给定一个字符串S ,任务是检查 S 是否包含一对长度为K的子字符串,它们是彼此的字谜并且其中不包含字符X。如果不存在这样的子字符串,则打印-1 。
例子:
Input: S = “geeksforgeeks”, X = ‘f’, K = 5
Output: geeks geeks
Explanation:
Substrings “geeks” and “geeks” are anagrams of each other and does not contain ‘f’.
Input: S = “rotator”, X = ‘a’, K = 3
Output: rot tor
Explanation:
Substrings “rot” and “tor” are anagrams of each other and does not contain ‘a’.
方法:
这个想法是根据字符生成前缀和。请按照以下步骤解决问题:
- 使用前缀 sum array迭代子字符串的字符串和生成频率。
- 如果 HashMap 中已经存在具有相同字符频率的子字符串。
- 否则,如果子字符串中字符X的频率为0 ,则将子字符串与当前子字符串的字符频率存储在HashMap 中。
下面是上述方法的实现:
Java
// Java Program to implement
// the above approach
import java.util.*;
// Class to represent a Substring
// in terms of frequency of
// characters present in it
class Substring {
int MOD = 1000000007;
// Store count of characters
int count[];
Substring() { count = new int[26]; }
public int hashCode()
{
int hash = 0;
for (int i = 0; i < 26; i++) {
hash += (i + 1) * count[i];
hash %= MOD;
}
return hash;
}
public boolean equals(Object o)
{
if (o == this)
return true;
if (!(o instanceof Substring))
return false;
Substring ob = (Substring)o;
for (int i = 0; i < 26; i++) {
if (ob.count[i] != count[i])
return false;
}
return true;
}
}
class GFG {
// Function to check anagrams
static void checkForAnagrams(String s, int n,
char X, int k)
{
boolean found = false;
// Prefix array to store frequencies
// of characters
int prefix[][] = new int[n + 1][26];
for (int i = 0; i < n; i++) {
prefix[i][s.charAt(i) - 97]++;
}
// Generate prefix sum
for (int i = 1; i < n; i++) {
for (int j = 0; j < 26; j++)
prefix[i][j] += prefix[i - 1][j];
}
// Map to store frequencies
HashMap map
= new HashMap<>();
// Check for anagrams
for (int i = 0; i < n; i++) {
if (i + k > n)
break;
// Generate frequencies of characters
// of substring starting from i
Substring sub = new Substring();
for (int j = 0; j < 26; j++) {
sub.count[j]
= prefix[i + k - 1][j]
- (i - 1 >= 0
? prefix[i - 1][j]
: 0);
}
// Check if forbidden character is
// present, then continue
if (sub.count[X - 97] != 0)
continue;
// If already present in HashMap
if (map.containsKey(sub)) {
found = true;
// Print the substrings
System.out.println(
s.substring(map.get(sub),
map.get(sub) + k)
+ " " + s.substring(i, i + k));
break;
}
else {
map.put(sub, i);
}
}
// If no such substring is found
if (!found)
System.out.println("-1");
}
// Driver Code
public static void main(String[] args)
{
String s = "rotator";
int n = s.length();
char X = 'a';
int k = 3;
checkForAnagrams(s, n, X, k);
}
}
输出:
rot tor
时间复杂度: O(N*26)
辅助空间: O(N*26)
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