Minimum sum of two numbers formed from digits of an array
Last Updated : 20 Jun, 2025
Given an array arr[] of digits (values are from 0 to 9), find the minimum possible sum of two numbers formed using all digits of the array.
Note: We need to return the minimum possible sum as a string.
Examples:
Input: arr[] = [6, 8, 4, 5, 2, 3, 0]
Output: "604"
Explanation: The minimum sum is formed by numbers 0358 and 246.
Input: arr[] = [5, 3, 0, 7, 4]
Output: "82"
Explanation: The minimum sum is formed by numbers 35 and 047.
[Naive Approach] Using Sorting - O(n logn) time and O(n) space
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 the first number by digits present in odd positions and second number by digits from even positions in the array. So, that the first two smallest digits are placed in the most significant positions of both numbers, the next two smallest digits are placed in the second most significant positions, and this pattern continues for the remaining digits.
Illustration:
C++ #include <algorithm> #include <iostream> #include <vector> using namespace std; // Function to add two strings and return the result string addString(string &s1, string &s2) { int i = s1.length() - 1; int j = s2.length() - 1; // initial carry is zero int carry = 0; // we will calculate and store the // resultant sum in reverse order in res string res = ""; while (i >= 0 || j >= 0 || carry > 0) { int sum = carry; if (i >= 0) sum += (s1[i] - '0'); if (j >= 0) sum += (s2[j] - '0'); res.push_back(sum % 10 + '0'); carry = sum / 10; i--; j--; } // remove leading zeroes which are currently // at the back due to reversed string res while (!res.empty() && res.back() == '0') res.pop_back(); // reverse our final string reverse(res.begin(), res.end()); return res; } // Function to find and return minimum sum of // two numbers formed from digits of the array. string minSum(vector<int> &arr) { sort(arr.begin(), arr.end()); // Two String for storing the two minimum numbers string s1 = "", s2 = ""; for (int i = 0; i < arr.size(); i++) { if (i % 2 == 0) s1.push_back(arr[i] + '0'); else s2.push_back(arr[i] + '0'); } return addString(s1, s2); } int main() { vector<int> arr = {6, 8, 4, 5, 2, 3, 0}; cout << minSum(arr); return 0; } import java.util.*; class GfG { // Function to add two numeric strings and return the sum as a string static String addString(String s1, String s2) { int i = s1.length() - 1; int j = s2.length() - 1; int carry = 0; String res = ""; while (i >= 0 || j >= 0 || carry > 0) { int sum = carry; if (i >= 0) sum += s1.charAt(i--) - '0'; if (j >= 0) sum += s2.charAt(j--) - '0'; res = (sum % 10) + res; carry = sum / 10; } return res; } // Function to find minimum sum of two numbers formed from digits static String minSum(int[] arr) { Arrays.sort(arr); String s1 = "", s2 = ""; for (int i = 0; i < arr.length; i++) { if (i % 2 == 0) // add to first number s1 += arr[i]; else // add to second number s2 += arr[i]; } return addString(s1, s2); } public static void main(String[] args) { int[] arr = {6, 8, 4, 5, 2, 3, 0}; System.out.println(minSum(arr)); } } # Function to add two numbers and return the result def addNumbers(l1, l2): i = len(l1) - 1 j = len(l2) - 1 # initial carry is zero carry = 0 # we will calculate and store the # resultant sum in reverse order in res res = [] while i >= 0 or j >= 0 or carry > 0: total = carry if i >= 0: total += l1[i] if j >= 0: total += l2[j] res.append(str(total % 10)) carry = total // 10 i -= 1 j -= 1 # remove leading zeroes which are currently # at the back due to reversed string res while len(res) > 0 and res[-1] == '0': res.pop() # reverse our final result res = res[::-1] return ''.join(res) # Function to find and return minimum sum of # two numbers formed from digits of the array. def minSum(arr): arr.sort() # Two Lists for storing the two minimum numbers l1 = [] l2 = [] for i in range(len(arr)): if i % 2 == 0: l1.append(arr[i]) else: l2.append(arr[i]) return addNumbers(l1, l2) if __name__ == '__main__': arr = [6, 8, 4, 5, 2, 3, 0] print(minSum(arr))
using System; class GfG{ // Function to add two numeric strings and return the result as a string static string addString(string s1, string s2){ int i = s1.Length - 1; int j = s2.Length - 1; int carry = 0; string res = ""; // Loop until both strings are processed and no carry remains while (i >= 0 || j >= 0 || carry > 0){ int sum = carry; // Add current digit from s1 if any if (i >= 0) sum += s1[i--] - '0'; // Add current digit from s2 if any if (j >= 0) sum += s2[j--] - '0'; // Add least significant digit of sum to result res = (sum % 10) + res; carry = sum / 10; } // Remove any leading zeros (except when result is "0") int k = 0; while (k < res.Length - 1 && res[k] == '0') k++; return res.Substring(k); } // Function to find and return minimum possible sum of two numbers // formed by using all digits in the array static string minSum(int[] arr){ // Sort the digits in ascending order Array.Sort(arr); string s1 = ""; string s2 = ""; // Distribute digits alternately between s1 and s2 for (int i = 0; i < arr.Length; i++) { if (i % 2 == 0) s1 += arr[i]; else s2 += arr[i]; } // Add the two numbers and return their sum as a string return addString(s1, s2); } static void Main(string[] args) { int[] arr = { 6, 8, 4, 5, 2, 3, 0 }; Console.WriteLine(minSum(arr)); } } // Function to add two lists and return the result function addNumbers(l1, l2) { let i = l1.length - 1; let j = l2.length - 1; // initial carry is zero let carry = 0; // we will calculate and store the // resultant sum in reverse order in res let res = []; while (i >= 0 || j >= 0 || carry > 0) { let sum = carry; if (i >= 0) sum += l1[i] - '0'; if (j >= 0) sum += l2[j] - '0'; res.push(sum % 10); carry = Math.floor(sum / 10); i--; j--; } // remove leading zeroes which are currently // at the back due to reversed list res while (res.length > 0 && res[res.length - 1] === 0) res.pop(); // reverse our final list and convert it to a string return res.reverse().join(''); } // Function to find and return minimum sum of // two numbers formed from digits of the array. function minSum(arr) { arr.sort((a, b) => a - b); // Two lists for storing the two minimum numbers let l1 = [], l2 = []; for (let i = 0; i < arr.length; i++) { if (i % 2 === 0) l1.push(arr[i]); else l2.push(arr[i]); } return addNumbers(l1, l2); } // Driver Code let arr = [6, 8, 4, 5, 2, 3, 0]; console.log(minSum(arr)); Time Complexity: O(n*logn) as we are sorting the input array.
Auxiliary Space: O(n) to store the result.
[Expected Approach] Using Count Array - O(n) time and O(n) space
The idea is to use a count array to track the frequency of each digit in the input array, eliminating the need for sorting. By iterating through the count array from 0 to 9 and alternately distributing digits to two strings while avoiding leading zeros, we can efficiently create the two smallest numbers that can be formed using all the input digits.
