给定一个由N个不同元素组成的数组arr [] ,任务是查找每个步骤中需要减少1对的最大对数,以使N – 1个数组元素减少为0,而其余的数组元素为a。非负整数。
例子:
Input: arr[] = {1, 2, 3}
Output: 3
Explanation:
Decrease arr[1] and arr[2] by 1 modifies arr[] = {1, 1, 2}
Decrease arr[1] and arr[2] by 1 modifies arr[] = {1, 0, 1}
Decrease arr[0] and arr[2] by 1 modifies arr[] = {0, 0, 0}
Therefore, the maximum number of decrements required is 3.
Input: arr[] = {1, 2, 3, 4, 5}
Output: 7
处理方法:可以贪婪地解决问题。请按照以下步骤解决问题:
- 初始化一个变量,例如cntOp ,以存储使数组的(N – 1)个元素等于0所需的最大步数。
- 创建一个优先级队列,例如PQ ,以存储数组元素。
- 遍历数组并将数组元素插入PQ 。
- 现在,从优先级队列中重复提取前2个元素,将这两个元素的值都减1 ,再次将这两个元素插入优先级队列,并将cntOp增1 。当PQ的(N – 1)元素等于0时,此过程继续进行。
- 最后,打印cntOp的值
下面是上述方法的实现:
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to count maximum number of steps
// to make (N - 1) array elements to 0
int cntMaxOperationToMakeN_1_0(int arr[], int N)
{
// Stores maximum count of steps to make
// (N - 1) elements equal to 0
int cntOp = 0;
// Stores array elements
priority_queue PQ;
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert arr[i] into PQ
PQ.push(arr[i]);
}
// Extract top 2 elements from the array
// while (N - 1) array elements become 0
while (PQ.size() > 1) {
// Stores top element
// of PQ
int X = PQ.top();
// Pop the top element
// of PQ.
PQ.pop();
// Stores top element
// of PQ
int Y = PQ.top();
// Pop the top element
// of PQ.
PQ.pop();
// Update X
X--;
// Update Y
Y--;
// If X is not equal to 0
if (X != 0) {
// Insert X into PQ
PQ.push(X);
}
// if Y is not equal
// to 0
if (Y != 0) {
// Insert Y
// into PQ
PQ.push(Y);
}
// Update cntOp
cntOp += 1;
}
return cntOp;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << cntMaxOperationToMakeN_1_0(arr, N);
return 0;
}
Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to count maximum number of steps
// to make (N - 1) array elements to 0
static int cntMaxOperationToMakeN_1_0(int[] arr, int N)
{
// Stores maximum count of steps to make
// (N - 1) elements equal to 0
int cntOp = 0;
// Stores array elements
PriorityQueue PQ = new PriorityQueue((a, b) -> b - a);
// Traverse the array
for(int i = 0; i < N; i++)
{
// Insert arr[i] into PQ
PQ.add(arr[i]);
}
// Extract top 2 elements from the array
// while (N - 1) array elements become 0
while (PQ.size() > 1)
{
// Stores top element
// of PQ
int X = PQ.peek();
// Pop the top element
// of PQ.
PQ.remove();
// Stores top element
// of PQ
int Y = PQ.peek();
// Pop the top element
// of PQ.
PQ.remove();
// Update X
X--;
// Update Y
Y--;
// If X is not equal to 0
if (X != 0)
{
// Insert X into PQ
PQ.add(X);
}
// if Y is not equal
// to 0
if (Y != 0)
{
// Insert Y
// into PQ
PQ.add(Y);
}
// Update cntOp
cntOp += 1;
}
return cntOp;
}
// Driver code
public static void main(String[] args)
{
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.length;
System.out.print(cntMaxOperationToMakeN_1_0(arr, N));
}
}
// This code is contributed by susmitakundugoaldanga
Python3
# Python3 program to implement
# the above approach
# Function to count maximum number of steps
# to make (N - 1) array elements to 0
def cntMaxOperationToMakeN_1_0(arr, N):
# Stores maximum count of steps to make
# (N - 1) elements equal to 0
cntOp = 0
# Stores array elements
PQ = []
# Traverse the array
for i in range(N):
# Insert arr[i] into PQ
PQ.append(arr[i])
PQ = sorted(PQ)
# Extract top 2 elements from the array
# while (N - 1) array elements become 0
while (len(PQ) > 1):
# Stores top element
# of PQ
X = PQ[-1]
# Pop the top element
# of PQ.
del PQ[-1]
# Stores top element
# of PQ
Y = PQ[-1]
# Pop the top element
# of PQ.
del PQ[-1]
# Update X
X -= 1
# Update Y
Y -= 1
# If X is not equal to 0
if (X != 0):
# Insert X into PQ
PQ.append(X)
# if Y is not equal
# to 0
if (Y != 0):
# Insert Y
# into PQ
PQ.append(Y)
# Update cntOp
cntOp += 1
PQ = sorted(PQ)
return cntOp
# Driver Code
if __name__ == '__main__':
arr = [1, 2, 3, 4, 5]
N = len(arr)
print (cntMaxOperationToMakeN_1_0(arr, N))
# This code is contributed by mohit kumar 29.
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to count maximum number of steps
// to make (N - 1) array elements to 0
static int cntMaxOperationToMakeN_1_0(int[] arr, int N)
{
// Stores maximum count of steps to make
// (N - 1) elements equal to 0
int cntOp = 0;
// Stores array elements
List PQ = new List();
// Traverse the array
for (int i = 0; i < N; i++) {
// Insert arr[i] into PQ
PQ.Add(arr[i]);
}
PQ.Sort();
PQ.Reverse();
// Extract top 2 elements from the array
// while (N - 1) array elements become 0
while (PQ.Count > 1) {
// Stores top element
// of PQ
int X = PQ[0];
// Pop the top element
// of PQ.
PQ.RemoveAt(0);
// Stores top element
// of PQ
int Y = PQ[0];
// Pop the top element
// of PQ.
PQ.RemoveAt(0);
// Update X
X--;
// Update Y
Y--;
// If X is not equal to 0
if (X != 0) {
// Insert X into PQ
PQ.Add(X);
PQ.Sort();
PQ.Reverse();
}
// if Y is not equal
// to 0
if (Y != 0) {
// Insert Y
// into PQ
PQ.Add(Y);
PQ.Sort();
PQ.Reverse();
}
// Update cntOp
cntOp += 1;
}
return cntOp;
}
// Driver code
static void Main() {
int[] arr = { 1, 2, 3, 4, 5 };
int N = arr.Length;
Console.WriteLine(cntMaxOperationToMakeN_1_0(arr, N));
}
}
// This code is contributed by divyesh072019
Javascript
输出:
7
时间复杂度: O(N * log(N))
辅助空间: O(N)