Minimum sum of two numbers formed from digits of an array

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.

Implementation:

C++

 // C++ program to find minimum sum of two numbers// formed from all digits in a given array.#includeusing namespace std; // Returns sum of two numbers formed// from all digits in a[]int minSum(int arr[], int n){    // min Heap    priority_queue , greater > pq;     // to store the 2 numbers formed by array elements to    // minimize the required sum    string num1, num2;     // Adding elements in Priority Queue    for(int i=0; i

Java

 // Java program to find minimum sum of two numbers// formed from all digits in a given array.import java.util.PriorityQueue; class MinSum{    // Returns sum of two numbers formed    // from all digits in a[]    public static long solve(int[] a)    {        // min Heap        PriorityQueue pq = new PriorityQueue();         // to store the 2 numbers formed by array elements to        // minimize the required sum        StringBuilder num1 = new StringBuilder();        StringBuilder num2 = new StringBuilder();         // Adding elements in Priority Queue        for (int x : a)            pq.add(x);         // checking if the priority queue is non empty        while (!pq.isEmpty())        {            num1.append(pq.poll()+ "");            if (!pq.isEmpty())                num2.append(pq.poll()+ "");        }         // the required sum calculated        long sum = Long.parseLong(num1.toString()) +                   Long.parseLong(num2.toString());         return sum;    }     // Driver code    public static void main (String[] args)    {        int arr[] = {6, 8, 4, 5, 2, 3};        System.out.println("The required sum is "+ solve(arr));    }}

Python3

 # Python3 program to find minimum # sum of two numbers formed from # all digits in a given array.from queue import PriorityQueue # Returns sum of two numbers formed# from all digits in a[]def solve(a):         # min Heap    pq = PriorityQueue()         # To store the 2 numbers     # formed by array elements to    # minimize the required sum    num1 = ""    num2 = ""     # Adding elements in     # Priority Queue    for x in a:        pq.put(x)     # Checking if the priority     # queue is non empty    while not pq.empty():        num1 += str(pq.get())        if not pq.empty():            num2 += str(pq.get())         # The required sum calculated    sum = int(num1) + int(num2)         return sum     # Driver codeif __name__=="__main__":         arr = [ 6, 8, 4, 5, 2, 3 ]    print("The required sum is ", solve(arr)) # This code is contributed by rutvik_56

C#

 // C# program to find minimum sum of two numbers// formed from all digits in a given array.using System;using System.Collections.Generic;class GFG{         // Returns sum of two numbers formed    // from all digits in a[]    public static long solve(int[] a)    {               // min Heap        List pq = new List();          // to store the 2 numbers formed by array elements to        // minimize the required sum        string num1 = "";        string num2 = "";          // Adding elements in Priority Queue        foreach(int x in a)            pq.Add(x);                     pq.Sort();          // checking if the priority queue is non empty        while (pq.Count > 0)        {            num1 = num1 + pq[0];            pq.RemoveAt(0);            if (pq.Count > 0)            {                num2 = num2 + pq[0];                pq.RemoveAt(0);            }        }          // the required sum calculated        int sum = Int32.Parse(num1) + Int32.Parse(num2);          return sum;    }       // Driver code  static void Main()  {    int[] arr = {6, 8, 4, 5, 2, 3};    Console.WriteLine("The required sum is "+ solve(arr));  }} // This code is contributed by divyesh072019.

Javascript



Output
604

Time Complexity: O(N * log N)
Auxiliary Space: O(N)

Another 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. First part is done now. Moving forward 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 an odd number for num2.

Below is the implementation:

C++

 // C++ program to find minimum sum of two numbers// formed from all digits in a given array.#include using namespace std; // Returns sum of two numbers formed// from all digits in a[]int minSum(int a[], int n){    // sort the elements    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;} // Driver codeint 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;} // This code is contributed by Sania Kumari Gupta

C

 // C program to find minimum sum of two numbers// formed from all digits in a given array.#include #include int cmpfunc(const void* a, const void* b){    return (*(int*)a - *(int*)b);} // Returns sum of two numbers formed// from all digits in a[]int minSum(int a[], int n){     // sort the elements    qsort(a, n, sizeof(int), cmpfunc);    //     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;} // Driver codeint main(){    int arr[] = { 5, 3, 0, 7, 4 };    int n = sizeof(arr) / sizeof(arr[0]);    printf("The required sum is %d", minSum(arr, n));    return 0;} // This code is contributed by Sania Kumari Gupta

Java

 import java.util.Arrays;// Java program to find minimum sum of two numbers// formed from all digits in a given array.public class AQRQ {     // Returns sum of two numbers formed    // from all digits in a[]    static int minSum(int a[], int n)    {        // sort the elements        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;    }     // Driver code    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));    }} // This code is contributed by Sania Kumari Gupta

Python3

 # 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 codearr = [5, 3, 0, 7, 4]n = len(arr)print("The required sum is",             minSum(arr, n))     # This code is contributed# by Mohit kumar 29

C#

 // C# program to find minimum sum of two numbers//formed from all digits in a given array. using System; public class GFG{         //Returns sum of two numbers formed    //from all digits in a[]    static int minSum(int []a, int n){         // sort the elements    Array.Sort(a);         int num1 = 0;    int num2 = 0;    for(int i = 0;i

PHP



Javascript



Output
The required sum is  82

Time Complexity: O(N * log N)
Auxiliary Space: O(1)