Given an array of digits (values are from 0 to 9), the task is to find the minimum possible sum of two numbers formed from digits of the array. Please note that all digits of the given array must be used to form the two numbers.
Examples:
Input: {6, 8, 4, 5, 2, 3}
Output: 604
Explanation: The minimum sum is formed by numbers 358 and 246Input: {5, 3, 0, 7, 4}
Output: 82
Explanation: The minimum sum is formed by numbers 35 and 047
Minimum sum of two numbers formed from digits of an array using Sorting:
A minimum number will be formed from set of digits when smallest digit appears at most significant position and next smallest digit appears at next most significant position and so on. The idea is to sort the array in increasing order and build two numbers by alternating picking digits from the array. So first number is formed by digits present in odd positions in the array and second number is formed by digits from even positions in the array.
Follow the given steps to solve the problem:
- Sort the array in increasing order
- Declare two variables a and b, representing the two numbers to be formed
- Traverse the array and if the index is odd then add this element into a, else add it to b
- Return the sum of two variables (a + b)
Below is the Implementation of the above approach:
// C++ program to find minimum sum of two numbers // formed from digits of the array. #include <bits/stdc++.h> using namespace std;
// Function to find and return minimum sum of // two numbers formed from digits of the array. int solve( int arr[], int N)
{ // Sort the array
sort(arr, arr + N);
// Let two numbers be a and b
int a = 0, b = 0;
for ( int i = 0; i < N; i++) {
// fill a and b with every alternate digit
// of input array
if (i & 1)
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// return the sum
return a + b;
} // Driver's code int main()
{ int arr[] = { 6, 8, 4, 5, 2, 3 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
cout << "Sum is " << solve(arr, N);
return 0;
} |
// C program to find minimum sum of two numbers // formed from digits of the array. #include <stdio.h> #include <stdlib.h> int cmpfunc( const void * a, const void * b)
{ return (*( int *)a - *( int *)b);
} // Function to find and return minimum sum of // two numbers formed from digits of the array. int solve( int arr[], int N)
{ // Sort the array
qsort (arr, N, sizeof ( int ), cmpfunc);
// Let two numbers be a and b
int a = 0, b = 0;
for ( int i = 0; i < N; i++) {
// fill a and b with every alternate digit
// of input array
if (i & 1)
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// return the sum
return a + b;
} // Driver's code int main()
{ int arr[] = { 6, 8, 4, 5, 2, 3 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
printf ( "Sum is %d" , solve(arr, N));
return 0;
} |
// Java program to find minimum sum of two numbers // formed from digits of the array. import java.util.Arrays;
class GFG {
// Function to find and return minimum sum of
// two numbers formed from digits of the array.
static int solve( int arr[], int N)
{
// sort the array
Arrays.sort(arr);
// let two numbers be a and b
int a = 0 , b = 0 ;
for ( int i = 0 ; i < N; i++) {
// fill a and b with every alternate
// digit of input array
if (i % 2 != 0 )
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// return the sum
return a + b;
}
// driver's code
public static void main(String[] args)
{
int arr[] = { 6 , 8 , 4 , 5 , 2 , 3 };
int N = arr.length;
System.out.print( "Sum is " + solve(arr, N));
}
} // This code is contributed by Anant Agarwal. |
# Python3 program to find minimum sum of two # numbers formed from digits of the array. # Function to find and return minimum sum of # two numbers formed from digits of the array. def solve(arr, N):
# sort the array
arr.sort()
# let two numbers be a and b
a = 0
b = 0
for i in range (N):
# Fill a and b with every alternate
# digit of input array
if (i % 2 ! = 0 ):
a = a * 10 + arr[i]
else :
b = b * 10 + arr[i]
# return the sum
return a + b
# Driver's code if __name__ = = '__main__' :
arr = [ 6 , 8 , 4 , 5 , 2 , 3 ]
N = len (arr)
# Function call
print ( "Sum is " , solve(arr, N))
# This code is contributed by Anant Agarwal.
|
// C# program to find minimum // sum of two numbers formed // from digits of the array. using System;
class GFG {
// Function to find and return
// minimum sum of two numbers
// formed from digits of the array.
