给定一个包含N 个元素的数组arr[] ，任务是执行以下操作：
- 从数组中选取两个最大的元素并删除这些元素。如果元素不相等，则将元素的绝对差值插入数组。
- 执行上述操作，直到数组中有 1 个元素或没有元素。如果数组只剩下一个元素，则打印该元素，否则打印“-1”。
例子：
Input: arr[] = { 3, 5, 2, 7 }
Output: 1
Explanation:
The two largest elements are 7 and 5. Discard them. Since both are not equal, insert 7 – 5 = 2 into the array. Hence, arr[] = { 3, 2, 2 }
The two largest elements are 3 and 2. Discard them. Since both are not equal, insert 3 – 2 = 1 into the array. Hence, arr[] = { 1, 2 }
The two largest elements are 2 and 1. Discard them. Since both are not equal, insert 2 – 1 = 1 into the array. Hence, arr[] = { 1 }
The only element left is 1. Print the value of the only element left.
Input: arr[] = { 3, 3 }
Output: -1
Explanation:
The two largest elements are 3 and 3. Discard them. Now the array has no elements. So, print -1.
方法：为了解决上述问题，我们将使用优先队列数据结构。以下是步骤：
- 将所有数组元素插入优先队列。
- 由于优先级队列是基于 Max-Heap 的实现。因此，从中删除元素会得到最大的元素。
- 直到优先级队列的大小不小于 2，从中删除两个元素（比如X & Y ）并执行以下操作：
- 如果X和Y都不一样然后将X和Y的绝对值为优先级队列。
- 否则返回步骤 3。
- 如果堆只有一个元素，则打印该元素。
- 否则打印“-1”。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the remaining element
int final_element(int arr[], int n)
{
// Priority queue can be used
// to construct max-heap
priority_queue heap;
// Insert all element of arr[]
// into priority queue
for (int i = 0; i < n; i++)
heap.push(arr[i]);
// Perform operation until heap
// size becomes 0 or 1
while (heap.size() > 1) {
// Remove largest element
int X = heap.top();
heap.pop();
// Remove 2nd largest element
int Y = heap.top();
heap.pop();
// If extracted element
// are not equal
if (X != Y) {
// Find X - Y and push
// it to heap
int diff = abs(X - Y);
heap.push(diff);
}
}
// If heap size is 1, then
// print the remaining element
if (heap.size() == 1) {
cout << heap.top();
}
// Else print "-1"
else {
cout << "-1";
}
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 3, 5, 2, 7 };
// Size of array arr[]
int n = sizeof(arr) / sizeof(arr[0]);
// Function Call
final_element(arr, n);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.util.Collections;
import java.util.*;
class GFG{
// Function to print the remaining element
static public int final_element(Integer[] arr, int n)
{
if(arr == null)
{
return 0;
}
// Priority queue can be used
// to construct max-heap
PriorityQueue heap = new
PriorityQueue(Collections.reverseOrder());
// Insert all element of arr[]
// into priority queue
for(int i = 0; i < n; i++)
{
heap.offer(arr[i]);
}
// Perform operation until heap
// size becomes 0 or 1
while (heap.size() > 1)
{
// Remove largest element
int X = heap.poll();
// Remove 2nd largest element
int Y = heap.poll();
// If extracted element
// are not equal
if (X != Y)
{
// Find X - Y and push
// it to heap
int diff = Math.abs(X - Y);
heap.offer(diff);
}
}
// If heap size is 1, then
// print the remaining element
// Else print "-1"
return heap.size() == 1 ? heap.poll() : -1;
}
// Driver code
public static void main (String[] args)
{
// Given array arr[]
Integer arr[] = new Integer[] { 3, 5, 2, 7};
// Size of array arr[]
int n = arr.length;
// Function Call
System.out.println(final_element(arr, n));
}
}
// This code is contributed by deepika_sharma Python3
# Python3 program for the above approach
from queue import PriorityQueue
# Function to print the remaining element
def final_element(arr, n):
# Priority queue can be used
# to construct max-heap
heap = PriorityQueue()
# Insert all element of
# arr[] into priority queue.
# Default priority queue in Python
# is min-heap so use -1*arr[i]
for i in range(n):
heap.put(-1 * arr[i])
# Perform operation until heap
# size becomes 0 or 1
while (heap.qsize() > 1):
# Remove largest element
X = -1 * heap.get()
# Remove 2nd largest element
Y = -1 * heap.get()
# If extracted elements
# are not equal
if (X != Y):
# Find X - Y and push
# it to heap
diff = abs(X - Y)
heap.put(-1 * diff)
# If heap size is 1, then
# print the remaining element
if (heap.qsize() == 1):
print(-1 * heap.get())
# Else print "-1"
else:
print("-1")
# Driver Code
if __name__ == '__main__':
# Given array arr[]
arr = [ 3, 5, 2, 7 ]
# Size of array arr[]
n = len(arr)
# Function call
final_element(arr, n)
# This code is contributed by himanshu77C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to print the remaining element
static void final_element(int[] arr, int n)
{
// Priority queue can be used
// to construct max-heap
List heap = new List();
// Insert all element of arr[]
// into priority queue
for(int i = 0; i < n; i++)
heap.Add(arr[i]);
// Perform operation until heap
// size becomes 0 or 1
while (heap.Count > 1)
{
// Remove largest element
heap.Sort();
heap.Reverse();
int X = heap[0];
heap.RemoveAt(0);
// Remove 2nd largest element
int Y = heap[0];
heap.RemoveAt(0);
// If extracted element
// are not equal
if (X != Y)
{
// Find X - Y and push
// it to heap
int diff = Math.Abs(X - Y);
heap.Add(diff);
}
}
// If heap size is 1, then
// print the remaining element
if (heap.Count == 1)
{
heap.Sort();
heap.Reverse();
Console.Write(heap[0]);
}
// Else print "-1"
else
{
Console.Write(-1);
}
}
// Driver code
static void Main()
{
// Given array arr[]
int[] arr = { 3, 5, 2, 7 };
// Size of array arr[]
int n = arr.Length;
// Function Call
final_element(arr, n);
}
}
// This code is contributed by divyeshrabadiya07 输出：
1时间复杂度： O(N*log(N))
辅助空间复杂度： O(N)
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