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 the given array must be used to form the two numbers.
Examples:
Input: arr[] = {6, 8, 4, 5, 2, 3}
Output: 604
246 + 358 = 604
Input: arr[] = {5, 3, 0, 7, 4}
Output: 82
Approach: A minimum number will be formed from the set of digits when smallest digit appears at the most significant position and next to smallest digit appears at next most significant position and so on…
The idea is to build two numbers by alternating picking digits from the array (assuming it is sorted in ascending). So the first number is formed by digits present in odd positions in the array and the second number is formed by digits from even positions in the array. Finally, we return the sum of the first and second number. In order to reduce the time complexity, the array can be sorted in O(n) using the frequency array of digits as every element of the original array is a single digit i.e. there can be at most 10 distinct elements.
Below is the implementation of the above approach:
// C++ implementation of above approach #include<bits/stdc++.h> using namespace std;
// Function to return the required minimum sum int minSum(vector< int > arr, int n)
{ // Array to store the
// frequency of each digit
int MAX = 10;
int *freq = new int [MAX];
for ( int i = 0; i < n; i++) {
// Store count of every digit
freq[arr[i]]++;
}
// Update arr[] such that it is
// sorted in ascending
int k = 0;
for ( int i = 0; i < MAX; i++) {
// Adding elements in arr[]
// in sorted order
for ( int j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0;
int num2 = 0;
// Generating numbers alternatively
for ( int i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
// Return the minimum possible sum
return num1 + num2;
} // Driver code int main( void )
{ vector< int >arr = { 6, 8, 4, 5, 2, 3 };
int n = arr.size();
cout << minSum(arr, n);
} // This code is contributed by ankush_953 |
// Java implementation of above approach public class GFG {
public static final int MAX = 10 ;
// Function to return the required minimum sum
static int minSum( int arr[], int n)
{
// Array to store the
// frequency of each digit
int freq[] = new int [MAX];
for ( int i = 0 ; i < n; i++) {
// Store count of every digit
freq[arr[i]]++;
}
// Update arr[] such that it is
// sorted in ascending
int k = 0 ;
for ( int i = 0 ; i < MAX; i++) {
// Adding elements in arr[]
// in sorted order
for ( int j = 0 ; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0 ;
int num2 = 0 ;
// Generating numbers alternatively
for ( int i = 0 ; i < n; i++) {
if (i % 2 == 0 )
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
// Return the minimum possible sum
return num1 + num2;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 6 , 8 , 4 , 5 , 2 , 3 };
int n = arr.length;
System.out.print(minSum(arr, n));
}
} |
# Python implementation of above approach # Function to return the required minimum sum def minSum(arr, n):
# Array to store the
# frequency of each digit
MAX = 10
freq = [ 0 ] * MAX
for i in range (n):
# Store count of every digit
freq[arr[i]] + = 1
# Update arr[] such that it is
# sorted in ascending
k = 0
for i in range ( MAX ):
# Adding elements in arr[]
# in sorted order
for j in range ( 0 ,freq[i]):
arr[k] = i
k + = 1
num1 = 0
num2 = 0
# Generating numbers alternatively
for i in range (n):
if i % 2 = = 0 :
num1 = num1 * MAX + arr[i]
else :
num2 = num2 * MAX + arr[i]
# Return the minimum possible sum
return num1 + num2
# Driver code arr = [ 6 , 8 , 4 , 5 , 2 , 3 ]
n = len (arr);
print (minSum(arr, n))
#This code is contributed by ankush_953 |
// C# implementation of above approach using System;
class GFG {
public static int MAX = 10;
// Function to return the required minimum sum
static int minSum( int [] arr, int n)
{
// Array to store the
// frequency of each digit
int [] freq = new int [MAX];
for ( int i = 0; i < n; i++) {
// Store count of every digit
freq[arr[i]]++;
}
// Update arr[] such that it is
// sorted in ascending
int k = 0;
for ( int i = 0; i < MAX; i++) {
// Adding elements in arr[]
// in sorted order
for ( int j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0;
int num2 = 0;
// Generating numbers alternatively
for ( int i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
// Return the minimum possible sum
return num1 + num2;
}
// Driver code
static public void Main()
{
int [] arr = { 6, 8, 4, 5, 2, 3 };
int n = arr.Length;
Console.WriteLine(minSum(arr, n));
}
} // This code is contributed by jit_t. |
<script> // Javascript implementation of above approach
let MAX = 10;
// Function to return the required minimum sum
function minSum(arr, n)
{
// Array to store the
// frequency of each digit
let freq = new Array(MAX);
freq.fill(0);
for (let i = 0; i < n; i++) {
// Store count of every digit
freq[arr[i]]++;
}
// Update arr[] such that it is
// sorted in ascending
let k = 0;
for (let i = 0; i < MAX; i++) {
// Adding elements in arr[]
// in sorted order
for (let j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
let num1 = 0;
let num2 = 0;
// Generating numbers alternatively
for (let i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
// Return the minimum possible sum
return num1 + num2;
}
let arr = [ 6, 8, 4, 5, 2, 3 ];
let n = arr.length;
document.write(minSum(arr, n));
</script> |
604
Time Complexity: O(n)
Space Complexity: O(n)