Minimum sum of two numbers formed from digits of an array in O(n)
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++
// C++ implementation of above approach#include<bits/stdc++.h>using namespace std;// Function to return the required minimum sumint 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 codeint 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
// Java implementation of above approachpublic 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)); }} |
Python3
# Python implementation of above approach# Function to return the required minimum sumdef 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 codearr = [ 6, 8, 4, 5, 2, 3 ]n = len(arr);print(minSum(arr, n))#This code is contributed by ankush_953 |
C#
// C# implementation of above approachusing 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. |
Javascript
<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)
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