检查是否可以通过替换少于一半的元素来使任何子数组成为回文

给定一个大小为N的数组arr[] ，任务是检查给定数组中的任何子数组是否可以通过将其少于一半的元素（即floor [length/2]）替换为子阵列。

例子：

 Input: arr[] = {2, 7, 4, 6, 7, 8}
Output: Yes
Explanation: Among all subarrays of this array, subarray {7, 4, 6, 7} requires only 1 operation to make it a palindrome i.e. replace arr[3] by 4 or arr[4] by 6, which is less than floor(4/2) ( = 2).

Input: arr[] = {3, 7, 19, 6}
Output: No

朴素的方法：解决这个问题的最简单的方法是生成给定数组的所有子数组，对于每个子数组，检查使该子数组成为回文所需的替换次数是否小于floor(length of subarray / 2) 。

时间复杂度： O(N ³ )
辅助空间： O(1)

高效的方法：上述方法可以基于以下观察进行优化：

If the array arr[] contains duplicate elements, then it is always possible to choose a subarray from initial occurrence to the next occurrence of that element. This subarray will require less than floor(length/2) operations as first and last element of subarray is already equal.

请按照以下步骤解决问题：

• 初始化一个 Map 来存储每个数组元素的频率。
• 迭代所有数组元素并计算每个数组元素的频率并将其插入到 Map 中。
• 检查是否发现任何数组元素的频率大于 1.. 如果发现为真，则打印“是” 。否则，打印“否” 。

下面是这个方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// A Utility Function to check if a subarray
// can be palindromic by replacing less than
// half of the elements present in it
bool isConsistingSubarrayUtil(int arr[], int n)
{
    // Stores frequency of array elements
    map mp;
 
    // Traverse the array
    for (int i = 0; i < n; ++i) {
 
        // Update frequency of
        // each array element
        mp[arr[i]]++;
    }
 
    // Iterator over the Map
    map::iterator it;
 
    for (it = mp.begin(); it != mp.end(); ++it) {
 
        // If frequency of any element exceeds 1
        if (it->second > 1) {
            return true;
        }
    }
 
    // If no repetition is found
    return false;
}
 
// Function to check and print if any subarray
// can be made palindromic by replacing less
// than half of its elements
void isConsistingSubarray(int arr[], int N)
{
    if (isConsistingSubarrayUtil(arr, N)) {
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4, 5, 1 };
 
    // Size of array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    isConsistingSubarray(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
     
// A Utility Function to check if a subarray
// can be palindromic by replacing less than
// half of the elements present in it
static boolean isConsistingSubarrayUtil(int arr[],
                                        int n)
{
     
    // Stores frequency of array elements
    TreeMap mp = new TreeMap();
  
    // Traverse the array
    for(int i = 0; i < n; ++i)
    {
         
        // Update frequency of
        // each array element
        mp.put(arr[i],
        mp.getOrDefault(arr[i], 0) + 1);
    }
     
    for(Map.Entry it : mp.entrySet())
    {
         
        // If frequency of any element exceeds 1
        if (it.getValue() > 1)
        {
            return true;
        }
    }
     
    // If no repetition is found
    return false;
}
  
// Function to check and print if any subarray
// can be made palindromic by replacing less
// than half of its elements
static void isConsistingSubarray(int arr[], int N)
{
    if (isConsistingSubarrayUtil(arr, N))
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
  
// Driver Code
public static void main(String args[])
{
     
    // Given array arr[]
    int arr[] = { 1, 2, 3, 4, 5, 1 };
     
    // Size of array
    int N = arr.length;
     
    // Function Call
    isConsistingSubarray(arr, N);
}
}
 
// This code is contributed by susmitakundugoaldanga

Python3

# Python3 program for the above approach
 
# A Utility Function to check if a subarray
# can be palindromic by replacing less than
# half of the elements present in it
def isConsistingSubarrayUtil(arr, n) :
 
    # Stores frequency of array elements
    mp = {};
 
    # Traverse the array
    for i in range(n) :
 
        # Update frequency of
        # each array element
        if arr[i] in mp :
            mp[arr[i]] += 1;           
        else :
            mp[arr[i]] = 1;
 
    # Iterator over the Map
    for it in mp :
 
        # If frequency of any element exceeds 1
        if (mp[it] > 1) :
            return True;
 
    # If no repetition is found
    return False;
 
# Function to check and print if any subarray
# can be made palindromic by replacing less
# than half of its elements
def isConsistingSubarray(arr, N) :
 
    if (isConsistingSubarrayUtil(arr, N)) :
        print("Yes");
    else :
        print("No");
 
# Driver Code
if __name__ == "__main__" :
 
    # Given array arr[]
    arr = [ 1, 2, 3, 4, 5, 1 ];
 
    # Size of array
    N = len(arr);
 
    # Function Call
    isConsistingSubarray(arr, N);
 
    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Linq;
     
class GFG{
     
// A Utility Function to check if a subarray
// can be palindromic by replacing less than
// half of the elements present in it
static bool isConsistingSubarrayUtil(int[] arr,
                                        int n)
{
      
    // Stores frequency of array elements
    Dictionary mp = new Dictionary();
   
    // Traverse the array
    for(int i = 0; i < n; ++i)
    {
          
        // Update frequency of
        // each array element
        if (mp.ContainsKey(arr[i]) == true)
        mp[arr[i]] += 1;
      else
        mp[arr[i]] = 1;
    }
     
    var val = mp.Keys.ToList();
    foreach(var key in val)
    {
        // If frequency of any element exceeds 1
        if (mp[key] > 1)
        {
            return true;
        }
    }
      
    // If no repetition is found
    return false;
}
   
// Function to check and print if any subarray
// can be made palindromic by replacing less
// than half of its elements
static void isConsistingSubarray(int[] arr, int N)
{
    if (isConsistingSubarrayUtil(arr, N))
    {
        Console.Write("Yes");
    }
    else
    {
        Console.Write("No");
    }
}
     
// Driver Code
public static void Main()
{
    // Given array arr[]
    int[] arr = { 1, 2, 3, 4, 5, 1 };
      
    // Size of array
    int N = arr.Length;
      
    // Function Call
    isConsistingSubarray(arr, N);
}
}
 
// This code is contributed by sanjoy62

Javascript

 
// Javascript program for the above approach
 
// A Utility Function to check if a subarray
// can be palindromic by replacing less than
// half of the elements present in it
function isConsistingSubarrayUtil(arr, n)
{
    // Stores frequency of array elements
    var mp = new Map();
 
    // Traverse the array
    for (var i = 0; i < n; ++i) {
 
        // Update frequency of
        // each array element
        if(mp.has(arr[i]))
        {
            mp.set(arr[i], mp.get(arr[i])+1);
        }
        else
        {
            mp.set(arr[i], 1);
        }
    }
 
    var ans = false;
    // Iterator over the Map
    mp.forEach((value, key) => {
         
        // If frequency of any element exceeds 1
        if (value > 1) {
            ans = true;
        }
    });
 
    if(ans)
        return true;
 
    // If no repetition is found
    return false;
}
 
// Function to check and print if any subarray
// can be made palindromic by replacing less
// than half of its elements
function isConsistingSubarray(arr, N)
{
    if (isConsistingSubarrayUtil(arr, N)) {
        document.write( "Yes");
    }
    else {
        document.write( "No" );
    }
}
 
// Driver Code
// Given array arr[]
var arr = [1, 2, 3, 4, 5, 1];
// Size of array
var N = arr.length;
// Function Call
isConsistingSubarray(arr, N);
 

输出：

Yes

时间复杂度： O(N)
辅助空间： O(N)

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