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.

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

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 ans 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. Finally, we return the sum of first and second number.

Below is C++ implementation of above idea.

// 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 code int main() { int arr[] = {6, 8, 4, 5, 2, 3}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Sum is " << solve(arr, n); return 0; }

Output :

Sum is 604

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