给定两个大小为N的数组，以及两个数字X和Y，任务是通过考虑以下几点来使总和最大化：
- 从第一个数组中选取x个值，从第二个数组中选取y个值，以使X + Y值的总和最大。
- 假定X + Y等于N。
例子：
Input: arr1[] = {1, 4, 1}, arr2[] = {2, 5, 3}, N = 3, X = 2, Y = 1
Output: 8
In order to maximize sum from 2 arrays,
pick 1st and 2nd element from first array and 3rd from second array.
Input: A[] = {1, 4, 1, 2}, B[] = {4, 3, 2, 5}, N = 4, X = 2, Y = 2
Output: 14
方法：贪婪的方法可以用来解决上述问题。以下是必需的步骤：
- 通过找到两个数组元素之间的最大差异，首先找到那些具有最大值的数组元素。
- 为此，找到第一个数组和第二个数组的值之间的绝对差，然后将其存储在另一个数组中。
- 以降序对数组进行排序。
- 排序时，跟踪数组中元素的原始位置。
- 现在比较两个数组的元素，然后将更大的值添加到maxAmount中。
- 如果两者具有相同的值，则如果X不为零，则添加第一个数组的元素，否则添加第二个数组的元素。
- 遍历数组后，将完全返回所计算的maxAmount。
下面是上述方法的实现：
C++
// C++ program to print the maximum
// possible sum from two arrays.
#include
using namespace std;
// class that store values of two arrays
// and also store their absolute difference
class triplet {
public:
int first;
int second;
int diff;
triplet(int f, int s, int d)
: first(f), second(s), diff(d)
{
}
};
// Compare function used to sort array in decreasing order
bool compare(triplet& a, triplet& b)
{
return a.diff > b.diff; // decreasing order
}
/// Function to find the maximum possible
/// sum that can be generated from 2 arrays
int findMaxAmount(int arr1[], int arr2[], int n, int x, int y)
{
// vector where each index stores 3 things:
// Value of 1st array
// Value of 2nd array
// Their absolute difference
vector v;
for (int i = 0; i < n; i++) {
triplet t(arr1[i], arr2[i], abs(arr1[i] - arr2[i]));
v.push_back(t);
}
// sort according to their absolute difference
sort(v.begin(), v.end(), compare);
// it will store maximum sum
int maxAmount = 0;
int i = 0;
// Run loop for N times or
// value of X or Y becomes zero
while (i < n && x > 0 && y > 0) {
// if 1st array element has greater
// value, add it to maxAmount
if (v[i].first > v[i].second) {
maxAmount += v[i].first;
x--;
}
// if 2nd array element has greater
// value, add it to maxAmount
if (v[i].first < v[i].second) {
maxAmount += v[i].second;
y--;
}
// if both have same value, add element
// of first array if X is not zero
// else add element of second array
if (v[i].first == v[i].second) {
if (x > 0) {
maxAmount += v[i].first;
x--;
}
else if (y > 0) {
maxAmount += v[i].second;
y--;
}
}
// increment after picking element
i++;
}
// add the remaining values
// of first array to maxAmount
while (i < v.size() && x--) {
maxAmount += v[i++].first;
}
// add the remaining values of
// second array to maxAmount
while (i < v.size() && y--) {
maxAmount += v[i++].second;
}
return maxAmount;
}
// Driver Code
int main()
{
int A[] = { 1, 4, 1, 2 };
int B[] = { 4, 3, 2, 5 };
int n = sizeof(A) / sizeof(A[0]);
int X = 2, Y = 2;
cout << findMaxAmount(A, B, n, X, Y) << "\n";
} Java
// Java program to print the maximum
// possible sum from two arrays.
