给定一个数组 A。让 x 成为数组中的一个元素。 x 具有阵列中的最大频率。找到数组的最小子段，其中 x 作为最大频率元素。
例子：
Input : arr[] = {4, 1, 1, 2, 2, 1, 3, 3}
Output : 1, 1, 2, 2, 1
The most frequent element is 1. The smallest
subarray that has all occurrences of it is
1 1 2 2 1
Input : A[] = {1, 2, 2, 3, 1}
Output : 2, 2
Note that there are two elements that appear
two times, 1 and 2. The smallest window for
1 is whole array and smallest window for 2 is
{2, 2}. Since window for 2 is smaller, this is
our output.方法：观察，如果 X 是我们子段的最大重复元素，那么太阳段应该看起来像这样 [X, ….., X]，因为如果子段以另一个元素结束或开始，我们可以删除它，这不会改变我们的回答。
为了解决这个问题，让我们为数组中的每个不同元素存储三个值，元素第一次出现的索引和元素最后一次出现的索引以及元素的频率。并且在最大重复元素的每一步最小化我们的子段的大小。
C++
// C++ implementation to find smallest
// subarray with all occurrences of
// a most frequent element
#include
using namespace std;
void smallestSubsegment(int a[], int n)
{
// To store left most occurrence of elements
unordered_map left;
// To store counts of elements
unordered_map count;
// To store maximum frequency
int mx = 0;
// To store length and starting index of
// smallest result window
int mn, strindex;
for (int i = 0; i < n; i++) {
int x = a[i];
// First occurrence of an element,
// store the index
if (count[x] == 0) {
left[x] = i;
count[x] = 1;
}
// increase the frequency of elements
else
count[x]++;
// Find maximum repeated element and
// store its last occurrence and first
// occurrence
if (count[x] > mx) {
mx = count[x];
mn = i - left[x] + 1; // length of subsegment
strindex = left[x];
}
// select subsegment of smallest size
else if (count[x] == mx && i - left[x] + 1 < mn) {
mn = i - left[x] + 1;
strindex = left[x];
}
}
// Print the subsegment with all occurrences of
// a most frequent element
for (int i = strindex; i < strindex + mn; i++)
cout << a[i] << " ";
}
// Driver code
int main()
{
int A[] = { 1, 2, 2, 2, 1 };
int n = sizeof(A) / sizeof(A[0]);
smallestSubsegment(A, n);
return 0;
} Java
// Java implementation to find smallest
// subarray with all occurrences of
// a most frequent element
import java.io.*;
import java.util.*;
class GfG {
static void smallestSubsegment(int a[], int n)
{
// To store left most occurrence of elements
HashMap left= new HashMap();
// To store counts of elements
HashMap count= new HashMap();
// To store maximum frequency
int mx = 0;
// To store length and starting index of
// smallest result window
int mn = -1, strindex = -1;
for (int i = 0; i < n; i++)
{
int x = a[i];
// First occurrence of an element,
// store the index
if (count.get(x) == null)
{
left.put(x, i) ;
count.put(x, 1);
}
// increase the frequency of elements
else
count.put(x, count.get(x) + 1);
// Find maximum repeated element and
// store its last occurrence and first
// occurrence
if (count.get(x) > mx)
{
mx = count.get(x);
// length of subsegment
mn = i - left.get(x) + 1;
strindex = left.get(x);
}
// select subsegment of smallest size
else if ((count.get(x) == mx) &&
(i - left.get(x) + 1 < mn))
{
mn = i - left.get(x) + 1;
strindex = left.get(x);
}
}
// Print the subsegment with all occurrences of
// a most frequent element
for (int i = strindex; i < strindex + mn; i++)
System.out.print(a[i] + " ");
}
// Driver program
public static void main (String[] args)
{
int A[] = { 1, 2, 2, 2, 1 };
int n = A.length;
smallestSubsegment(A, n);
}
}
// This code is contributed by Gitanjali. Python3
# Python3 implementation to find smallest
# subarray with all occurrences of
# a most frequent element
def smallestSubsegment(a, n):
# To store left most occurrence of elements
left = dict()
# To store counts of elements
count = dict()
# To store maximum frequency
mx = 0
# To store length and starting index of
# smallest result window
mn, strindex = 0, 0
for i in range(n):
x = a[i]
# First occurrence of an element,
# store the index
if (x not in count.keys()):
left[x] = i
count[x] = 1
# increase the frequency of elements
else:
count[x] += 1
# Find maximum repeated element and
# store its last occurrence and first
# occurrence
if (count[x] > mx):
mx = count[x]
mn = i - left[x] + 1 # length of subsegment
strindex = left[x]
# select subsegment of smallest size
elif (count[x] == mx and
i - left[x] + 1 < mn):
mn = i - left[x] + 1
strindex = left[x]
# Print the subsegment with all occurrences of
# a most frequent element
for i in range(strindex, strindex + mn):
print(a[i], end = " ")
# Driver code
A = [1, 2, 2, 2, 1]
n = len(A)
smallestSubsegment(A, n)
# This code is contributed by Mohit KumarC#
// C# implementation to find smallest
// subarray with all occurrences of
// a most frequent element
using System;
using System.Collections.Generic;
class GfG
{
static void smallestSubsegment(int []a, int n)
{
// To store left most occurrence of elements
Dictionary left = new Dictionary();
// To store counts of elements
Dictionary count = new Dictionary();
// To store maximum frequency
int mx = 0;
// To store length and starting index of
// smallest result window
int mn = -1, strindex = -1;
for (int i = 0; i < n; i++)
{
int x = a[i];
// First occurrence of an element,
// store the index
if (!count.ContainsKey(x))
{
left.Add(x, i) ;
count.Add(x, 1);
}
// increase the frequency of elements
else
count[x] = count[x] + 1;
// Find maximum repeated element and
// store its last occurrence and first
// occurrence
if (count[x] > mx)
{
mx = count[x];
// length of subsegment
mn = i - left[x] + 1;
strindex = left[x];
}
// select subsegment of smallest size
else if ((count[x] == mx) &&
(i - left[x] + 1 < mn))
{
mn = i - left[x] + 1;
strindex = left[x];
}
}
// Print the subsegment with all occurrences of
// a most frequent element
for (int i = strindex; i < strindex + mn; i++)
Console.Write(a[i] + " ");
}
// Driver code
public static void Main (String[] args)
{
int []A = { 1, 2, 2, 2, 1 };
int n = A.Length;
smallestSubsegment(A, n);
}
}
// This code is contributed by PrinciRaj1992 Javascript
输出：
2 2 2 时间复杂度： O(n)
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