Minimum sum of two numbers formed from digits of a number

Given an array of digits (values are from 0 to 9), find the minimum possible sum of two numbers formed from digits of the array. All digits of given array must be used to form the two numbers.

Examples:

Input: [6, 8, 4, 5, 2, 3] Output: 604 The minimum sum is formed by numbers 358 and 246 Input: [5, 3, 0, 7, 4] Output: 82 The minimum sum is formed by numbers 35 and 047

Since we want to minimize the sum of two numbers to be formed, we must divide all digits in two halves and assign half-half digits to them. We also need to make sure that the leading digits are smaller. We build a Min Heap with the elements of the given array, which takes O(n) worst time. Now we retrieve min values (2 at a time) of array, by polling from the Priority Queue and append these two min values to our numbers, till the heap becomes empty, i.e., all the elements of array get exhausted. We return the sum of two formed numbers, which is our required answer. Overall complexity is O(nlogn) as push() operation takes O(logn) and it’s repeated n times.

 
#include<bits/stdc++.h> 
using namespace std; 
int minSum(int arr[], int n) 
{ 
    priority_queue <int, vector<int>, greater<int> > pq; 
    string num1, num2; 
    for(int i=0; i<n; i++) 
        pq.push(arr[i]); 
    while(!pq.empty()) 
    { 
        num1+=(48 + pq.top()); 
        pq.pop(); 
        if(!pq.empty()) 
        { 
            num2+=(48 + pq.top()); 
            pq.pop(); 
        } 
    } 
    int a = atoi(num1.c_str()); 
    int b = atoi(num2.c_str()); 
    return a+b; 
} 
int main() 
{ 
    int arr[] = {6, 8, 4, 5, 2, 3}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    cout<<minSum(arr, n)<<endl; 
    return 0; 
} 

 
import java.util.PriorityQueue; 



class MinSum 
{ 
    public static long solve(int[] a) 
    { 
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>(); 
        StringBuilder num1 = new StringBuilder(); 
        StringBuilder num2 = new StringBuilder(); 
        for (int x : a) 
            pq.add(x); 
        while (!pq.isEmpty()) 
        { 
            num1.append(pq.poll()+ ""); 
            if (!pq.isEmpty()) 
                num2.append(pq.poll()+ ""); 
        } 
        long sum = Long.parseLong(num1.toString()) + 
                   Long.parseLong(num2.toString()); 
        return sum; 
    } 
    public static void main (String[] args) 
    { 
        int arr[] = {6, 8, 4, 5, 2, 3}; 
        System.out.println("The required sum is "+ solve(arr)); 
    } 
} 

/div>

Output: The required sum is 604

Anothor method: We can follow another approach also like this, as we need two numbers such that their sum is minimum, then we would also need two minimum numbers. If we arrange our array in ascending order then we can two digits that will form the smallest numbers, e.g, 2 3 4 5 6 8, now we can get two numbers starting from 2 and 3. Now first part is done. Now we have to form such that they would contain small digits, i.e pick digits alternatively from array extend your two numbers.

i.e 246, 358. Now if we see analyze this, then we can pick even indexed numbers for num1 and odd number for num2.

Below is the implementation:

 
#include<bits/stdc++.h> 
using namespace std; 
int minSum(int a[], int n){ 
    sort(a,a+n); 



    int num1 = 0; 
    int num2 = 0; 
    for(int i = 0;i<n;i++){ 
        if(i%2==0) 
            num1 = num1*10+a[i]; 
        else num2 = num2*10+a[i]; 
    } 
    return num2+num1; 
} 
int main() 
{ 
    int arr[] = {5, 3, 0, 7, 4};  
    int n = sizeof(arr)/sizeof(arr[0]);  
    cout<<"The required sum is  "<<minSum(arr, n)<<endl;  
    return 0; 
} 

import java.util.Arrays; 
public class AQRQ { 
    static int minSum(int a[], int n){ 
     Arrays.sort(a); 
     int num1 = 0; 
     int num2 = 0; 
     for(int i = 0;i<n;i++){ 
         if(i%2==0) 
             num1 = num1*10+a[i]; 
         else num2 = num2*10+a[i]; 
     } 
     return num2+num1; 
    } 
    public static void main(String[] args) { 
         int arr[] = {5, 3, 0, 7, 4};  
         int n = arr.length; 
         System.out.println("The required sum is  " + minSum(arr, n)); 



    } 
} 

# Python 3 program to find minimum # sum of two numbers formed

# from all digits in an given array

# Returns sum of two numbers formed # from all digits in a[]

def minSum(a, n):

# sorted the elements a = sorted(a)

num1, num2 = 0, 0

for i in range(n): if i % 2 == 0: num1 = num1 * 10 + a[i] else:

num2 = num2 * 10 + a[i]

return num2 + num1

# Driver code arr = [5, 3, 0, 7, 4] n = len(arr) print(“The required sum is”,

minSum(arr, n))

# This code is contributed
# by Mohit kumar 29

 
using System; 
public class GFG{ 
    static int minSum(int []a, int n){ 
    Array.Sort(a); 
    int num1 = 0; 
    int num2 = 0; 
    for(int i = 0;i<n;i++){ 
        if(i%2==0) 
            num1 = num1*10+a[i]; 
        else num2 = num2*10+a[i]; 
    } 
    return num2+num1; 
    } 



    static public void Main (){ 
        int []arr = {5, 3, 0, 7, 4};  
        int n = arr.Length; 
        Console.WriteLine("The required sum is " + minSum(arr, n)); 
    } 
} 

<?php 
function minSum($a, $n) 
{  
    sort($a);  
    $num1 = 0;  
    $num2 = 0;  
    for($i = 0; $i < $n; $i++) 
    {  
        if($i % 2 == 0)  
            $num1 = $num1 * 10 + $a[$i];  
        else $num2 = $num2 * 10 + $a[$i];  
    }  
    return ($num2 + $num1);  
}  
$arr = array(5, 3, 0, 7, 4);  
$n = sizeof($arr);  
echo "The required sum is ",  
     minSum($arr, $n), "
";  
?> 