static int solve( int [] arr, int N)
{
// Sort the array
Array.Sort(arr);
// Let two numbers be a and b
int a = 0, b = 0;
for ( int i = 0; i < N; i++) {
// Fill a and b with every alternate digit
// of input array
if (i % 2 != 0)
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// Return the sum
return a + b;
}
// Driver's code
public static void Main()
{
int [] arr = { 6, 8, 4, 5, 2, 3 };
int N = arr.Length;
// Function call
Console.WriteLine( "Sum is " + solve(arr, N));
}
} // This code is contributed by Anant Agarwal. |
// Javascript program to find minimum sum of two numbers // formed from digits of the array. // Function to find and return minimum sum of
// two numbers formed from digits of the array.
function solve(arr, n)
{
// sort the array
arr.sort();
// let two numbers be a and b
let a = 0, b = 0;
for (let i = 0; i < n; i++)
{
// fill a and b with every alternate
// digit of input array
if (i % 2 != 0)
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// return the sum
return a + b;
}
// Driver Code let arr = [6, 8, 4, 5, 2, 3];
let n = arr.length;
document.write( "Sum is "
+ solve(arr, n));
|
<?php // PHP program to find minimum // sum of two numbers formed // from digits of the array. // Function to find and return // minimum sum of two numbers // formed from digits of the array. function solve( $arr , $N )
{ // sort the array
sort( $arr ); sort( $arr , $N );
// let two numbers be a and b
$a = 0; $b = 0;
for ( $i = 0; $i < $N ; $i ++)
{
// fill a and b with every
// alternate digit of input array
if ( $i & 1)
$a = $a * 10 + $arr [ $i ];
else
$b = $b * 10 + $arr [ $i ];
}
// return the sum
return $a + $b ;
} // Driver's code $arr = array (6, 8, 4, 5, 2, 3);
$N = sizeof( $arr );
// Function call echo "Sum is " , solve( $arr , $N );
// This code is contributed by nitin mittal. ?> |
Sum is 604
Time Complexity: O(Nlog2N) because of we are using the sort function in all the code snipets
Auxiliary Space: O(1)
Minimum sum of two numbers formed from digits of an array for large numbers using Strings:
The basic idea of approaching the question is the same as above, but instead of using numbers, strings will be used to handle sum of two large numbers
Follow the given steps to solve the problem:
- Sort the array in increasing order
- Declare two strings a and b, representing the two numbers to be formed
- Traverse the array and if the index is odd then add this element into string a, else add it to the string b
- Return the sum of two strings, in the form of a string
Below is the Implementation of the above approach:
// C++ code for the above approach #include <bits/stdc++.h> using namespace std;
string solve( int arr[], int n)
{ sort(arr, arr + n);
// Two String for storing our two minimum numbers
string a = "" , b = "" ;
for ( int i = 0; i < n; i += 2) {
a += (arr[i] + '0' );
}
for ( int i = 1; i < n; i += 2) {
b += (arr[i] + '0' );
}
int j = a.length() - 1;
int k = b.length() - 1;
// as initial carry is zero
int carry = 0;
string ans = "" ;
while (j >= 0 && k >= 0) {
int sum = 0;
sum += (a[j] - '0' ) + (b[k] - '0' ) + carry;
ans += to_string(sum % 10);
carry = sum / 10;
j--;
k--;
}
// If string b is over and string a is left
// here we don't need to put here while condition
// as it would run at max one time. Because the
// difference between both the strings could be at
// max 1.