C++ #include <iostream> #include <vector> #include <string> #include <algorithm> using namespace std; // Function to add two strings and return the result string addString(string &s1, string &s2) { int i = s1.length() - 1; int j = s2.length() - 1; // initial carry is zero int carry = 0; // we will calculate and store the // resultant sum in reverse order in res string res = ""; while (i >= 0 || j >= 0 || carry > 0) { int sum = carry; if (i >= 0) sum += (s1[i] - '0'); if (j >= 0) sum += (s2[j] - '0'); res.push_back(sum % 10 + '0'); carry = sum / 10; i--; j--; } // remove leading zeroes which are currently // at the back due to reversed string res while (!res.empty() && res.back() == '0') res.pop_back(); // reverse our final string reverse(res.begin(), res.end()); return res; } // Function to find minimum sum using count array approach string minSum(vector<int> &arr) { // Count array to store frequency of each digit vector<int> count(10, 0); // Count occurrences of each digit for (int num : arr) { count[num]++; } // Two strings for storing the two minimum numbers string s1 = "", s2 = ""; // Flag to alternate between s1 and s2 bool firstNum = true; // Traverse count array from 0 to 9 for (int digit = 0; digit < 10; digit++) { // Add digit to appropriate string // while it has occurrences while (count[digit]--) { if (firstNum) { // Avoid leading zeros for both numbers if (!(s1.empty() && digit == 0)) s1.push_back(digit + '0'); firstNum = false; } else { if (!(s2.empty() && digit == 0)) s2.push_back(digit + '0'); firstNum = true; } } } // Handle case where s1 or s2 might be empty if (s1.empty()) s1 = "0"; if (s2.empty()) s2 = "0"; return addString(s1, s2); } int main() { vector<int> arr = {6, 8, 4, 5, 2, 3, 0}; cout << minSum(arr) << endl; return 0; } import java.util.*; class GfG { // Function to add two strings and return the result static String addString(String s1, String s2) { int i = s1.length() - 1; int j = s2.length() - 1; // initial carry is zero int carry = 0; // we will calculate and store the // resultant sum in reverse order in res StringBuilder res = new StringBuilder(); while (i >= 0 || j >= 0 || carry > 0) { int sum = carry; if (i >= 0) sum += (s1.charAt(i) - '0'); if (j >= 0) sum += (s2.charAt(j) - '0'); res.append(sum % 10); carry = sum / 10; i--; j--; } // remove leading zeroes which are currently // at the back due to reversed string res while (res.length() > 0 && res.charAt(res.length() - 1) == '0') res.deleteCharAt(res.length() - 1); // reverse our final string res.reverse(); return res.length() == 0 ? "0" : res.toString(); } // Function to find minimum sum using count array approach static String minSum(int[] arr) { // Count array to store frequency of each digit int[] count = new int[10]; // Count occurrences of each digit for (int num : arr) { count[num]++; } // Two strings for storing the two minimum numbers StringBuilder s1 = new StringBuilder(); StringBuilder s2 = new StringBuilder(); // Flag to alternate between s1 and s2 boolean firstNum = true; // Traverse count array from 0 to 9 for (int digit = 0; digit < 10; digit++) { // Add digit to appropriate string // while it has occurrences while (count[digit]-- > 0) { if (firstNum) { // Avoid leading zeros for both numbers if (!(s1.length() == 0 && digit == 0)) s1.append(digit); firstNum = false; } else { if (!(s2.length() == 0 && digit == 0)) s2.append(digit); firstNum = true; } } } // Handle case where s1 or s2 might be empty if (s1.length() == 0) s1.append("0"); if (s2.length() == 0) s2.append("0"); return addString(s1.toString(), s2.toString()); } public static void main(String[] args) { int[] arr = {6, 8, 4, 5, 2, 3, 0}; System.out.println(minSum(arr)); } } # Function to add two strings and return the result def addString(s1, s2): i = len(s1) - 1 j = len(s2) - 1 # initial carry is zero carry = 0 # we will calculate and store the # resultant sum in reverse order in res res = [] while i >= 0 or j >= 0 or carry > 0: total = carry if i >= 0: total += int(s1[i]) if j >= 0: total += int(s2[j]) res.append(str(total % 10)) carry = total // 10 i -= 1 j -= 1 # remove leading zeroes which are currently # at the back due to reversed string res while res and res[-1] == '0': res.pop() # reverse our final string return ''.join(reversed(res)) if res