给定 N 个二进制字符串b1、b2、b3…。十亿任务是找到可以通过任意次数交换任何字符对来回文的二进制字符串的最大数量。字符可以来自同一个字符串，也可以来自不同的字符串
例子：
Input: N=3
1110
100110
010101
Output: 2
Explanation:
b1 = 1110
b2 = 100110 - > 110010
b3 = 010101 -> 110010
Now swap last 0 in s2 and s3
with 1's in s1
Final string become
b1 = 1000
b2 = 110011
b3 = 110011
where b1 and b2 are a palindrome
Input: N=3
1
1000
111110
Output: 3方法：
字符串的长度不会改变。同样重要的是要注意，如果我们得到一个奇数长度的二进制字符串，那么我们总是可以交换字符以将该字符串转换为回文字符串。这是因为如果长度是奇数，那么我们将有（偶数个零和奇数个 1）或（偶数个零和奇数个零）。所以它总是可以以这样的方式放置，使字符串回文。
现在我们的 ans 可以是 N 或 N-1。我们必须考虑我们的 ans 是 N 的情况。所以，我们得到了 N 个二进制字符串。如果至少有 1 个长度为奇数的字符串，那么我们的答案肯定是 N。
- 抓住所有的 0 和 1 并将它们从它们的位置上移除。然后我们将至少有一对 0 或 1，然后将它们对称地放入它们的空闲位置（跳过奇数长度的中间）。所以到现在为止，所有偶数长度的字符串都被填充了，奇数长度的字符串在中间有一个空闲点，可以很容易地用剩余的字符填充。所以在这种情况下，我们的 ans 将是 N。
- 现在，另一个案例。如果我们单独拥有所有 N 个偶数长度的字符串，并且 1 和 0 的总数是偶数（即 1 的总数是偶数，0 的总数是偶数），那么在这种情况下，我们的 ans 也将是 N。这是因为 1 和 0 可以对称地放置在所有 N 个字符串，使它们成为回文。否则，我们的答案将是 N-1。
下面是上述方法的实现：
C++
// C++ program for the
// above approach
#include
using namespace std;
int max_palindrome(string s[], int n)
{
int flag = 0;
for (int i = 0; i < n; i++) {
// To check if there is
// any string of odd length
if (s[i].size() % 2 != 0) {
flag = 1;
}
}
// If there is at least
// 1 string of odd
// length.
if (flag == 1) {
return n;
}
int z = 0, o = 0;
// If all the strings are
// of even length.
for (int i = 0; i < n; i++) {
for (int j = 0; j < s[i].size(); j++) {
// Count of 0's in all
// the strings
if (s[i][j] == '0')
z++;
// Count of 1's in
// all the strings
else
o++;
}
}
// If z is even
// and o is even
// then ans will be N.
if (o % 2 == 0 && z % 2 == 0) {
return n;
}
// Otherwise ans will be N-1.
else {
return n - 1;
}
}
// Driver code
int main()
{
int n = 3;
string s[n] = { "1110", "100110", "010101" };
cout << max_palindrome(s, n);
return 0;
} Java
// Java program for the above approach
class GFG
{
static int max_palindrome(String []s, int n)
{
int flag = 0;
for (int i = 0; i < n; i++)
{
// To check if there is
// any string of odd length
if (s[i].length() % 2 != 0)
{
flag = 1;
}
}
// If there is at least
// 1 string of odd
// length.
if (flag == 1)
{
return n;
}
int z = 0;
int o = 0;
// If all the strings are
// of even length.
for (int i = 0; i < n; i++)
{
for (int j = 0; j < s[i].length(); j++)
{
// Count of 0's in all
// the strings
if (s[i].charAt(j) == '0')
z += 1;
// Count of 1's in
// all the strings
else
o += 1;
}
}
// If z is even
// and o is even
// then ans will be N.
if (o % 2 == 0 && z % 2 == 0)
{
return n;
}
// Otherwise ans will be N-1.
else
{
return n - 1;
}
}
// Driver code
public static void main (String[] args)
{
int n = 3;
String s[] = {"1110", "100110", "010101" };
System.out.println(max_palindrome(s, n));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 program for the above approach
def max_palindrome(s, n) :
flag = 0;
for i in range(n) :
# To check if there is
# any string of odd length
if (len(s[i]) % 2 != 0) :
flag = 1;
# If there is at least
# 1 string of odd
# length.
if (flag == 1) :
return n;
z = 0; o = 0;
# If all the strings are
# of even length.
for i in range(n) :
for j in range(len(s[i])) :
# Count of 0's in all
# the strings
if (s[i][j] == '0') :
z += 1;
# Count of 1's in
# all the strings
else :
o += 1;
# If z is even
# and o is even
# then ans will be N.
if (o % 2 == 0 and z % 2 == 0) :
return n;
# Otherwise ans will be N-1.
else :
return n - 1;
# Driver code
if __name__ == "__main__" :
n = 3;
s = [ "1110", "100110", "010101" ];
print(max_palindrome(s, n));
# This code is contributed by AnkitRai01C#
// C# program for the above approach
using System;
class GFG
{
static int max_palindrome(string []s, int n)
{
int flag = 0;
for (int i = 0; i < n; i++)
{
// To check if there is
// any string of odd length
if (s[i].Length % 2 != 0)
{
flag = 1;
}
}
// If there is at least
// 1 string of odd
// length.
if (flag == 1)
{
return n;
}
int z = 0;
int o = 0;
// If all the strings are
// of even length.
for (int i = 0; i < n; i++)
{
for (int j = 0; j < s[i].Length; j++)
{
// Count of 0's in all
// the strings
if (s[i][j] == '0')
z += 1;
// Count of 1's in
// all the strings
else
o += 1;
}
}
// If z is even
// and o is even
// then ans will be N.
if (o % 2 == 0 && z % 2 == 0)
{
return n;
}
// Otherwise ans will be N-1.
else
{
return n - 1;
}
}
// Driver code
public static void Main ()
{
int n = 3;
string []s = {"1110", "100110", "010101" };
Console.WriteLine(max_palindrome(s, n));
}
}
// This code is contributed by AnkitRai01Javascript
输出：
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