给定由N个整数组成的两个数组arr []和q [] ，任务是针对每个查询q [i]确定子数组[q [i]中最大数组元素出现的次数arr [N – 1]] 。
例子：
Input: arr[] = {5, 4, 5, 3, 2}, q[] = {1, 2, 3, 4, 5}
Output: 2 1 1 1 1
Explanation:
For the first query index start from 1. The subarray is [5, 4, 5, 3, 2], the maximum value is 5 and it’s occurrence in subarray is 2.
For the second query index start from 2. The subarray is [4, 5, 3, 2], the maximum value is 5 and it’s occurrence in subarray is 1.
For the third query index start from 3. The subarray is [5, 3, 2], the maximum value is 5 and it’s occurrence in subarray is 1.
For the forth query index start from 4. The subarray is [3, 2], the maximum value is 3 and it’s occurrence in subarray is 1.
For the fifth query index start from 5. The subarray is [2], the maximum value is 2 and it’s occurrence in subarray is 1.
Input: arr[] = {2, 1, 2}, q[] = {1, 2, 3}
Output: 2 1 1
天真的方法：最简单的方法是遍历数组并找到最大的数组元素。现在，对于每个查询q [i]，遍历子数组[q [i]，arr [N – 1]]并打印子数组中最大元素的出现次数。
时间复杂度： O(N 2 )
辅助空间： O(1)
高效方法：为了优化上述方法，我们的想法是使用哈希。步骤如下：
- 创建一个数组maxFreqVec []来存储从给定索引q [i]到N的max元素的出现。
- 从右侧遍历数组arr []并跟踪该数组中的最大元素，并在出现该最大元素时在该索引处更新该数组maxFreqVec [] 。
- 完成上述步骤后，遍历数组q []并输出值maxFreqVec [q [i] – 1]作为每个查询的每个子数组中的最大元素。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find occurrence of
// max element in given subarray
void FreqOfMaxElement(vector arr,
vector q)
{
// Store the frequency of maximum
// element
vector maxFreqVec(arr.size());
int maxSoFar = INT_MIN;
int maxFreq = 0;
// Traverse over the array arr[]
// from right to left
for(int i = arr.size() - 1; i >= 0; i--)
{
// If curr element is greater
// than maxSofar
if (arr[i] > maxSoFar)
{
// Reset maxSofar and maxFreq
maxSoFar = arr[i];
maxFreq = 1;
}
// If curr is equal to maxSofar
else if (arr[i] == maxSoFar)
{
// Increment the maxFreq
maxFreq++;
}
// Update maxFreqVec[i]
maxFreqVec[i] = maxFreq;
}
// Print occurrence of maximum
// element for each query
for(int k : q)
{
cout << maxFreqVec[k - 1] << " ";
}
}
// Driver Code
int main()
{
vector arr = { 5, 4, 5, 3, 2 };
vector q = { 1, 2, 3, 4, 5 };
// Function Call
FreqOfMaxElement(arr, q);
}
// This code is contributed by mohit kumar 29 Java
// Java program for the above approach
import java.util.*;
import java.lang.*;
class GFG {
// Function to find occurrence of
// max element in given subarray
static void FreqOfMaxElement(
int[] arr, int[] q)
{
// Store the frequency of maximum
// element
int[] maxFreqVec
= new int[arr.length];
int maxSoFar = Integer.MIN_VALUE;
int maxFreq = 0;
// Traverse over the array arr[]
// from right to left
for (int i = arr.length - 1;
i >= 0; i--) {
// If curr element is greater
// than maxSofar
if (arr[i] > maxSoFar) {
// Reset maxSofar and maxFreq
maxSoFar = arr[i];
maxFreq = 1;
}
// If curr is equal to maxSofar
else if (arr[i] == maxSoFar) {
// Increment the maxFreq
maxFreq++;
}
// Update maxFreqVec[i]
maxFreqVec[i] = maxFreq;
}
// Print occurrence of maximum
// element for each query
for (int k : q) {
System.out.print(
maxFreqVec[k - 1] + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 5, 4, 5, 3, 2 };
int[] q = { 1, 2, 3, 4, 5 };
// Function Call
FreqOfMaxElement(arr, q);
}
}Python3
# Python3 program for
# the above approach
import sys;
# Function to find occurrence
# of max element in given
# subarray
def FreqOfMaxElement(arr, q):
# Store the frequency of
# maximum element
maxFreqVec = [0] * (len(arr));
maxSoFar = -sys.maxsize;
maxFreq = 0;
# Traverse over the array
# arr from right to left
for i in range(len(arr)-1,
-1, -1):
# If curr element is
# greater than maxSofar
if (arr[i] > maxSoFar):
# Reset maxSofar
# and maxFreq
maxSoFar = arr[i];
maxFreq = 1;
# If curr is equal to
# maxSofar
elif (arr[i] == maxSoFar):
# Increment the
# maxFreq
maxFreq += 1;
# Update maxFreqVec[i]
maxFreqVec[i] = maxFreq;
# Proccurrence of maximum
# element for each query
for i in range(0, len(q)):
print(maxFreqVec[q[i] - 1],
end = " ");
# Driver Code
if __name__ == '__main__':
arr = [5, 4, 5, 3, 2];
q = [1, 2, 3, 4, 5];
# Function Call
FreqOfMaxElement(arr, q);
# This code is contributed by shikhasingrajputC#
// C# program for the
// above approach
using System;
class GFG{
// Function to find occurrence of
// max element in given subarray
static void FreqOfMaxElement(int[] arr,
int[] q)
{
// Store the frequency of
// maximum element
int[] maxFreqVec = new int[arr.Length];
int maxSoFar = Int32.MinValue;
int maxFreq = 0;
// Traverse over the array arr[]
// from right to left
for (int i = arr.Length - 1;
i >= 0; i--)
{
// If curr element is greater
// than maxSofar
if (arr[i] > maxSoFar)
{
// Reset maxSofar and maxFreq
maxSoFar = arr[i];
maxFreq = 1;
}
// If curr is equal to maxSofar
else if (arr[i] == maxSoFar)
{
// Increment the maxFreq
maxFreq++;
}
// Update maxFreqVec[i]
maxFreqVec[i] = maxFreq;
}
// Print occurrence of maximum
// element for each query
foreach (int k in q)
{
Console.Write(maxFreqVec[k - 1] +
" ");
}
}
// Driver code
static void Main()
{
int[] arr = {5, 4, 5, 3, 2};
int[] q = {1, 2, 3, 4, 5};
// Function Call
FreqOfMaxElement(arr, q);
}
}
// This code is contributed by divyeshrabadiya07输出：
2 1 1 1 1
时间复杂度： O(N)
辅助空间： O(N)