Largest number in given Array formed by repeatedly combining two same elements
Given an array arr[], the task is to find the largest number in given Array, formed by repeatedly combining two same elements. If there are no same elements in the array initially, then print the output as -1.
Examples:
Input: arr = {1, 1, 2, 4, 7, 8}
Output: 16
Explanation:
Repetition 1: Combine 1s from the array and insert the sum 2 in it. Updated Array = {2, 2, 4, 7, 8}
Repetition 2: Combine 2s from the array and insert the sum 4 in it. Updated Array = {4, 4, 7, 8}
Repetition 3: Combine 4s from the array and insert the sum 8 in it. Updated Array = {8, 7, 8}
Repetition 4: Combine 8s from the array and insert the sum 16 in it. Updated Array = {7, 16}
Largest sum = 16
Input: arr = {1, 2, 3}
Output: -1
Explanation:
There are no duplicate elements in the array initially. Hence no combination can be performed.
Approach: This problem can be solved using the frequency of elements in the given array.
- Find and store the frequencies of each element of the given array in a map.
- Upon traversing the frequency map for each distinct element, if there is any duplicate element K in the map with a frequency more than 1, then increase the frequency of element 2*K by half the frequency of the K element.
- Finally, find the maximum number from the map.
Below is the implementation of the above approach:
CPP
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std; // Function to return the largest sum int largest_sum( int arr[], int n) { // Variable to store the largest sum int maximum = -1; // Map to store the frequencies // of each element map< int , int > m; // Store the Frequencies for ( int i = 0; i < n; i++) { m[arr[i]]++; } // Loop to combine duplicate elements // and update the sum in the map for ( auto j : m) { // If j is a duplicate element if (j.second > 1) { // Update the frequency of 2*j m[2 * j.first] = m[2 * j.first] + j.second / 2; // If the new sum is greater than // maximum value, Update the maximum if (2 * j.first > maximum) maximum = 2 * j.first; } } // Returns the largest sum return maximum; } // Driver Code int main() { int arr[] = { 1, 1, 2, 4, 7, 8 }; int n = sizeof (arr) / sizeof (arr[0]); // Function Calling cout << largest_sum(arr, n); return 0; } |
Java
// Java implementation of the above approach import java.util.*; class GFG { // Function to return the largest sum static int largest_sum( int arr[], int n) { // Variable to store the largest sum int maximum = - 1 ; // Map to store the frequencies // of each element HashMap<Integer, Integer> m = new HashMap<Integer, Integer>(); // Store the Frequencies for ( int i = 0 ; i < n; i++) { if (m.containsKey(arr[i])){ m.put(arr[i], m.get(arr[i]) + 1 ); } else { m.put(arr[i], 1 ); } } // Loop to combine duplicate elements // and update the sum in the map for ( int i = 0 ; i < n; i++){ // If j is a duplicate element if (m.get(arr[i]) > 1 ) { if (m.containsKey( 2 *arr[i])) { // Update the frequency of 2*j m.put( 2 *arr[i],m.get( 2 * arr[i])+ m.get(arr[i]) / 2 ); } else { m.put( 2 *arr[i],m.get(arr[i]) / 2 ); } // If the new sum is greater than // maximum value, Update the maximum if ( 2 * arr[i] > maximum) maximum = 2 * arr[i]; } } // Returns the largest sum return maximum; } // Driver Code public static void main (String[] args) { int arr[] = { 1 , 1 , 2 , 4 , 7 , 8 }; int n = arr.length; // Function Calling System.out.println(largest_sum(arr, n)); } } // This code is contributed by Yash_R |
Python3
# Python3 implementation of the above approach # Function to return the largest sum def largest_sum(arr, n): # Variable to store the largest sum maximum = - 1 # Map to store the frequencies # of each element m = dict () # Store the Frequencies for i in arr: m[i] = m.get(i, 0 ) + 1 # Loop to combine duplicate elements # and update the sum in the map for j in list (m): # If j is a duplicate element if ((j in m) and m[j] > 1 ): # Update the frequency of 2*j x, y = 0 , 0 if 2 * j in m: m[ 2 * j] = m[ 2 * j] + m[j] / / 2 else : m[ 2 * j] = m[j] / / 2 # If the new sum is greater than # maximum value, Update the maximum if ( 2 * j > maximum): maximum = 2 * j # Returns the largest sum return maximum # Driver Code if __name__ = = '__main__' : arr = [ 1 , 1 , 2 , 4 , 7 , 8 ] n = len (arr) # Function Calling print (largest_sum(arr, n)) # This code is contributed by mohit kumar 29 |
C#
// C# implementation of the above approach using System; using System.Collections.Generic; class GFG { // Function to return the largest sum static int largest_sum( int []arr, int n) { // Variable to store the largest sum int maximum = -1; // Map to store the frequencies // of each element Dictionary< int , int > m = new Dictionary< int , int >(); // Store the Frequencies for ( int i = 0; i < n; i++) { if (m.ContainsKey(arr[i])){ m[arr[i]]++; } else { m.Add(arr[i] , 1); } } // Loop to combine duplicate elements // and update the sum in the map for ( int i = 0; i < n; i++){ // If j is a duplicate element //Console.Write(m[arr[i]]); if (m[arr[i]] > 1) { if (m.ContainsKey(2*arr[i])) { // Update the frequency of 2*j m[2*arr[i]] = m[2 * arr[i]]+ m[arr[i]] / 2; } else { m.Add(2*arr[i],m[arr[i]] / 2); } // If the new sum is greater than // maximum value, Update the maximum if (2 * arr[i] > maximum) maximum = 2 * arr[i]; } } // Returns the largest sum return maximum; } // Driver Code public static void Main () { int [] arr = { 1, 1, 2, 4, 7, 8 }; int n = arr.Length; // Function Calling Console.Write(largest_sum(arr, n)); } } // This code is contributed by chitranayal |
Javascript
<script> // Javascript implementation of the above approach // Function to return the largest sum function largest_sum(arr, n) { // Variable to store the largest sum let maximum = -1; // Map to store the frequencies // of each element let m = new Map(); // Store the Frequencies for (let i = 0; i < n; i++) { if (m.has(arr[i])){ m.set(arr[i], m.get(arr[i]) + 1); } else { m.set(arr[i], 1); } } // Loop to combine duplicate elements // and update the sum in the map for (let i = 0; i < n; i++){ // If j is a duplicate element if (m.get(arr[i]) > 1) { if (m.has(2*arr[i])) { // Update the frequency of 2*j m.set(2*arr[i],m.get(2 * arr[i])+ m.get(arr[i]) / 2); } else { m.set(2*arr[i],m.get(arr[i]) / 2); } // If the new sum is greater than // maximum value, Update the maximum if (2 * arr[i] > maximum) maximum = 2 * arr[i]; } } // Returns the largest sum return maximum; } // Driver code let arr = [ 1, 1, 2, 4, 7, 8 ]; let n = arr.length; // Function Calling document.write(largest_sum(arr, n)); </script> |
16
Time Complexity: O(n)
Auxiliary Space: O(n)