给定一个字符串str ，任务是找到子字符串的最小长度，以使str中具有该长度的所有子字符串都至少包含一个公共字符。如果无法获得这样的长度，请打印-1 。
例子：
Input: str = “saad”
Output: 2
Explanation:
All the substrings of length two of a given string are {“sa”, “aa”, “ad”}.
Since ‘a’ is common among all the substring, the minimum possible length of substrings having at least one common character is 2.
Input: str = “geeksforgeeks”
Output: 7
Explanation:
All possible substrings of length 7 are as follows:
- geeksfo
- eeksfor
- eksforg
- ksforge
- sforgee
- forgeek
- orgeeks
Since all the substrings have a common character e, the minimum possible length of substrings having at least one common character is 7.
方法：使用二进制搜索技术可以轻松解决此问题。请按照以下步骤解决问题：
- 初始化低为1和高为字符串str的长度。
- 在高低之间找到中间。
- 如果给定字符串的所有子字符串中都有一个公共字符，则将其更新为高-1。
- 如果不是这种情况，则将Low更新为mid + 1 。
- 对字母表中的所有26个可能的字符重复上述步骤。
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 and return if
// substrings of length mid has a
// common character a
bool check(string& str, int mid, char a)
{
// Length of the string
int n = str.size();
// Initialise the first
// occurrence of character a
int previous = -1, i;
for (i = 0; i < n; ++i) {
if (str[i] == a) {
// Check that distance b/w
// the current and previous
// occurrence of character a
// is less than or equal to mid
if (i - previous > mid) {
return false;
}
previous = i;
}
}
// If distance exceeds mid
if (i - previous > mid)
return false;
else
return true;
}
// Function to check for all the
// alphabets, if substrings of
// length mid have a character common
bool possible(string& str, int mid)
{
// Check for all 26 alphabets
for (int i = 0; i < 26; ++i) {
// Check that char i + a is
// common in all the substrings
// of length mid
if (check(str, mid, i + 'a'))
return true;
}
// If no characters is common
return false;
}
// Function to calculate and return
// the minm length of substrings
int findMinLength(string& str)
{
// Initialise low and high
int low = 1, high = str.length();
// Perform binary search
while (low <= high) {
// Update mid
int mid = (low + high) / 2;
// Check if one common character is
// present in the length of the mid
if (possible(str, mid))
high = mid - 1;
else
low = mid + 1;
}
// Returns the minimum length
// that contain one
// common character
return high + 1;
}
// Function to check if all
// characters are distinct
bool ifAllDistinct(string str)
{
set s;
for (char c : str) {
s.insert(c);
}
return s.size() == str.size();
}
// Driver Code
int main()
{
string str = "geeksforgeeks";
if (ifAllDistinct(str))
cout << -1 << endl;
else
cout << findMinLength(str);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG{
// Function to check and return if
// subStrings of length mid has a
// common character a
static boolean check(char[] str, int mid,
char a)
{
// Length of the String
int n = str.length;
// Initialise the first
// occurrence of character a
int previous = -1, i;
for(i = 0; i < n; ++i)
{
if (str[i] == a)
{
// Check that distance b/w
// the current and previous
// occurrence of character a
// is less than or equal to mid
if (i - previous > mid)
{
return false;
}
previous = i;
}
}
// If distance exceeds mid
if (i - previous > mid)
return false;
else
return true;
}
// Function to check for all the
// alphabets, if subStrings of
// length mid have a character common
static boolean possible(char[] str, int mid)
{
// Check for all 26 alphabets
for(int i = 0; i < 26; ++i)
{
// Check that char i + a is
// common in all the subStrings
// of length mid
if (check(str, mid, (char)(i + 'a')))
return true;
}
// If no characters is common
return false;
}
// Function to calculate and return
// the minm length of subStrings
static int findMinLength(char[] str)
{
// Initialise low and high
int low = 1, high = str.length;
// Perform binary search
while (low <= high)
{
// Update mid
int mid = (low + high) / 2;
// Check if one common character is
// present in the length of the mid
if (possible(str, mid))
high = mid - 1;
else
low = mid + 1;
}
// Returns the minimum length
// that contain one
// common character
return high + 1;
