给定一个正整数M和两个分别由N和K 个正整数组成的数组arr[]和value[] ，任务是将value[]中的每个元素与arr[] 中的一个元素相加，这样在执行所有加法之后，数组中的最大元素最多为M 。如果可以这样做，则打印“是” 。否则，打印“否” 。
例子：
Input: arr[] = {5, 9, 3, 8, 7}, value[] = {1, 2, 3, 4}, M = 9
Output: Yes
Explanation:
Add 1 & 3 to arr[0] maximizing it to 9.
Add 2 & 4 to arr[2] maximizes it to 9.
Hence, the final arr becomes {9, 9, 9, 8, 7}.
Input: arr[] = {5, 8, 3, 8, 7}, value[] = {1, 2, 3, 4}, M = 8
Output: No
Explanation:
Adding 1 to arr[4], 3 to arr[0] and 4 to arr[2], the array is modified to {8, 8, 7, 8, 8}.
The current maximum element in arr[] is 8.
Hence, only 2 needs to be added from value[] to any element of arr[].
But, on adding 2 to any element in arr[], the maximum element in the array exceeds 8.
天真的方法：
最简单的方法是从给定的数组arr[] 中选择任意K 个元素，并将K 个值添加到value[]数组中，并选择这些K 个值。这些K值可以添加到K中数组arr[]的K 个选定数字中！方式(在最坏的情况下)。
时间复杂度： O( N P K )
有效的方法：
请按照以下步骤解决问题：
- 按降序对value[]数组中的元素进行排序。
- 将arr[] 的所有元素存储在最小优先级队列中。
- 现在从优先级队列中提取最小元素(比如X )并将数组value[] 中的元素添加到X 。
- 当将数组value[] 中的值添加到X超过M 时，将元素X插入优先级队列，并为优先级队列中的下一个最小值重复上述步骤。
- 如果value[]中的所有元素都添加到arr[] 中的某些元素，则为“Yes” ，否则打印“No” 。
下面是上述方法的实现：
C++
// C++ Program to implement the
// above approach
#include
using namespace std;
// Function which checks if all
// additions are possible
void solve(int ar[], int values[],
int N, int K, int M)
{
// Sorting values[] in
// decreasing order
sort(values, values + K,
greater());
// Minimum priority queue which
// contains all the elements
// of array arr[]
priority_queue,
greater >
pq;
for (int x = 0; x < N; x++) {
pq.push(ar[x]);
}
// poss stores whether all the
// additions are possible
bool poss = true;
for (int x = 0; x < K; x++) {
// Minium value in the
// priority queue
int mini = pq.top();
pq.pop();
int val = mini + values[x];
// If on addition it exceeds
// M then not possible
if (val > M) {
poss = false;
break;
}
pq.push(val);
}
// If all elements are added
if (poss) {
cout << "Yes" << endl;
}
else {
cout << "No" << endl;
}
}
// Driver Code
int main()
{
int ar[] = { 5, 9, 3, 8, 7 };
int N = 5;
int values[] = { 1, 2, 3, 4 };
int K = 4;
int M = 9;
solve(ar, values, N, K, M);
return 0;
} Java
// Java program to implement the
// above approach
import java.io.*;
import java.util.*;
class GFG{
// Function which checks if all
// additions are possible
static void solve(Integer ar[], Integer values[],
int N, int K, int M)
{
// Sorting values[] in
// decreasing order
Arrays.sort(values, (a, b) -> b - a);
// Minimum priority queue which
// contains all the elements
// of array arr[]
PriorityQueue pq = new PriorityQueue<>();
for(int x = 0; x < N; x++)
{
pq.add(ar[x]);
}
// poss stores whether all the
// additions are possible
boolean poss = true;
for(int x = 0; x < K; x++)
{
// Minium value in the
// priority queue
int mini = pq.peek();
pq.poll();
int val = mini + values[x];
// If on addition it exceeds
// M then not possible
if (val > M)
{
poss = false;
break;
}
pq.add(val);
}
// If all elements are added
if (poss)
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
// Driver Code
public static void main(String args[])
{
Integer ar[] = { 5, 9, 3, 8, 7 };
int N = 5;
Integer values[] = { 1, 2, 3, 4 };
int K = 4;
int M = 9;
solve(ar, values, N, K, M);
}
}
// This code is contributed by offbeat Python3
# Python3 program to implement the
# above approach
from queue import PriorityQueue
# Function which checks if all
# additions are possible
def solve(ar, values, N, K, M):
# Sorting values[] in
# decreasing order
values.sort(reverse = True)
# Minimum priority queue which
# contains all the elements
# of array arr[]
pq = PriorityQueue()
for x in range(N):
pq.put(ar[x]);
# poss stores whether all the
# additions are possible
poss = True;
for x in range(K):
# Minium value in the
# priority queue
mini = pq.get();
val = mini + values[x];
# If on addition it exceeds
# M then not possible
if (val > M):
poss = False;
break;
pq.put(val);
# If all elements are added
if (poss):
print("Yes");
else:
print("No");
# Driver Code
if __name__=='__main__':
ar = [ 5, 9, 3, 8, 7 ]
N = 5;
values = [ 1, 2, 3, 4 ]
K = 4;
M = 9;
solve(ar, values, N, K, M);
# This code is contributed by rutvik_56C#
// C# Program to implement the
// above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function which checks if all
// additions are possible
static void solve(int[] ar, int[] values,
int N, int K, int M)
{
// Sorting values[] in
// decreasing order
Array.Sort(values);
Array.Reverse(values);
// Minimum priority queue which
// contains all the elements
// of array arr[]
List pq = new List();
for (int x = 0; x < N; x++)
{
pq.Add(ar[x]);
}
pq.Sort();
// poss stores whether all the
// additions are possible
bool poss = true;
for (int x = 0; x < K; x++)
{
// Minium value in the
// priority queue
int mini = pq[0];
pq.RemoveAt(0);
int val = mini + values[x];
// If on addition it exceeds
// M then not possible
if (val > M)
{
poss = false;
break;
}
pq.Add(val);
pq.Sort();
}
// If all elements are added
if (poss)
{
Console.WriteLine("Yes");
}
else
{
Console.WriteLine("No");
}
}
// Driver code
static void Main()
{
int[] ar = { 5, 9, 3, 8, 7 };
int N = 5;
int[] values = { 1, 2, 3, 4 };
int K = 4;
int M = 9;
solve(ar, values, N, K, M);
}
}
// This code is contributed by divyeshrabadiya07. 输出：
Yes时间复杂度： O((N+K)*log(N))
辅助空间： O(N)
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