给定一个字符串str ，其子字符串的形成方式是，以该字符串的第一个字符开头的所有子字符串将按其长度的排序顺序首先出现，然后是所有以第二个字符开头的子字符串字符串的字符，按其长度的排序顺序等。
例如，对于字符串abc ，其子字符串按要求的顺序为a ， ab ， abc ， b ， bc和c 。
现在给定整数k ，任务是按所需顺序查找第k个子字符串。
例子：
Input: str = abc, k = 4
Output: b
The required order is “a”, “ab”, “abc”, “b”, “bc” and “c”
Input: str = abc, k = 9
Output: -1
Only 6 sub-strings are possible.
方法：想法是使用二进制搜索。数组子字符串将用于存储以ith字符+ substring [i – 1]开头的子字符串的数量。现在，对数组子字符串使用二进制搜索，找到所需子字符串的起始索引，然后以end = length_of_string –(substring [start] – k)查找同一子字符串的结束索引。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to prints kth sub-string
void Printksubstring(string str, int n, int k)
{
// Total sub-strings possible
int total = (n * (n + 1)) / 2;
// If k is greater than total
// number of sub-strings
if (k > total) {
printf("-1\n");
return;
}
// To store number of sub-strings starting
// with ith character of the string
int substring[n + 1];
substring[0] = 0;
// Compute the values
int temp = n;
for (int i = 1; i <= n; i++) {
// substring[i - 1] is added
// to store the cumulative sum
substring[i] = substring[i - 1] + temp;
temp--;
}
// Binary search to find the starting index
// of the kth sub-string
int l = 1;
int h = n;
int start = 0;
while (l <= h) {
int m = (l + h) / 2;
if (substring[m] > k) {
start = m;
h = m - 1;
}
else if (substring[m] < k)
l = m + 1;
else {
start = m;
break;
}
}
// To store the ending index of
// the kth sub-string
int end = n - (substring[start] - k);
// Print the sub-string
for (int i = start - 1; i < end; i++)
cout << str[i];
}
// Driver code
int main()
{
string str = "abc";
int k = 4;
int n = str.length();
Printksubstring(str, n, k);
return 0;
} Java
// Java implementation of the approach
class GFG
{
// Function to prints kth sub-string
static void Printksubstring(String str, int n, int k)
{
// Total sub-strings possible
int total = (n * (n + 1)) / 2;
// If k is greater than total
// number of sub-strings
if (k > total)
{
System.out.printf("-1\n");
return;
}
// To store number of sub-strings starting
// with ith character of the string
int substring[] = new int[n + 1];
substring[0] = 0;
// Compute the values
int temp = n;
for (int i = 1; i <= n; i++)
{
// substring[i - 1] is added
// to store the cumulative sum
substring[i] = substring[i - 1] + temp;
temp--;
}
// Binary search to find the starting index
// of the kth sub-string
int l = 1;
int h = n;
int start = 0;
while (l <= h)
{
int m = (l + h) / 2;
if (substring[m] > k)
{
start = m;
h = m - 1;
}
else if (substring[m] < k)
{
l = m + 1;
}
else
{
start = m;
break;
}
}
// To store the ending index of
// the kth sub-string
int end = n - (substring[start] - k);
// Print the sub-string
for (int i = start - 1; i < end; i++)
{
System.out.print(str.charAt(i));
}
}
// Driver code
public static void main(String[] args)
{
String str = "abc";
int k = 4;
int n = str.length();
Printksubstring(str, n, k);
}
}
// This code has been contributed by 29AjayKumarPython3
# Python3 implementation of the approach
# Function to prints kth sub-string
def Printksubstring(str1, n, k):
# Total sub-strings possible
total = int((n * (n + 1)) / 2)
# If k is greater than total
# number of sub-strings
if (k > total):
print("-1")
return
# To store number of sub-strings starting
# with ith character of the string
substring = [0 for i in range(n + 1)]
substring[0] = 0
# Compute the values
temp = n
for i in range(1, n + 1, 1):
# substring[i - 1] is added
# to store the cumulative sum
substring[i] = substring[i - 1] + temp
temp -= 1
# Binary search to find the starting index
# of the kth sub-string
l = 1
h = n
start = 0
while (l <= h):
m = int((l + h) / 2)
if (substring[m] > k):
start = m
h = m - 1
elif (substring[m] < k):
l = m + 1
else:
start = m
break
# To store the ending index of
# the kth sub-string
end = n - (substring[start] - k)
# Print the sub-string
for i in range(start - 1, end):
print(str1[i], end = "")
# Driver code
if __name__ == '__main__':
str1 = "abc"
k = 4
n = len(str1)
Printksubstring(str1, n, k)
# This code is contributed by
# Surendra_GangwarC#
// C# implementation of the approach
using System;
class GFG
{
// Function to prints kth sub-string
static void Printksubstring(String str, int n, int k)
{
// Total sub-strings possible
int total = (n * (n + 1)) / 2;
// If k is greater than total
// number of sub-strings
if (k > total)
{
Console.Write("-1\n");
return;
}
// To store number of sub-strings starting
// with ith character of the string
int []substring = new int[n + 1];
substring[0] = 0;
// Compute the values
int temp = n;
for (int i = 1; i <= n; i++)
{
// substring[i - 1] is added
// to store the cumulative sum
substring[i] = substring[i - 1] + temp;
temp--;
}
// Binary search to find the starting index
// of the kth sub-string
int l = 1;
int h = n;
int start = 0;
while (l <= h)
{
int m = (l + h) / 2;
if (substring[m] > k)
{
start = m;
h = m - 1;
}
else if (substring[m] < k)
{
l = m + 1;
}
else
{
start = m;
break;
}
}
// To store the ending index of
// the kth sub-string
int end = n - (substring[start] - k);
// Print the sub-string
for (int i = start - 1; i < end; i++)
{
Console.Write(str[i]);
}
}
// Driver code
public static void Main(String[] args)
{
String str = "abc";
int k = 4;
int n = str.Length;
Printksubstring(str, n, k);
}
}
// This code contributed by Rajput-JiPHP
$total)
{
printf("-1\n");
return;
}
// To store number of sub-strings starting
// with ith character of the string
$substring = array();
$substring[0] = 0;
// Compute the values
$temp = $n;
for ($i = 1; $i <= $n; $i++)
{
// substring[i - 1] is added
// to store the cumulative sum
$substring[$i] = $substring[$i - 1] + $temp;
$temp--;
}
// Binary search to find the starting index
// of the kth sub-string
$l = 1;
$h = $n;
$start = 0;
while ($l <= $h)
{
$m = floor(($l + $h) / 2);
if ($substring[$m] > $k)
{
$start = $m;
$h = $m - 1;
}
else if ($substring[$m] < $k)
$l = $m + 1;
else
{
$start = $m;
break;
}
}
// To store the ending index of
// the kth sub-string
$end = $n - ($substring[$start] - $k);
// Print the sub-string
for ($i = $start - 1; $i < $end; $i++)
print($str[$i]);
}
// Driver code
$str = "abc";
$k = 4;
$n = strlen($str);
Printksubstring($str, $n, $k);
// This code is contributed by Ryuga
?>输出：
b