while (j >= 0) {
int sum = 0;
sum += (a[j] - '0' ) + carry;
ans += to_string(sum % 10);
carry = sum / 10;
j--;
}
// If string a is over and string b is left
while (k >= 0) {
int sum = 0;
sum += (b[k] - '0' ) + carry;
ans += to_string(sum % 10);
carry = sum / 10;
k--;
}
// if carry is left
if (carry) {
ans += to_string(carry);
}
// to remove leading zeroes as they will be ahead of our
// sum
while (!ans.empty() and ans.back() == '0' )
ans.pop_back();
// reverse our final string because we were storing sum
// from left to right
reverse(ans.begin(), ans.end());
return ans;
} // Driver's Code int main()
{ int arr[] = { 6, 8, 4, 5, 2, 3 };
int N = sizeof (arr) / sizeof (arr[0]);
// Function call
cout << "Sum is " << solve(arr, N);
return 0;
} // Driver Code Ends
|
// Java code for the above approach import java.util.Arrays;
import java.util.Collections;
class Main {
public static String reverseString(String str)
{
StringBuilder sb = new StringBuilder(str);
sb.reverse();
return sb.toString();
}
public static String solve( int [] arr, int N)
{
Arrays.sort(arr);
// Two String for storing our two minimum numbers
String a = "" , b = "" ;
for ( int i = 0 ; i < N; i += 2 ) {
a += Integer.toString(arr[i]);
}
for ( int i = 1 ; i < N; i += 2 ) {
b += Integer.toString(arr[i]);
}
int j = a.length() - 1 ;
int k = b.length() - 1 ;
// As initial carry is zero
int carry = 0 ;
String ans = "" ;
while (j >= 0 && k >= 0 ) {
int sum = 0 ;
sum += (a.charAt(j) - '0' ) + (b.charAt(k) - '0' )
+ carry;
int x = sum % 10 ;
ans += Integer.toString(sum % 10 );
carry = sum / 10 ;
j--;
k--;
}
// If string b is over and string a is left
// here we dont need to put here while condition
// as it would run at max one time. Because the
// difference between both the strings could be at
// max 1.
while (j >= 0 ) {
int sum = 0 ;
sum += (a.charAt(j) - '0' ) + carry;
ans += Integer.toString(sum % 10 );
carry = sum / 10 ;
j--;
}
// If string a is over and string b is left
while (k >= 0 ) {
int sum = 0 ;
sum += (b.charAt(k) - '0' ) + carry;
ans += Integer.toString(sum % 10 );
carry = sum / 10 ;
k--;
}
// if carry is left
if (carry != 0 ) {
ans += Integer.toString(carry);
}
// to remove leading zeroes as they will be ahead of
// our sum
while (ans.isEmpty() == false
&& ans.charAt(ans.length() - 1 ) == '0' )
ans = ans.substring( 0 , ans.length() - 1 );
// reverse our final string because we were storing
// sum from left to right
ans = reverseString(ans);
return ans;
}
// driver's code
public static void main(String[] args)
{
int [] arr = { 6 , 8 , 4 , 5 , 2 , 3 };
int N = arr.length;
// Function call
System.out.print( "Sum is " + solve(arr, N));
}
} // This code is contributed by Chhavi |
# Python3 code for the approach def solve(arr, N):
arr.sort()
# Two String for storing our two minimum numbers
a, b = " ", " "
for i in range ( 0 , N, 2 ):
a + = str (arr[i])
for i in range ( 1 , N, 2 ):
b + = str (arr[i])
j = len (a) - 1
k = len (b) - 1
# As initial carry is zero
carry = 0
ans = ""
while (j > = 0 and k > = 0 ):
sum = 0
sum + = ( ord (a[j]) - ord ( '0' ) + ord (b[k]) - ord ( '0' )) + carry
ans + = str ( sum % 10 )
