给定大小为N的等长字符串数组arr [] ，任务是检查是否可以通过重复交换给定数组中相同或不同字符串的任意一对字符使所有字符串相等。如果发现是真的，则打印“是” 。否则，打印“否” 。
例子：
Input: arr[] = { “acbdd”, “abcee” }
Output: YES
Explanation:
Swapping arr[0][1] and arr[0][2] modifies arr[] to { “abcdd”, “abcee” }
Swapping arr[0][3] and arr[1][4] modifies arr[] to { “abced”, “abced” }
Therefore, the required output is “YES”.
Input: arr[] = { “abb”, “acc”, “abc” }
Output: YES
Explanation:
Swapping arr[0][2] with arr[1][1] modifies arr[] to { “abc”, “abc”, “abc” }.
Therefore, the required output is “YES”.
方法：我们的想法是计算所有字符串的每个不同字符的频率，并检查是否每个不同字符的频率是整除N或不是。如果发现是真的，则打印“是” 。否则，打印“否” 。请按照以下步骤解决问题：
- 初始化一个数组，例如cntFreq [] ，以存储所有字符串中每个不同字符的频率。
- 遍历阵列和用于遇到的每一个字符串，计数字符串的每个不同的字符的频率。
- 使用变量i遍历数组cntFreq [] 。对于每个第i个字符，检查cntFreq [i]是否可被N整除。如果发现错误，则打印“否” 。
- 否则，打印“是” 。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to check if all strings can be
// made equal by swapping any pair of
// characters from the same or different strings
bool isEqualStrings(string arr[], int N)
{
// Stores length of string
int M = arr[0].length();
// Store frequency of each distinct
// character of the strings
int cntFreq[256] = { 0 };
// Traverse the array
for (int i = 0; i < N; i++) {
// Iterate over the characters
for (int j = 0; j < M; j++) {
// Update frequency
// of arr[i][j]
cntFreq[arr[i][j] - 'a']++;
}
}
// Traverse the array cntFreq[]
for (int i = 0; i < 256; i++) {
// If cntFreq[i] is
// divisible by N or not
if (cntFreq[i] % N != 0) {
return false;
}
}
return true;
}
// Driver Code
int main()
{
string arr[] = { "aab", "bbc", "cca" };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
if (isEqualStrings(arr, N)) {
cout << "YES";
}
else {
cout << "NO";
}
return 0;
} Java
// Java program to implement
// the above approach
class GFG{
// Function to check if all strings can be
// made equal by swapping any pair of
// characters from the same or different strings
static boolean isEqualStrings(String[] arr, int N)
{
// Stores length of string
int M = arr[0].length();
// Store frequency of each distinct
// character of the strings
int[] cntFreq = new int[256];
for(int i = 0; i < N; i++)
{
cntFreq[i] = 0;
}
// Traverse the array
for(int i = 0; i < N; i++)
{
// Iterate over the characters
for(int j = 0; j < M; j++)
{
// Update frequency
// of arr[i].charAt(j)
cntFreq[arr[i].charAt(j) - 'a'] += 1;
}
}
// Traverse the array cntFreq[]
for(int i = 0; i < 256; i++)
{
// If cntFreq[i] is
// divisible by N or not
if (cntFreq[i] % N != 0)
{
return false;
}
}
return true;
}
// Driver Code
public static void main(String[] args)
{
String[] arr = { "aab", "bbc", "cca" };
int N = arr.length;
// Function Call
if (isEqualStrings(arr, N))
{
System.out.println("YES");
}
else
{
System.out.println("NO");
}
}
}
// This code is contributed by AnkThonPython3
# Python3 program to implement
# the above approach
# Function to check if all strings can
# be made equal by swapping any pair of
# characters from the same or different
# strings
def isEqualStrings(arr, N):
# Stores length of string
M = len(arr[0])
# Store frequency of each distinct
# character of the strings
cntFreq = [0] * 256
# Traverse the array
for i in range(N):
# Iterate over the characters
for j in range(M):
# Update frequency
# of arr[i][j]
cntFreq[ord(arr[i][j]) - ord('a')] += 1
# Traverse the array cntFreq[]
for i in range(256):
# If cntFreq[i] is
# divisible by N or not
if (cntFreq[i] % N != 0):
return False
return True
# Driver Code
if __name__ == "__main__" :
arr = [ "aab", "bbc", "cca" ]
N = len(arr)
# Function Call
if isEqualStrings(arr, N):
print("YES")
else:
print("NO")
# This code is contributed by jana_sayantanC#
// C# program to implement
// the above approach
using System;
class GFG{
// Function to check if all strings can be
// made equal by swapping any pair of
// characters from the same or different strings
static bool isEqualStrings(string[] arr, int N)
{
// Stores length of string
int M = arr[0].Length;
// Store frequency of each distinct
// character of the strings
int[] cntFreq = new int[256];
for(int i = 0; i < N; i++)
{
cntFreq[i] = 0;
}
// Traverse the array
for(int i = 0; i < N; i++)
{
// Iterate over the characters
for(int j = 0; j < M; j++)
{
// Update frequency
// of arr[i][j]
cntFreq[arr[i][j] - 'a'] += 1;
}
}
// Traverse the array cntFreq[]
for(int i = 0; i < 256; i++)
{
// If cntFreq[i] is
// divisible by N or not
if (cntFreq[i] % N != 0)
{
return false;
}
}
return true;
}
// Driver Code
public static void Main()
{
string[] arr = { "aab", "bbc", "cca" };
int N = arr.Length;
// Function Call
if (isEqualStrings(arr, N))
{
Console.WriteLine("YES");
}
else
{
Console.WriteLine("NO");
}
}
}
// This code is contributed by susmitakundugoaldanga输出：
YES时间复杂度：O(N * M + 256)，其中M是字符串的长度
辅助空间： O(256)