给定一个由 n 个元素组成的数组,找到具有最大算术平均值的最长子数组。子数组的长度必须大于 1,且均值应仅作为整数计算。
例子:
Input : arr[] = {3, 2, 1, 2}
Output : 2
sub-array 3, 2 has greatest arithmetic mean
Input :arr[] = {3, 3, 3, 2}
Output : 3
这个想法是首先从数组中找到两个连续元素的最大平均值。再次迭代数组并尝试找到最长的序列,其中每个元素必须大于或等于计算的最大平均值。
由于以下关键点,上述方法有效:
- 最小可能的序列长度为 2,因此两个连续元素的最大平均值将始终是结果的一部分。
- 任何等于或大于计算平均值的元素都可以是最长序列的一部分。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include
using namespace std;
// Function to find maximum distance
// between unequal elements
int longestSubarray(int arr[], int n)
{
// Calculate maxMean
int maxMean = 0;
for (int i = 1; i < n; i++)
maxMean = max(maxMean,
(arr[i] + arr[i - 1]) / 2);
// Iterate over array and calculate largest subarray
// with all elements greater or equal to maxMean
int ans = 0;
int subarrayLength = 0;
for (int i = 0; i < n; i++)
if (arr[i] >= maxMean)
ans = max(ans, ++subarrayLength);
else
subarrayLength = 0;
return ans;
}
// Driver code
int main()
{
int arr[] = { 4, 3, 3, 2, 1, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << longestSubarray(arr, n);
return 0;
}
Java
// Java implementation of the above approach
import java.io.*;
class GFG
{
// Function to find maximum distance
// between unequal elements
static int longestSubarray(int arr[], int n)
{
// Calculate maxMean
int maxMean = 0;
for (int i = 1; i < n; i++)
maxMean = Math.max(maxMean,
(arr[i] + arr[i - 1]) / 2);
// Iterate over array and calculate largest subarray
// with all elements greater or equal to maxMean
int ans = 0;
int subarrayLength = 0;
for (int i = 0; i < n; i++)
if (arr[i] >= maxMean)
ans = Math.max(ans, ++subarrayLength);
else
subarrayLength = 0;
return ans;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 4, 3, 3, 2, 1, 4 };
int n = arr.length;
System.out.println (longestSubarray(arr, n));
}
}
// This code is contributed by ajit_00023
Python3
# Python implementation of the above approach
# Function to find maximum distance
# between unequal elements
def longestSubarray(arr, n):
# Calculate maxMean
maxMean = 0;
for i in range(1, n):
maxMean = max(maxMean,
(arr[i] + arr[i - 1]) // 2);
# Iterate over array and calculate largest subarray
# with all elements greater or equal to maxMean
ans = 0;
subarrayLength = 0;
for i in range(n):
if (arr[i] >= maxMean):
subarrayLength += 1;
ans = max(ans, subarrayLength);
else:
subarrayLength = 0;
return ans;
# Driver code
arr = [ 4, 3, 3, 2, 1, 4 ];
n = len(arr);
print(longestSubarray(arr, n));
# This code contributed by PrinciRaj1992
C#
// C# program for the above approach
using System;
class GFG
{
// Function to find maximum distance
// between unequal elements
static int longestSubarray(int []arr,
int n)
{
// Calculate maxMean
int maxMean = 0;
for (int i = 1; i < n; i++)
maxMean = Math.Max(maxMean,
(arr[i] + arr[i - 1]) / 2);
// Iterate over array and calculate
// largest subarray with all elements
// greater or equal to maxMean
int ans = 0;
int subarrayLength = 0;
for (int i = 0; i < n; i++)
if (arr[i] >= maxMean)
ans = Math.Max(ans, ++subarrayLength);
else
subarrayLength = 0;
return ans;
}
// Driver code
public static void Main ()
{
int []arr = { 4, 3, 3, 2, 1, 4 };
int n = arr.Length;
Console.WriteLine(longestSubarray(arr, n));
}
}
// This code is contributed by AnkitRai01
Javascript
输出:
3
时间复杂度: O(N)
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