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Sort the numbers according to their product of digits

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  • Last Updated : 10 Mar, 2022

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 n
int 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 elements
void 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 code
int main()
{
    int arr[] = { 12, 10, 102, 31, 15 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    sortArr(arr, n);
 
    return 0;
}

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

Javascript




<script>
// Javascript implementation of the
// above approach
 
// Function to return the product
// of the digits of n
function 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 elements
function 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 code
var 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)
 


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