i]), arr[i])); # Sort the vector, this will sort the pair # according to the product of the digits vp.sort(); # Print the sorted vector content for i in range(len(vp)) : print(vp[i][1], end = " "); # Driver code if __name__ == "__main__" : arr = [ 12, 10, 102, 31, 15 ]; n = len(arr); sortArr(arr, n); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the// above approachusing System;using System.Linq;using System.Collections.Generic;class GFG { // Function to return the product // of the digits of n static int productOfDigit(int n) { int product = 1; while (n > 0) { product *= n % 10; n /= 10; } return product; } // Function to sort the array according to // the product of the digits of elements static void sortArr(int[] arr, int n) { // Vector to store the digit product // with respective elements List<List<int> > vp = new List<List<int> >(); // Loop to create 2D array using 1D array for (var i = 0; i < n; i++) { vp.Add(new List<int>()); } // Inserting digit product with elements // in the vector pair for (var i = 0; i < n; i++) { vp[i].Add(productOfDigit(arr[i])); vp[i].Add(arr[i]); } // Sort the vector, this will sort the pair // according to the product of the digits vp = vp.OrderBy(x => x[0]) .ThenBy(x => x[1]) .ToList(); // Print the sorted vector content for (int i = 0; i < n; i++) Console.Write(vp[i][1] + " "); } // Driver code public static void Main(string[] args) { int[] arr = { 12, 10, 102, 31, 15 }; int n = arr.Length; sortArr(arr, n); }}// This code is contributed by phasing17 |
Javascript
<script>// Javascript implementation of the// above approach// Function to return the product// of the digits of nfunction productOfDigit(n){ var product = 1; while (n > 0) { product *= n % 10; n = Math.floor(n / 10); } return product;} // Function to sort the array according to// the product of the digits of elementsfunction sortArr(arr, n){ // Vector to store the digit product // with respective elements var vp = new Array(n); // Loop to create 2D array using 1D array for (var i = 0; i < vp.length; i++) { vp[i] = []; } // Inserting digit product with elements // in the vector pair for (var i = 0; i < n; i++) { vp[i].push(productOfDigit(arr[i])); vp[i].push(arr[i]); } // Sort the vector, this will sort the pair // according to the product of the digits vp.sort(); // Print the sorted vector content for (var i = 0; i < n; i++) document.write(vp[i][1] + " ");}// Driver codevar arr = [ 12, 10, 102, 31, 15];var n = arr.length;sortArr(arr, n);// This code is contributed by ShubhamSingh10</script> |
Output:
10 102 12 31 15
Time Complexity: O(nlogn)
Auxiliary Space: O(n), since n extra space has been taken.
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