给定数组arr []由[1，N]范围内的N个整数组成(允许重复)，任务是为每个可能的子数组长度找到最小的公共元素。如果子数组的任何特定长度都没有这样的元素，则打印-1 。
例子：
Input: arr[] = {1, 3, 4, 5, 6, 7}
Output: -1 -1 -1 4 3 1
Explanation:
K = 1: No common element exists. Therefore, print -1.
K = 2: No common element exists. Therefore, print -1.
K = 3: No common element exists. Therefore, print -1.
K = 4: Since 4 is common in all subarrays of size 4, print 4.
K = 5: Since 3 and 4 is common in all subarrays of size 5, print 3 as it is the minimum.
K = 6: Print 1 as it is the minimum element in the array.
Input: arr[]: {1, 2, 2, 2, 1}
Output: -1 2 2 1 1
方法：请按照以下步骤解决问题：
- 遍历数组并存储Map中每个元素的最后一次出现。
- 初始化一个数组temp []并为每个值存储数组中该数组的任何一对连续重复之间的最大距离。
- 完成上述步骤后，通过将temp [i]与最后一次出现的i与数组末尾的距离进行比较来更新temp [] 。
- 现在，存储所有长度为1到N的子数组的最小comment元素，并打印它们。
下面是上述方法的实现：
C++
// C++ Program to implement the
// above approach
#include
using namespace std;
// Function to find maximum distance
// between every two element
void max_distance(int a[], int temp[], int n)
{
// Stores index of last occurence
// of each array element
map mp;
// Initialize temp[] with -1
for (int i = 1; i <= n; i++) {
temp[i] = -1;
}
// Traverse the array
for (int i = 0; i < n; i++) {
// If array element has
// not occurred previously
if (mp.find(a[i]) == mp.end())
// Update index in temp
temp[a[i]] = i + 1;
// Otherwise
else
// Compare temp[a[i]] with distance
// from its previous occurence and
// store the maximum
temp[a[i]] = max(temp[a[i]],
i - mp[a[i]]);
mp[a[i]] = i;
}
for (int i = 1; i <= n; i++) {
// Compare temp[i] with distance
// of its last occurence from the end
// of the array and store the maximum
if (temp[i] != -1)
temp[i] = max(temp[i], n - mp[i]);
}
}
// Function to find the minimum common
// element in subarrays of all possible lengths
void min_comm_ele(int a[], int ans[],
int temp[], int n)
{
// Function call to find a the maximum
// distance between every pair of repetition
max_distance(a, temp, n);
// Initialize ans[] to -1
for (int i = 1; i <= n; i++) {
ans[i] = -1;
}
for (int i = 1; i <= n; i++) {
// Check if subarray of length
// temp[i] contains i as one
// of the common elements
if (ans[temp[i]] == -1)
ans[temp[i]] = i;
}
for (int i = 1; i <= n; i++) {
// Find the minimum of all
// common elements
if (i > 1 && ans[i - 1] != -1) {
if (ans[i] == -1)
ans[i] = ans[i - 1];
else
ans[i] = min(ans[i],
ans[i - 1]);
}
cout << ans[i] << " ";
}
}
// Driver Code
int main()
{
int N = 6;
int a[] = { 1, 3, 4, 5, 6, 7 };
int temp[100], ans[100];
min_comm_ele(a, ans, temp, N);
return 0;
} Java
// Java program to implement the
// above approach
import java.util.*;
class GFG{
// Function to find maximum distance
// between every two element
static void max_distance(int a[], int temp[], int n)
{
// Stores index of last occurence
// of each array element
Map mp = new HashMap();
// Initialize temp[] with -1
for(int i = 1; i <= n; i++)
{
temp[i] = -1;
}
// Traverse the array
for(int i = 0; i < n; i++)
{
// If array element has
// not occurred previously
if (mp.get(a[i]) == null)
// Update index in temp
temp[a[i]] = i + 1;
// Otherwise
else
// Compare temp[a[i]] with distance
// from its previous occurence and
// store the maximum
temp[a[i]] = Math.max(temp[a[i]],
i - mp.getOrDefault(a[i], 0));
mp.put(a[i], i);
}
for(int i = 1; i <= n; i++)
{
// Compare temp[i] with distance
// of its last occurence from the end
// of the array and store the maximum
if (temp[i] != -1)
temp[i] = Math.max(temp[i],
n - mp.getOrDefault(i, 0));
}
}
// Function to find the minimum common
// element in subarrays of all possible lengths
static void min_comm_ele(int a[], int ans[],
int temp[], int n)
{
// Function call to find a the maximum
// distance between every pair of repetition
max_distance(a, temp, n);
// Initialize ans[] to -1
for(int i = 1; i <= n; i++)
{
ans[i] = -1;
}
for(int i = 1; i <= n; i++)
{
// Check if subarray of length
// temp[i] contains i as one
// of the common elements
if (temp[i] >= 0 && ans[temp[i]] == -1)
ans[temp[i]] = i;
}
for(int i = 1; i <= n; i++)
{
// Find the minimum of all
// common elements
if (i > 1 && ans[i - 1] != -1)
{
if (ans[i] == -1)