else "0" # Function to find minimum sum using count array approach def minSum(arr): # Count array to store frequency of each digit count = [0] * 10 # Count occurrences of each digit for num in arr: count[num] += 1 # Two strings for storing the two minimum numbers s1 = [] s2 = [] # Flag to alternate between s1 and s2 firstNum = True # Traverse count array from 0 to 9 for digit in range(10): while count[digit] > 0: if firstNum: if not (len(s1) == 0 and digit == 0): s1.append(str(digit)) firstNum = False else: if not (len(s2) == 0 and digit == 0): s2.append(str(digit)) firstNum = True count[digit] -= 1 # Handle case where s1 or s2 might be empty if not s1: s1.append("0") if not s2: s2.append("0") return addString(''.join(s1), ''.join(s2)) if __name__ == "__main__": arr = [6, 8, 4, 5, 2, 3, 0] print(minSum(arr)) using System; using System.Collections.Generic; class GfG { // Function to add two strings and return the result static string addString(string s1, string s2) { int i = s1.Length - 1; int j = s2.Length - 1; // initial carry is zero int carry = 0; // we will calculate and store the // resultant sum in reverse order in res List<char> res = new List<char>(); while (i >= 0 || j >= 0 || carry > 0) { int sum = carry; if (i >= 0) sum += (s1[i] - '0'); if (j >= 0) sum += (s2[j] - '0'); res.Add((char)(sum % 10 + '0')); carry = sum / 10; i--; j--; } // remove leading zeroes which are currently // at the back due to reversed string res while (res.Count > 0 && res[res.Count - 1] == '0') res.RemoveAt(res.Count - 1); // reverse our final string res.Reverse(); return res.Count == 0 ? "0" : new string(res.ToArray()); } // Function to find minimum sum using count array approach static string minSum(int[] arr) { // Count array to store frequency of each digit int[] count = new int[10]; // Count occurrences of each digit foreach (int num in arr) { count[num]++; } // Two strings for storing the two minimum numbers string s1 = ""; string s2 = ""; // Flag to alternate between s1 and s2 bool firstNum = true; // Traverse count array from 0 to 9 for (int digit = 0; digit < 10; digit++) { // Add digit to appropriate string // while it has occurrences while (count[digit]-- > 0) { if (firstNum) { // Avoid leading zeros for both numbers if (!(s1.Length == 0 && digit == 0)) s1 += digit.ToString(); firstNum = false; } else { if (!(s2.Length == 0 && digit == 0)) s2 += digit.ToString(); firstNum = true; } } } // Handle case where s1 or s2 might be empty if (s1.Length == 0) s1 = "0"; if (s2.Length == 0) s2 = "0"; return addString(s1, s2); } static void Main() { int[] arr = { 6, 8, 4, 5, 2, 3, 0 }; Console.WriteLine(minSum(arr)); } } // Function to add two strings and return the result function addString(s1, s2) { let i = s1.length - 1; let j = s2.length - 1; // initial carry is zero let carry = 0; // we will calculate and store the // resultant sum in reverse order in res let res = []; while (i >= 0 || j >= 0 || carry > 0) { let total = carry; if (i >= 0) total += parseInt(s1[i]); if (j >= 0) total += parseInt(s2[j]); res.push(total % 10); carry = Math.floor(total / 10); i--; j--; } // remove leading zeroes which are currently // at the back due to reversed string res while (res.length > 0 && res[res.length - 1] === 0) { res.pop(); } // reverse our final string return res.reverse().join('') || "0"; } // Function to find minimum sum using count array approach function minSum(arr) { // Count array to store frequency of each digit let count = new Array(10).fill(0); // Count occurrences of each digit for (let num of arr) { count[num]++; } // Two strings for storing the two minimum numbers let s1 = []; let s2 = []; // Flag to alternate between s1 and s2 let firstNum = true; // Traverse count array from 0 to 9 for (let digit = 0; digit < 10; digit++) { while (count[digit]-- > 0) { if (firstNum) { if (!(s1.length === 0 && digit === 0)) s1.push(digit); firstNum = false; } else { if (!(s2.length === 0 && digit === 0)) s2.push(digit); firstNum = true; } } } // Handle case where s1 or s2 might be empty if (s1.length === 0) s1.push(0); if (s2.length === 0) s2.push(0); return addString(s1.join(''), s2.join('')); } // Driver Code let arr = [6, 8, 4, 5, 2, 3, 0]; console.log(minSum(arr)); Similar Reads
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