给定一个大小为N的排序数组arr ，任务是减少数组，使得每个元素最多可以出现两次。
例子：
Input: arr[] = {1, 2, 2, 2, 3}
Output: {1, 2, 2, 3}
Explanation:
Remove 2 once, as it occurs more than 2 times.
Input: arr[] = {3, 3, 3}
Output: {3, 3}
Explanation:
Remove 3 once, as it occurs more than 2 times.
方法：这可以借助两个指针算法来解决。
- 从左边开始遍历数组并保留两个指针。
- 一个指针(假设为 i)用于迭代数组。
- 第二个指针(假设 st)向前移动以查找下一个唯一元素。 i 中的元素出现两次以上。
下面是上述方法的实现：
CPP
// C++ program to reduce the array
// such that each element appears
// at most 2 times
#include
using namespace std;
// Function to remove duplicates
void removeDuplicates(int arr[], int n)
{
// Initialise 2nd pointer
int st = 0;
// Iterate over the array
for (int i = 0; i < n; i++) {
if (i < n - 2
&& arr[i] == arr[i + 1]
&& arr[i] == arr[i + 2])
continue;
// Updating the 2nd pointer
else {
arr[st] = arr[i];
st++;
}
}
cout << "{";
for (int i = 0; i < st; i++) {
cout << arr[i];
if (i != st - 1)
cout << ", ";
}
cout << "}";
}
// Driver code
int main()
{
int arr[]
= { 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 };
int n = sizeof(arr)
/ sizeof(arr[0]);
// Function call
removeDuplicates(arr, n);
return 0;
} Java
// Java program to reduce the array
// such that each element appears
// at most 2 times
class GFG
{
// Function to remove duplicates
static void removeDuplicates(int arr[], int n)
{
// Initialise 2nd pointer
int st = 0;
// Iterate over the array
for (int i = 0; i < n; i++) {
if (i < n - 2
&& arr[i] == arr[i + 1]
&& arr[i] == arr[i + 2])
continue;
// Updating the 2nd pointer
else {
arr[st] = arr[i];
st++;
}
}
System.out.print("{");
for (int i = 0; i < st; i++) {
System.out.print(arr[i]);
if (i != st - 1)
System.out.print(", ");
}
System.out.print("}");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 };
int n = arr.length;
// Function call
removeDuplicates(arr, n);
}
}
// This code is contributed by sapnasingh4991Python3
# Python3 program to reduce the array
# such that each element appears
# at most 2 times
# Function to remove duplicates
def removeDuplicates(arr, n) :
# Initialise 2nd pointer
st = 0;
# Iterate over the array
for i in range(n) :
if (i < n - 2 and arr[i] == arr[i + 1]
and arr[i] == arr[i + 2]) :
continue;
# Updating the 2nd pointer
else :
arr[st] = arr[i];
st += 1;
print("{",end="")
for i in range(st) :
print(arr[i],end="");
if (i != st - 1) :
print(", ",end="");
print("}",end="");
# Driver code
if __name__ == "__main__" :
arr = [ 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 ];
n = len(arr);
# Function call
removeDuplicates(arr, n);
# This code is contributed by Yash_RC#
// C# program to reduce the array
// such that each element appears
// at most 2 times
using System;
class GFG
{
// Function to remove duplicates
static void removeDuplicates(int []arr, int n)
{
// Initialise 2nd pointer
int st = 0;
// Iterate over the array
for (int i = 0; i < n; i++) {
if (i < n - 2
&& arr[i] == arr[i + 1]
&& arr[i] == arr[i + 2])
continue;
// Updating the 2nd pointer
else {
arr[st] = arr[i];
st++;
}
}
Console.Write("{");
for (int i = 0; i < st; i++) {
Console.Write(arr[i]);
if (i != st - 1)
Console.Write(", ");
}
Console.Write("}");
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5 };
int n = arr.Length;
// Function call
removeDuplicates(arr, n);
}
}
// This code is contributed by sapnasingh4991Javascript
Python3
# Python3 program to reduce the array
# such that each element appears
# at most 2 times
from collections import Counter
# Function to remove duplicates
def removeDuplicates(arr, n):
freq = Counter(arr)
# Taking empty list
l = []
for i in range(n):
if(freq[arr[i]] >= 2):
# Making frequency to 1
freq[arr[i]] = 1
l.append(arr[i])
elif(freq[arr[i]] == 1):
# Making frequency to 0
# and appending to list
l.append(arr[i])
freq[arr[i]] = 0
# Printing the list
for i in l:
print(i, end=" ")
# Driver code
if __name__ == "__main__":
arr = [1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5]
n = len(arr)
# Function call
removeDuplicates(arr, n)
# This code is contributed by vikkycirus输出
{1, 1, 2, 2, 3, 3, 4, 5}时间复杂度： O(N)
空间复杂度： O(1)
另一种方法：使用 Counter()函数
- 使用计数器函数计算所有元素的频率。
- 取一个空列表。
- 遍历数组。
- 如果任何元素的频率大于或等于 2，则将其频率设为 1 并将其附加到列表中。
- 如果任何元素的频率等于 1，则取其频率 0 并将其附加到列表中。
- 打印列表。
下面是上述方法的实现：
蟒蛇3
# Python3 program to reduce the array
# such that each element appears
# at most 2 times
from collections import Counter
# Function to remove duplicates
def removeDuplicates(arr, n):
freq = Counter(arr)
# Taking empty list
l = []
for i in range(n):
if(freq[arr[i]] >= 2):
# Making frequency to 1
freq[arr[i]] = 1
l.append(arr[i])
elif(freq[arr[i]] == 1):
# Making frequency to 0
# and appending to list
l.append(arr[i])
freq[arr[i]] = 0
# Printing the list
for i in l:
print(i, end=" ")
# Driver code
if __name__ == "__main__":
arr = [1, 1, 1, 2,
2, 2, 3, 3,
3, 3, 3, 3,
4, 5]
n = len(arr)
# Function call
removeDuplicates(arr, n)
# This code is contributed by vikkycirus
输出
1 1 2 2 3 3 4 5 时间复杂度： O(N)
空间复杂度： O(N)
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