}
// Function to check if all
// characters are distinct
static boolean ifAllDistinct(String str)
{
HashSet s = new HashSet();
for(char c : str.toCharArray())
{
s.add(c);
}
return s.size() == str.length();
}
// Driver Code
public static void main(String[] args)
{
String str = "geeksforgeeks";
if (ifAllDistinct(str))
System.out.print(-1 + "\n");
else
System.out.print(findMinLength(
str.toCharArray()));
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 program to implement
# the above approach
# Function to check and return if
# substrings of length mid has a
# common character a
def check(st, mid, a):
# Length of the string
n = len(st)
# Initialise the first
# occurrence of character a
previous = -1
for i in range(n):
if (st[i] == chr(a)):
# Check that distance b/w
# the current and previous
# occurrence of character a
# is less than or equal to mid
if (i - previous > mid):
return False
previous = i
# If distance exceeds mid
if (i - previous > mid):
return False
else:
return True
# Function to check for all the
# alphabets, if substrings of
# length mid have a character common
def possible(st, mid):
# Check for all 26 alphabets
for i in range(26):
# Check that char i + a is
# common in all the substrings
# of length mid
if (check(st, mid, i + ord('a'))):
return True
# If no characters is common
return False
# Function to calculate and return
# the minm length of substrings
def findMinLength(st):
# Initialise low and high
low = 1
high = len(st)
# Perform binary search
while (low <= high):
# Update mid
mid = (low + high) // 2
# Check if one common character is
# present in the length of the mid
if (possible(st, mid)):
high = mid - 1
else:
low = mid + 1
# Returns the minimum length
# that contain one
# common character
return high + 1
# Function to check if all
# characters are distinct
def ifAllDistinct( st):
s = []
for c in st:
s.append(c)
return len(set(s)) == len(st)
# Driver Code
if __name__ == "__main__":
st = "geeksforgeeks"
if (ifAllDistinct(st)):
print("-1")
else:
print(findMinLength(st))
# This code is contributed by chitranayalC#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to check and return if
// subStrings of length mid has a
// common character a
static bool check(char[] str, int mid,
char a)
{
// Length of the String
int n = str.Length;
// Initialise the first
// occurrence of character a
int previous = -1, i;
for(i = 0; i < n; ++i)
{
if (str[i] == a)
{
// Check that distance b/w
// the current and previous
// occurrence of character a
// is less than or equal to mid
if (i - previous > mid)
{
return false;
}
previous = i;
}
}
// If distance exceeds mid
if (i - previous > mid)
return false;
else
return true;
}
// Function to check for all the
// alphabets, if subStrings of
// length mid have a character common
static bool possible(char[] str, int mid)
{
// Check for all 26 alphabets
for(int i = 0; i < 26; ++i)
{
// Check that char i + a is
// common in all the subStrings
// of length mid
if (check(str, mid, (char)(i + 'a')))
return true;
}
// If no characters is common
return false;
}
// Function to calculate and return
// the minm length of subStrings
static int findMinLength(char[] str)
{
// Initialise low and high
int low = 1, high = str.Length;
// Perform binary search
while (low <= high)
{
// Update mid
int mid = (low + high) / 2;
// Check if one common character is
// present in the length of the mid
if (possible(str, mid))
high = mid - 1;
else
low = mid + 1;
}
// Returns the minimum length
// that contain one
// common character
return high + 1;
}
// Function to check if all
// characters are distinct
static bool ifAllDistinct(String str)
{
HashSet s = new HashSet();
foreach(char c in str.ToCharArray())
{
s.Add(c);
}
return s.Count == str.Length;
}
// Driver Code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
if (ifAllDistinct(str))
Console.Write(-1 + "\n");
else
Console.Write(findMinLength(
str.ToCharArray()));
}
}
// This code is contributed by gauravrajput1 输出：
7
时间复杂度： O(26 * N * log(N))
辅助空间： O(1)