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 sumint 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 Codeint 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 approachimport 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 sumdef 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 Codeif __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 approachusing 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> |
Output:
16
Time Complexity: O(n)
Auxiliary Space: O(n)