carry = sum / / 10
j - = 1
k - = 1
# If string b is over and string a is left
# here we dont need to put here while condition
# as it would run at max one time. Because the difference
# between both the strings could be at max 1.
while (j > = 0 ):
sum = 0
sum + = (a[j] - '0' ) + carry
ans + = ( sum % 10 ).toString()
carry = sum / / 10
j - = 1
# If string a is over and string b is left
while (k > = 0 ):
sum = 0
sum + = ord (b[k]) - ord ( '0' ) + carry
ans + = str ( sum % 10 )
carry = ( sum / / 10 )
k - = 1
# If carry is left
if (carry):
ans + = str (carry)
# To remove leading zeroes as they will be
# ahead of our sum
while ( len (ans) and ans[ len (ans) - 1 ] = = '0' ):
ans.pop()
# Reverse our final string because we were
# storing sum from left to right
ans = ans[:: - 1 ]
return ans
# Driver's Code if __name__ = = '__main__' :
arr = [ 6 , 8 , 4 , 5 , 2 , 3 ]
N = len (arr)
# Function call
print ( "Sum is " + solve(arr, N))
# This code is contributed by shinjanpatra |
// Include namespace system using System;
using System.Text;
using System.Linq;
using System.Collections;
public class GFG
{ public static string reverseString( string str) {
if (str.Length <= 1) return str;
else return reverseString(str.Substring(1)) + str[0];
} public static String solve( int [] arr, int N)
{
Array.Sort(arr);
// Two String for storing our two minimum numbers
var a = "" ;
var b = "" ;
for ( int i = 0; i < N;
i += 2)
{
a += Convert.ToString(arr[i]);
}
for ( int i = 1; i < N;
i += 2)
{
b += Convert.ToString(arr[i]);
}
var j = a.Length - 1;
var k = b.Length - 1;
// As initial carry is zero
var carry = 0;
var ans = "" ;
while (j >= 0 && k >= 0)
{
var sum = 0;
sum += (( int )(a[j]) - ( int )( '0' )) + (( int )(b[k]) - ( int )( '0' )) + carry;
ans += Convert.ToString(sum % 10);
carry = ( int )(sum / 10);
j--;
k--;
}
// If string b is over and string a is left
// here we dont need to put here while condition
// as it would run at max one time. Because the
// difference between both the strings could be at
// max 1.
while (j >= 0)
{
var sum = 0;
sum += (( int )(a[j]) - ( int )( '0' )) + carry;
ans += Convert.ToString(sum % 10);
carry = ( int )(sum / 10);
j--;
}
// If string a is over and string b is left
while (k >= 0)
{
var sum = 0;
sum += (( int )(b[k]) - ( int )( '0' )) + carry;
ans += Convert.ToString(sum % 10);
carry = ( int )(sum / 10);
k--;
}
// if carry is left
if (carry != 0)
{
ans += Convert.ToString(carry);
}
// to remove leading zeroes as they will be ahead of
// our sum
while (ans.Length == 0 == false && ans[ans.Length - 1] == '0' )
{ans = ans.Substring(0,ans.Length - 1-0);
}
// reverse our final string because we were storing
// sum from left to right
ans = GFG.reverseString(ans);
return ans;
}
// driver's code
public static void Main(String[] args)
{
int [] arr = {6, 8, 4, 5, 2, 3};
var N = arr.Length;
// Function call
Console.Write( "Sum is " + GFG.solve(arr, N));
}
} // This code is contributed by sourabhdalal0001. |
function solve(arr, N)
{ arr.sort();
// Two String for storing our two minimum numbers
let a = "" , b = "" ;
// string string alternatively
for (let i = 0; i < N; i += 2)
{
a += arr[i];
}
for (let i = 1; i < N; i += 2)
{
b += arr[i];
}
let j = a.length - 1;
let k = b.length - 1;
// as initial carry is zero
let carry = 0;
let ans = "" ;
while (j >= 0 && k >= 0)
{
let sum = 0;
sum += (a.charCodeAt(j) - '0' .charCodeAt(0)) + (b.charCodeAt(k) - '0' .charCodeAt(0)) + carry;
ans += (sum % 10).toString();
carry = Math.floor(sum / 10);
j--;
k--;
}
// if string b is over and string a is left
// here we dont need to put here while condition
// as it would run at max one time. Because the difference
// between both the strings could be at max 1.
while (j >= 0)
{
let sum = 0;
sum += (a[j] - '0' ) + carry;
ans += (sum % 10).toString();
carry = Math.floor(sum / 10);
j--;
}
// if string a is over and string b is left
while (k >= 0)
{
let sum = 0;
sum += (b.charCodeAt(k) - '0' .charCodeAt(0)) + carry;
ans += (sum % 10).toString();
carry = Math.floor(sum / 10);
k--;
}
// if carry is left
if (carry)
{
ans += carry.toString();
}
// to remove leading zeroes as they will be ahead of our sum
while (ans.length && ans[ans.length-1] == '0' )
ans.pop();
// reverse our final string because we were storing sum from left to right
ans = ans.split( '' ).reverse().join( '' );
return ans;
} // Driver Code Starts. let arr = [6, 8, 4, 5, 2, 3]; let N = arr.length; document.write( "Sum is " + solve(arr, N));
// This code is contributed by shinjanpatra |
Sum is 604
Time complexity: O(Nlog2N) because we are sorting the given array.
Auxiliary Space: O(1)