给定大小为N的字符串str ，任务是打印在最大可能分区之后形成的子字符串数，以使两个子字符串都不具有相同的字符。
例子：
Input : str = “ababcbacadefegdehijhklij”
Output : 3
Explanation:
Partitioning at the index 8 and at 15 produces three substrings “ababcbaca”, “defegde” and “hijhklij” such that none of them have a common character. So, the maximum partitions is 3.
Input: str = “aaa”
Output: 1
Explanation:
Since, the string consists of a single character, no partition can be performed.
方法：
- 从字符串末尾查找每个唯一字符的最后一个索引，并将其存储在映射中。
- 从索引0到最后一个索引遍历数组，并创建一个单独的变量来存储分区的结束索引。
- 遍历字符串中的每个变量，并检查分区的末端是否大于上一个末端，该末端由存储在映射中的str [i]的索引表示。如果是这样，请对其进行更新。
- 检查当前变量是否超过分区末尾，这意味着找到了第一个分区。将分区的末尾更新为max(k，mp [str [i]]) 。
- 遍历整个字符串，并使用类似的过程查找下一个分区，依此类推。
下面是上述方法的实现：
C++
// C++ program to find the
// maximum number of
// partitions possible such
// that no two substrings
// have common character
#include
using namespace std;
// Function to calculate and return
// the maximum number of partitions
int maximum_partition(string str)
{
// r: Stores the maximum number
// of partitions
// k: Stores the ending index
// of the partition
int i = 0, j = 0, k = 0;
int c = 0, r = 0;
// Stores the last index
// of every unique character
// of the string
unordered_map m;
// Traverse the string and
// store the last index
// of every character
for (i = str.length() - 1;
i >= 0;
i--) {
if (m[str[i]] == 0) {
m[str[i]] = i;
}
}
i = 0;
// Store the last index of the
// first character from map
k = m[str[i]];
for (i = 0; i < str.length(); i++) {
if (i <= k) {
c = c + 1;
// Update K to find
// the end of partition
k = max(k, m[str[i]]);
}
// Otherwise, the end of
// partition is found
else {
// Increment r
r = r + 1;
c = 1;
// Update k for the
// next partition
k = max(k, m[str[i]]);
}
}
// Add the last partition
if (c != 0) {
r = r + 1;
}
return r;
}
// Driver Program
int main()
{
string str
= "ababcbacadefegdehijhklij";
cout << maximum_partition(str);
}Java
// Java program to find the
// maximum number of
// partitions possible such
// that no two subStrings
// have common character
import java.util.*;
class GFG{
// Function to calculate and return
// the maximum number of partitions
static int maximum_partition(String str)
{
// r: Stores the maximum number
// of partitions
// k: Stores the ending index
// of the partition
int i = 0, j = 0, k = 0;
int c = 0, r = 0;
// Stores the last index
// of every unique character
// of the String
HashMap m = new HashMap<>();
// Traverse the String and
// store the last index
// of every character
for (i = str.length() - 1;
i >= 0; i--)
{
if (!m.containsKey(str.charAt(i)))
{
m.put(str.charAt(i), i);
}
}
i = 0;
// Store the last index of the
// first character from map
k = m.get(str.charAt(i));
for (i = 0; i < str.length(); i++)
{
if (i <= k)
{
c = c + 1;
// Update K to find
// the end of partition
k = Math.max(k, m.get(str.charAt(i)));
}
// Otherwise, the end of
// partition is found
else
{
// Increment r
r = r + 1;
c = 1;
// Update k for the
// next partition
k = Math.max(k, m.get(str.charAt(i)));
}
}
// Add the last
// partition
if (c != 0)
{
r = r + 1;
}
return r;
}
// Driver code
public static void main(String[] args)
{
String str = "ababcbacadefegdehijhklij";
System.out.print(maximum_partition(str));
}
}
// This code is contributed by Princi Singh Python 3
# Python3 program to find the
# maximum number of
# partitions possible such
# that no two substrings
# have common character
# Function to calculate and return
# the maximum number of partitions
def maximum_partition(strr):
# r: Stores the maximum number
# of partitions
# k: Stores the ending index
# of the partition
i = 0
j = 0
k = 0
c = 0
r = 0
# Stores the last index
# of every unique character
# of the strring
m = {}
# Traverse the and
# store the last index
# of every character
for i in range(len(strr) - 1, -1, -1):
if (strr[i] not in m):
m[strr[i]] = i
i = 0
# Store the last index of the
# first character from map
k = m[strr[i]]
for i in range(len(strr)):
if (i <= k):
c = c + 1
# Update K to find
# the end of partition
k = max(k, m[strr[i]])
# Otherwise, the end of
# partition is found
else:
# Increment r
r = r + 1
c = 1
# Update k for the
# next partition
k = max(k, m[strr[i]])
# Add the last partition
if (c != 0):
r = r + 1
return r
# Driver Code
if __name__ == '__main__':
strr = "ababcbacadefegdehijhklij"
print(maximum_partition(strr))
# This code is contributed by Mohit KumarC#
// C# program to find the
// maximum number of
// partitions possible such
// that no two subStrings
// have common character
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate and return
// the maximum number of partitions
static int maximum_partition(String str)
{
// r: Stores the maximum number
// of partitions
// k: Stores the ending index
// of the partition
int i = 0, j = 0, k = 0;
int c = 0, r = 0;
// Stores the last index
// of every unique character
// of the String
Dictionary m = new Dictionary();
// Traverse the String and
// store the last index
// of every character
for (i = str.Length - 1;
i >= 0; i--)
{
if (!m.ContainsKey(str[i]))
{
m.Add(str[i], i);
}
}
i = 0;
// Store the last index of the
// first character from map
k = m[str[i]];
for (i = 0; i < str.Length; i++)
{
if (i <= k)
{
c = c + 1;
// Update K to find
// the end of partition
k = Math.Max(k, m[str[i]]);
}
// Otherwise, the end of
// partition is found
else
{
// Increment r
r = r + 1;
c = 1;
// Update k for the
// next partition
k = Math.Max(k, m[str[i]]);
}
}
// Add the last
// partition
if (c != 0)
{
r = r + 1;
}
return r;
}
// Driver code
public static void Main(String[] args)
{
String str = "ababcbacadefegdehijhklij";
Console.Write(maximum_partition(str));
}
}
// This code is contributed by Amit Katiyar 输出：
3
时间复杂度： O(N)
辅助空间： O(1)