给定一个由N 个整数组成的数组arr[] ,任务是在子数组的两半上找到由相同类型元素组成的子数组的最大长度。此外,两半上的元素彼此不同。
例子:
Input: arr[] = {2, 3, 4, 4, 5, 5, 6, 7, 8, 10}
Output: 4
Explanation:
{2, 3}, {3, 4}, {4, 4, 5, 5}, {5, 6}, etc, are the valid sub-arrays where both halves have only one type of element.
{4, 4, 5, 5} is the sub-array having maximum length.
Hence, the output is 4.
Input: arr[] = {1, 7, 7, 10, 10, 7, 7, 7, 8, 8, 8, 9}
Output: 6
Explanation:
{1, 7}, {7, 7, 10, 10}, {7, 7, 7, 8, 8, 8}, {8, 9}, etc, are the valid sub-arrays where both halves have only one type of element.
{7, 7, 7, 8, 8, 8} is the sub-array having maximum length.
Hence, the output is 6.
朴素的方法:朴素的想法是生成所有可能的子数组,并检查任何最大长度的子数组是否可以分为两半,使得两半中的所有元素都相同。
时间复杂度: O(N 3 )
辅助空间: O(1)
高效的方法:解决这个问题的想法是使用前缀和的概念。请按照以下步骤解决问题:
- 从头开始向前遍历数组,并将每个索引的连续出现的整数存储在数组forward[] 中。
- 类似地,从数组的末尾反向遍历,并将连续出现的每个索引的整数存储在一个数组向后[]中。
- 存储min(forward[i],backward[i+1])*2的最大值,用于所有索引,其中arr[i]!=arr[i+1] 。
- 打印上一步得到的值。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function that finds the maximum
// length of the sub-array that
// contains equal element on both
// halves of sub-array
void maxLengthSubArray(int A[], int N)
{
// To store continuous occurence
// of the element
int forward[N], backward[N];
// To store continuous
// forward occurence
for (int i = 0; i < N; i++) {
if (i == 0
|| A[i] != A[i - 1]) {
forward[i] = 1;
}
else
forward[i] = forward[i - 1] + 1;
}
// To store continuous
// backward occurence
for (int i = N - 1; i >= 0; i--) {
if (i == N - 1
|| A[i] != A[i + 1]) {
backward[i] = 1;
}
else
backward[i] = backward[i + 1] + 1;
}
// To store the maximum length
int ans = 0;
// Find maximum length
for (int i = 0; i < N - 1; i++) {
if (A[i] != A[i + 1])
ans = max(ans,
min(forward[i],
backward[i + 1])
* 2);
}
// Print the result
cout << ans;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 2, 3, 4, 4,
4, 6, 6, 6, 9 };
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
maxLengthSubArray(arr, N);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function that finds the maximum
// length of the sub-array that
// contains equal element on both
// halves of sub-array
static void maxLengthSubArray(int A[], int N)
{
// To store continuous occurence
// of the element
int forward[] = new int[N];
int backward[] = new int[N];
// To store continuous
// forkward occurence
for(int i = 0; i < N; i++)
{
if (i == 0 || A[i] != A[i - 1])
{
forward[i] = 1;
}
else
forward[i] = forward[i - 1] + 1;
}
// To store continuous
// backward occurence
for(int i = N - 1; i >= 0; i--)
{
if (i == N - 1 || A[i] != A[i + 1])
{
backward[i] = 1;
}
else
backward[i] = backward[i + 1] + 1;
}
// To store the maximum length
int ans = 0;
// Find maximum length
for(int i = 0; i < N - 1; i++)
{
if (A[i] != A[i + 1])
ans = Math.max(ans,
Math.min(forward[i],
backward[i + 1]) * 2);
}
// Print the result
System.out.println(ans);
}
// Driver Code
public static void main(String[] args)
{
// Given array
int arr[] = { 1, 2, 3, 4, 4,
4, 6, 6, 6, 9 };
// Size of the array
int N = arr.length;
// Function call
maxLengthSubArray(arr, N);
}
}
// This code is contributed by rutvik_56
Python3
# Python3 program for the above approach
# Function that finds the maximum
# length of the sub-array that
# contains equal element on both
# halves of sub-array
def maxLengthSubArray(A, N):
# To store continuous occurence
# of the element
forward = [0] * N
backward = [0] * N
# To store continuous
# forward occurence
for i in range(N):
if i == 0 or A[i] != A[i - 1]:
forward[i] = 1
else:
forward[i] = forward[i - 1] + 1
# To store continuous
# backward occurence
for i in range(N - 1, -1, -1):
if i == N - 1 or A[i] != A[i + 1]:
backward[i] = 1
else:
backward[i] = backward[i + 1] + 1
# To store the maximum length
ans = 0
# Find maximum length
for i in range(N - 1):
if (A[i] != A[i + 1]):
ans = max(ans,
min(forward[i],
backward[i + 1]) * 2);
# Print the result
print(ans)
# Driver Code
# Given array
arr = [ 1, 2, 3, 4, 4, 4, 6, 6, 6, 9 ]
# Size of the array
N = len(arr)
# Function call
maxLengthSubArray(arr, N)
# This code is contributed by yatinagg
C#
// C# program for the above approach
using System;
class GFG{
// Function that finds the maximum
// length of the sub-array that
// contains equal element on both
// halves of sub-array
static void maxLengthSubArray(int []A, int N)
{
// To store continuous occurence
// of the element
int []forward = new int[N];
int []backward = new int[N];
// To store continuous
// forkward occurence
for(int i = 0; i < N; i++)
{
if (i == 0 || A[i] != A[i - 1])
{
forward[i] = 1;
}
else
forward[i] = forward[i - 1] + 1;
}
// To store continuous
// backward occurence
for(int i = N - 1; i >= 0; i--)
{
if (i == N - 1 || A[i] != A[i + 1])
{
backward[i] = 1;
}
else
backward[i] = backward[i + 1] + 1;
}
// To store the maximum length
int ans = 0;
// Find maximum length
for(int i = 0; i < N - 1; i++)
{
if (A[i] != A[i + 1])
ans = Math.Max(ans,
Math.Min(forward[i],
backward[i + 1]) * 2);
}
// Print the result
Console.WriteLine(ans);
}
// Driver Code
public static void Main(String[] args)
{
// Given array
int []arr = { 1, 2, 3, 4, 4,
4, 6, 6, 6, 9 };
// Size of the array
int N = arr.Length;
// Function call
maxLengthSubArray(arr, N);
}
}
// This code is contributed by Princi Singh
Javascript
6
时间复杂度: O(N)
辅助空间: O(N)
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