import java.util.*;
// class that store values of two arrays
// and also store their absolute difference
class Triplet implements Comparable
{
int first;
int second;
int diff;
Triplet(int f, int s, int d)
{
first = f;
second = s;
diff = d;
}
// CompareTo function used to sort
// array in decreasing order
public int compareTo(Triplet o)
{
return o.diff - this.diff;
}
}
class GFG{
// Function to find the maximum possible
// sum that can be generated from 2 arrays
public static int findMaxAmount(int arr1[],
int arr2[],
int n, int x,
int y)
{
// Vector where each index
// stores 3 things:
// Value of 1st array
// Value of 2nd array
// Their absolute difference
Vector v = new Vector<>();
for(int i = 0; i < n; i++)
{
v.add(new Triplet(arr1[i], arr2[i],
Math.abs(arr1[i] -
arr2[i])));
}
// Sort according to their
// absolute difference
Collections.sort(v);
// It will store maximum sum
int maxAmount = 0;
int i = 0;
// Run loop for N times or
// value of X or Y becomes zero
while (i < n && x > 0 && y > 0)
{
// If 1st array element has greater
// value, add it to maxAmount
if (v.get(i).first > v.get(i).second)
{
maxAmount += v.get(i).first;
x--;
}
// If 2nd array element has greater
// value, add it to maxAmount
if (v.get(i).first < v.get(i).second)
{
maxAmount += v.get(i).second;
y--;
}
// If both have same value, add element
// of first array if X is not zero
// else add element of second array
if (v.get(i).first == v.get(i).second)
{
if (x > 0)
{
maxAmount += v.get(i).first;
x--;
}
else if (y > 0)
{
maxAmount += v.get(i).second;
y--;
}
}
// Increment after picking element
i++;
}
// Add the remaining values
// of first array to maxAmount
while (i < v.size() && x-- > 0)
{
maxAmount += v.get(i++).first;
}
// Add the remaining values of
// second array to maxAmount
while (i < v.size() && y-- > 0)
{
maxAmount += v.get(i++).second;
}
return maxAmount;
}
// Driver Code
public static void main(String []args)
{
int A[] = { 1, 4, 1, 2 };
int B[] = { 4, 3, 2, 5 };
int n = A.length;
int X = 2, Y = 2;
System.out.println(findMaxAmount(A, B, n, X, Y));
}
}
// This code is contributed by jrishabh99 Python3
# Python3 program to print the maximum
# possible sum from two arrays.
# Class that store values of two arrays
# and also store their absolute difference
class triplet:
def __init__(self, f, s, d):
self.first = f
self.second = s
self.diff = d
# Function to find the maximum possible
# sum that can be generated from 2 arrays
def findMaxAmount(arr1, arr2, n, x, y):
# vector where each index stores 3 things:
# Value of 1st array
# Value of 2nd array
# Their absolute difference
v = []
for i in range(0, n):
t = triplet(arr1[i], arr2[i],
abs(arr1[i] - arr2[i]))
v.append(t)
# sort according to their absolute difference
v.sort(key = lambda x: x.diff, reverse = True)
# it will store maximum sum
maxAmount, i = 0, 0
# Run loop for N times or
# value of X or Y becomes zero
while i < n and x > 0 and y > 0:
# if 1st array element has greater
# value, add it to maxAmount
if v[i].first > v[i].second:
maxAmount += v[i].first
x -= 1
# if 2nd array element has greater
# value, add it to maxAmount
if v[i].first < v[i].second:
maxAmount += v[i].second
y -= 1
# if both have same value, add element
# of first array if X is not zero
# else add element of second array
if v[i].first == v[i].second:
if x > 0:
maxAmount += v[i].first
x -= 1
elif y > 0:
maxAmount += v[i].second
y -= 1
# increment after picking element
i += 1
# add the remaining values
# of first array to maxAmount
while i < len(v) and x > 0:
maxAmount += v[i].first
i, x = i + 1, x - 1
# add the remaining values of
# second array to maxAmount
while i < len(v) and y > 0:
maxAmount += v[i].second
i, y = i + 1, y - 1
return maxAmount
# Driver Code
if __name__ == "__main__":
A = [1, 4, 1, 2]
B = [4, 3, 2, 5]
n = len(A)
X, Y = 2, 2
print(findMaxAmount(A, B, n, X, Y))
# This code is contributed by Rituraj Jain输出：
14时间复杂度： O(N log N)
辅助空间： O(N)