ans[i] = ans[i - 1];
else
ans[i] = Math.min(ans[i],
ans[i - 1]);
}
System.out.print(ans[i] + " ");
}
}
// Driver Code
public static void main(String args[])
{
int N = 6;
int a[] = { 1, 3, 4, 5, 6, 7 };
int []temp = new int[100];
Arrays.fill(temp, 0);
int []ans = new int[100];
Arrays.fill(ans, 0);
min_comm_ele(a, ans, temp, N);
}
}
// This code is contributed by SURENDRA_GANGWAR Python3
# Python3 Program to implement
# the above approach
# Function to find maximum
# distance between every
# two element
def max_distance(a, temp, n):
# Stores index of last
# occurence of each
# array element
mp = {}
# Initialize temp[]
# with -1
for i in range(1, n + 1):
temp[i] = -1
# Traverse the array
for i in range(n):
# If array element has
# not occurred previously
if (a[i] not in mp):
# Update index in temp
temp[a[i]] = i + 1
# Otherwise
else:
# Compare temp[a[i]] with
# distance from its previous
# occurence and store the maximum
temp[a[i]] = max(temp[a[i]],
i - mp[a[i]])
mp[a[i]] = i
for i in range(1, n + 1):
# Compare temp[i] with
# distance of its last
# occurence from the end
# of the array and store
# the maximum
if (temp[i] != -1):
temp[i] = max(temp[i],
n - mp[i])
# Function to find the minimum
# common element in subarrays
# of all possible lengths
def min_comm_ele(a, ans,
temp, n):
# Function call to find
# a the maximum distance
# between every pair of
# repetition
max_distance(a, temp, n)
# Initialize ans[] to -1
for i in range(1, n + 1):
ans[i] = -1
for i in range(1, n + 1):
# Check if subarray of length
# temp[i] contains i as one
# of the common elements
if (ans[temp[i]] == -1):
ans[temp[i]] = i
for i in range(1, n + 1):
# Find the minimum of all
# common elements
if (i > 1 and
ans[i - 1] != -1):
if (ans[i] == -1):
ans[i] = ans[i - 1]
else:
ans[i] = min(ans[i],
ans[i - 1])
print(ans[i], end = " ")
# Driver Code
if __name__ == "__main__":
N = 6
a = [1, 3, 4, 5, 6, 7]
temp = [0] * 100
ans = [0] * 100
min_comm_ele(a, ans,
temp, N)
# This code is contributed by ChitranayalC#
// C# program to implement the
// above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to find maximum distance
// between every two element
static void max_distance(int[] a, int[] temp, int n)
{
// Stores index of last occurence
// of each array element
Dictionary mp = new Dictionary();
// Initialize temp[] with -1
for(int i = 1; i <= n; i++)
{
temp[i] = -1;
}
// Traverse the array
for(int i = 0; i < n; i++)
{
// If array element has
// not occurred previously
if (!mp.ContainsKey(a[i]))
// Update index in temp
temp[a[i]] = i + 1;
// Otherwise
else
// Compare temp[a[i]] with distance
// from its previous occurence and
// store the maximum
temp[a[i]] = Math.Max(temp[a[i]], i - mp[a[i]]);
if(mp.ContainsKey(a[i]))
{
mp[a[i]] = i;
}
else{
mp.Add(a[i], i);
}
}
for(int i = 1; i <= n; i++)
{
// Compare temp[i] with distance
// of its last occurence from the end
// of the array and store the maximum
if (temp[i] != -1)
{
if(mp.ContainsKey(i))
{
temp[i] = Math.Max(temp[i], n - mp[i]);
}
else{
temp[i] = Math.Max(temp[i], n);
}
}
}
}
// Function to find the minimum common
// element in subarrays of all possible lengths
static void min_comm_ele(int[] a, int[] ans,
int[] temp, int n)
{
// Function call to find a the maximum
// distance between every pair of repetition
max_distance(a, temp, n);
// Initialize ans[] to -1
for(int i = 1; i <= n; i++)
{
ans[i] = -1;
}
for(int i = 1; i <= n; i++)
{
// Check if subarray of length
// temp[i] contains i as one
// of the common elements
if (temp[i] >= 0 && ans[temp[i]] == -1)
ans[temp[i]] = i;
}
for(int i = 1; i <= n; i++)
{
// Find the minimum of all
// common elements
if (i > 1 && ans[i - 1] != -1)
{
if (ans[i] == -1)
ans[i] = ans[i - 1];
else
ans[i] = Math.Min(ans[i],
ans[i - 1]);
}
Console.Write(ans[i] + " ");
}
}
// Driver code
static void Main()
{
int N = 6;
int[] a = { 1, 3, 4, 5, 6, 7 };
int[] temp = new int[100];
Array.Fill(temp, 0);
int[] ans = new int[100];
Array.Fill(ans, 0);
min_comm_ele(a, ans, temp, N);
}
}
// This code is contributed by divyeshrabadiya07 输出：
-1 -1 -1 4 3 1时间复杂度： O(N)
辅助空间： O(N)