给定一个包含小写拉丁字母的字符串S。如果长度为ATLEASTķS的每个子字符串包含此字符c字符C被称为K-惊人。找到最小可能的 K，使得至少存在一个 K-amazing字符。
例子：
Input : S = “abcde”
Output :3
Explanation : Every substring of length atleast 3 contains character ‘c’, i.e.
{“abc”, “bcd”, “cde”, “abcd”, “bcde”, “abcde”}
Input :S = “aaaa”
Output :1
先决条件：二分搜索
天真的解决方案：一个简单的方法是迭代所有可能的子字符串长度，即从 1 到 N(字符串的大小)，并为每个当前长度的子字符串检查是否某个字符出现在所有这些子字符串中。
有效的解决方案：关键思想是对答案K执行二分搜索，因为如果某个字符c 出现在长度为X 的所有子字符串中，那么它将始终出现在长度为(X + 1) 的所有子字符串中。因此，我们可以检查当前长度并尝试使用分治算法将其最小化。为了检查某个字符出现在所有长度为 X 的子字符串中，迭代从 ‘a’ 到 ‘z’ 的所有字符，并在另一个循环中迭代地存储最后一个字符的最后一次出现。
设当前位置为j，所以最后一个长度为X的子串是从(j-X)到X。检查当前K-amazing字符最后出现的位置是否大于(j-X)。如果它更大，那么该子字符串是一个有效的字符串。
下面是上述方法的实现。
C++
// CPP Program to find minimum K such that
// every substring of length atleast K
// contains some character c
#include
using namespace std;
// This function checks if there exists some
// character which appears in all K length
// substrings
int check(string s, int K)
{
// Iterate over all possible characters
for (int ch = 0; ch < 26; ch++) {
char c = 'a' + ch;
// stores the last occurrence
int last = -1;
// set answer as true;
bool found = true;
for (int i = 0; i < K; i++)
if (s[i] == c)
last = i;
// No occurrence found of current
// character in first substring
// of length K
if (last == -1)
continue;
// Check for every last substring
// of length K where last occurr-
// ence exists in substring
for (int i = K; i < s.size(); i++) {
if (s[i] == c)
last = i;
// If last occ is not
// present in substring
if (last <= (i - K)) {
found = false;
break;
}
}
// current character is K amazing
if (found)
return 1;
}
return 0;
}
// This function performs binary search over the
// answer to minimise it
int binarySearch(string s)
{
int low = 1, high = (int)s.size();
int ans;
while (low <= high) {
int mid = (high + low) >> 1;
// Check if answer is found try
// to minimise it
if (check(s, mid)) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
// Driver Code to test above functions
int32_t main()
{
string s = "abcde";
cout << binarySearch(s) << endl;
s = "aaaa";
cout << binarySearch(s) << endl;
return 0;
} Java
// Java Program to find minimum K such that
// every substring of length atleast K
// contains some character c
class GFG
{
// This function checks if there exists some
// character which appears in all K length
// substrings
static int check(String s, int K)
{
// Iterate over all possible characters
for (int ch = 0; ch < 26; ch++) {
char c = (char)( 'a' + ch);
// stores the last occurrence
int last = -1;
// set answer as true;
boolean found = true;
for (int i = 0; i < K; i++)
if (s.charAt(i) == c)
last = i;
// No occurrence found of current
// character in first substring
// of length K
if (last == -1)
continue;
// Check for every last substring
// of length K where last occurr-
// ence exists in substring
for (int i = K; i < s.length(); i++) {
if (s.charAt(i) == c)
last = i;
// If last occ is not
// present in substring
if (last <= (i - K)) {
found = false;
break;
}
}
// current character is K amazing
if (found)
return 1;
}
return 0;
}
// This function performs binary search over the
// answer to minimise it
static int binarySearch(String s)
{
int low = 1, high = s.length();
int ans=0;
while (low <= high) {
int mid = (high + low) >> 1;
// Check if answer is found try
// to minimise it
if (check(s, mid)==1) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
// Driver Code to test above functions
public static void main(String args[])
{
String s = "abcde";
System.out.println(binarySearch(s));
s = "aaaa";
System.out.println(binarySearch(s));
}
}
// This article is contributed
// by ihritikPython3
# Python3 Program to find minimum K such
# that every substring of length atleast
# K contains some character c
# This function checks if there exists
# some character which appears in all
# K length substrings
def check(s, K):
# Iterate over all possible characters
for ch in range(0, 26):
c = chr(97 + ch) # Ascii value of 'a' => 97
# stores the last occurrence
last = -1
# set answer as true
found = True
for i in range(0, K):
if s[i] == c:
last = i
# No occurrence found of current character
# in first substring of length K
if last == -1:
continue
# Check for every last substring
# of length K where last occurr-
# ence exists in substring
for i in range(K, len(s)):
if s[i] == c:
last = i
# If last occ is not
# present in substring
if last <= (i - K):
found = False
break
# current character is K amazing
if found:
return 1
return 0
# This function performs binary search
# over the answer to minimise it
def binarySearch(s):
low, high, ans = 1, len(s), None
while low <= high:
mid = (high + low) >> 1
# Check if answer is found
# try to minimise it
if check(s, mid):
ans, high = mid, mid - 1
else:
low = mid + 1
return ans
# Driver Code
if __name__ == "__main__":
s = "abcde"
print(binarySearch(s))
s = "aaaa"
print(binarySearch(s))
# This code is contributed by Rituraj JainC#
// C# Program to find minimum K such that
// every substring of length atleast K
// contains some character c
using System;
class GFG
{
// This function checks if there exists some
// character which appears in all K length
// substrings
static int check(String s, int K)
{
// Iterate over all possible characters
for (int ch = 0; ch < 26; ch++) {
char c = (char)( 'a' + ch);
// stores the last occurrence
int last = -1;
// set answer as true;
bool found = true;
for (int i = 0; i < K; i++)
if (s[i] == c)
last = i;
// No occurrence found of current
// character in first substring
// of length K
if (last == -1)
continue;
// Check for every last substring
// of length K where last occurr-
// ence exists in substring
for (int i = K; i < s.Length; i++) {
if (s[i] == c)
last = i;
// If last occ is not
// present in substring
if (last <= (i - K)) {
found = false;
break;
}
}
// current character is K amazing
if (found)
return 1;
}
return 0;
}
// This function performs binary search over the
// answer to minimise it
static int binarySearch(String s)
{
int low = 1, high = s.Length;
int ans=0;
while (low <= high) {
int mid = (high + low) >> 1;
// Check if answer is found try
// to minimise it
if (check(s, mid)==1) {
ans = mid;
high = mid - 1;
}
else
low = mid + 1;
}
return ans;
}
// Driver Code to test above functions
public static void Main()
{
String s = "abcde";
Console.WriteLine(binarySearch(s));
s = "aaaa";
Console.WriteLine(binarySearch(s));
}
}
// This article is contributed
// by ihritikPHP
> 1;
// Check if answer is found try
// to minimise it
if (check($s, $mid))
{
$ans = $mid;
$high = $mid - 1;
}
else
$low = $mid + 1;
}
return $ans;
}
// Driver Code
$s = "abcde";
echo binarySearch($s) . "\n";
$s = "aaaa";
echo binarySearch($s) . "\n";
// This code is contributed by Ryuga
?>Javascript
输出：
3
1时间复杂度： O(N * logN * 26)，其中 N 是给定字符串的大小。
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