给定一个数组arr[] ，任务是在给定的 Array 中找到最大的数，由两个相同的元素重复组合而成。如果最初数组中没有相同的元素，则将输出打印为 -1。
例子：
Input: arr = {1, 1, 2, 4, 7, 8}
Output: 16
Explanation:
Repitition 1: Combine 1s from the array and insert the sum 2 in it. Updated Array = {2, 2, 4, 7, 8}
Repitition 2: Combine 2s from the array and insert the sum 4 in it. Updated Array = {4, 4, 7, 8}
Repitition 3: Combine 4s from the array and insert the sum 8 in it. Updated Array = {8, 7, 8}
Repitition 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.
方法：这个问题可以使用给定数组中元素的频率来解决。
- 在地图中查找并存储给定数组的每个元素的频率。
- 在遍历每个不同元素的频率图时，如果图中存在频率大于 1 的任何重复元素K ，则将元素2*K的频率增加K元素频率的一半。
- 最后，从地图中找到最大数量。
下面是上述方法的实现：
CPP
// C++ implementation of the above approach
#include
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 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 m = new HashMap();
// 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 29C#
// 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 m = new Dictionary();
// 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
输出：
16