最小化包含至少一个公共字符的子串的长度

给定一个字符串str ，任务是找到最小长度的子字符串，使得str中该长度的所有子字符串至少包含一个公共字符。如果无法获得这样的长度，则打印-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：这个问题可以通过使用二分搜索技术轻松解决。请按照以下步骤解决问题：

• 将低初始化为1 ，将高初始化为字符串str的长度。
• 在低和高之间找到中间。
• 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high更新为mid -1。
• 如果不是这种情况，则将低更新为mid + 1 。
• 对字母表中所有 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 chitranayal

C#

// 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

Javascript

 
// JavaScript program to implement
// the above approach
 
// Function to check and return if
// subStrings of length mid has a
// common character a
function check(str, mid, a)
{
     
    // Length of the String
    var n = str.length;
     
    // Initialise the first
    // occurrence of character a
    var 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
function possible(str, mid)
{
     
    // Check for all 26 alphabets
    for(var i = 0; i < 26; ++i)
    {
        // Check that char i + a is
        // common in all the subStrings
        // of length mid
         
        if (check(str, mid,
            String.fromCharCode(
            i + "a".charCodeAt(0))))
            return true;
    }
     
    // If no characters is common
    return false;
}
 
// Function to calculate and return
// the minm length of subStrings
function findMinLength(str)
{
     
    // Initialise low and high
    var low = 1,
    high = str.length;
     
    // Perform binary search
    while (low <= high)
    {
         
        // Update mid
        var mid = parseInt((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
function ifAllDistinct(str)
{
    var s = new Array();
    var temp = str.split("");
     
    for(const c of temp)
    {
        s.push(c);
    }
    var set = new Set(s);
    return set.size === str.length;
}
 
// Driver Code
var str = "geeksforgeeks";
 
if (ifAllDistinct(str))
    document.write(-1 + "<br>");
else
    document.write(
        findMinLength(str.split("")));
         
// This code is contributed by rdtank 
 

输出：

7

时间复杂度： O(26 * N * log(N))
辅助空间： O(1)

如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live