Given an array arr[] of N positive elements, the task is to find a pair in the array such that the digit sum of elements in the pair are equal and their sum is maximum. Print -1 if it is not possible to find the pair else print the maximum sum.
Examples:
Input: arr[] = {55, 23, 32, 46, 88}
Output: 101
(23, 32) with digit sum 5 and (55, 46)
with digit sum 10 are the only pairs
with equal digit sum. The pair with maximum
sum is (55, 46) i.e. 55 + 46 = 101Input: arr[] = {18, 19, 23, 15}
Output: -1
Since there are no two elements
whose digit sums are equal.
Approach: Traverse through the array and use a unordered_map to store the maximum number present for a particular digit sum. For every element arr[i], find the digit sum of arr[i] and check if there is a value present in the map for this digit sum, if yes then update the answer by ans = max(ans, arr[i] + map[digSum]). Also, update the map if the value of arr[i] is greater than the value present in the map.
Below is the implementation of the above approach:
CPP
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the // digit sum of n long digitSum(long n) { long sum = 0; while (n) { sum += (n % 10); n /= 10; } return sum; } // Function to return the maximum // sum possible for a pair whose // digit sum is equal int findSum(vector<long> arr, int n) { unordered_map<long, long> mp; long ans = -1; // Traversing through all the // elements of the array for (long i = 0; i < n; i++) { // Finding the digit sum of arr[i] long dSum = digitSum(arr[i]); if (mp[dSum] != 0) { ans = max(ans, mp[dSum] + arr[i]); } mp[dSum] = max(mp[dSum], arr[i]); } return ans; } // Driver code int main() { vector<long> arr = { 55, 23, 32, 46, 88 }; int n = arr.size(); cout << findSum(arr, n); return 0; } |
Java
// Java implementation of the approach import java.util.*; class GFG { // Function to return the // digit sum of n static int digitSum(int n) { int sum = 0; while (n != 0) { sum += (n % 10); n /= 10; } return sum; } // Function to return the maximum // sum possible for a pair whose // digit sum is equal static int findSum(int arr[], int n) { HashMap<Integer, Integer> mp = new HashMap<Integer,Integer>(); for (int i = 0; i < 1000 ; i++) { mp.put(i, 0); } int ans = -1; // Traversing through all the // elements of the array for (int i = 0; i < n; i++) { // Finding the digit sum of arr[i] int dSum = digitSum(arr[i]); if ((int)mp.get(dSum) != 0) { ans = Math.max(ans, (int)mp.get(dSum) + arr[i]); } mp.put(dSum,Math.max((int)mp.get(dSum), arr[i])) ; } return ans; } // Driver code public static void main (String[] args) { int arr[] = { 55, 23, 32, 46, 88 }; int n = arr.length; System.out.println(findSum(arr, n)); } } // This code is contributed by AnkitRai01 |
Python
# Python3 implementation of the approach # Function to return the # digit sum of n def digitSum(n): sum = 0 while (n): sum += (n % 10) n //= 10 return sum # Function to return the maximum # sum possible for a pair whose # digit sum is equal def findSum(arr, n): mp = dict() ans = -1 # Traversing through all the # elements of the array for i in range(n): # Finding the digit sum of arr[i] dSum = digitSum(arr[i]) if (dSum in mp): ans = max(ans, mp[dSum] + arr[i]) mp[dSum] = mp.get(dSum, 0) + arr[i] return ans # Driver code arr = [55, 23, 32, 46, 88 ] n = len(arr) print(findSum(arr, n)) # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approach using System; using System.Collections.Generic; class GFG { // Function to return the // digit sum of n static int digitSum(int n) { int sum = 0; while (n != 0) { sum += (n % 10); n /= 10; } return sum; } // Function to return the maximum // sum possible for a pair whose // digit sum is equal static int findSum(int[] arr, int n) { Dictionary<int,int> mp = new Dictionary<int,int>(); for (int i = 0; i < 1000 ; i++) { mp.Add(i, 0); } int ans = -1; // Traversing through all the // elements of the array for (int i = 0; i < n; i++) { // Finding the digit sum of arr[i] int dSum = digitSum(arr[i]); if ((int)mp[dSum] != 0) { ans = Math.Max(ans, (int)mp[dSum] + arr[i]); } mp[dSum]=Math.Max((int)mp[dSum], arr[i]); } return ans; } // Driver code public static void Main () { int[] arr = { 55, 23, 32, 46, 88 }; int n = arr.Length; Console.Write(findSum(arr, n)); } } // This code is contributed by chitranayal |
101
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