给定一个由小写英文字母和整数K组成的字符串S ，任务是为 S 中的每个不同字符找到恰好包含该字符K次的最大索引。如果不存在这样的字符，则打印 -1。按字典顺序打印结果。
注意：考虑 S 中基于 0 的索引。
例子：
Input: S = “cbaabaacbcd”, K = 2
Output: { {a 4}, {b 7}, {c 8} }
Explanation:
For ‘a’, the largest index having 2 a’s is “cbaab”.
For ‘b’, the largest index having 2 b’s is “cbaabaac”.
For ‘c’, the largest index having 2 c’s is “cbaabaacb”.
For ‘d’, the is no index up to which we have 2 d’s
Input: P = “acbacbacbaba”, K = 3
Output: { {a 8}, {b 9}, {c 11} }
方法：思路是先找出字符串S 中所有不同的字符，然后对于每个小写英文字符，检查它是否存在于 S 中，并从 S 的开头运行一个 for 循环并保持该字符的计数发生到现在。当计数变得等于 K 时，相应地更新索引答案。最后，在向量结果中附加这个字符及其对应的索引。
下面是上述方法的实现。
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to find largest index for each
// distinct character occuring exactly K times.
void maxSubstring(string& S, int K, int N)
{
// finding all characters present in S
int freq[26];
memset(freq, 0, sizeof freq);
// Finding all distinct characters in S
for (int i = 0; i < N; ++i) {
freq[S[i] - 'a'] = 1;
}
// vector to store result for each character
vector > answer;
// loop through each lower case English character
for (int i = 0; i < 26; ++i) {
// if current character is absent in s
if (freq[i] == 0)
continue;
// getting current character
char ch = i + 97;
// finding count of character ch in S
int count = 0;
// to store max Index encountred so far
int index = -1;
for (int j = 0; j < N; ++j) {
if (S[j] == ch)
count++;
if (count == K)
index = j;
}
answer.push_back({ ch, index });
}
int flag = 0;
// printing required result
for (int i = 0; i < (int)answer.size(); ++i) {
if (answer[i].second > -1) {
flag = 1;
cout << answer[i].first << " "
<< answer[i].second << endl;
}
}
// If no such character exists, print -1
if (flag == 0)
cout << "-1" << endl;
}
// Driver code
int main()
{
string S = "cbaabaacbcd";
int K = 2;
int N = S.length();
maxSubstring(S, K, N);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG{
static class pair
{ char first;
int second;
public pair(char first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find largest
// index for each distinct
// character occuring exactly K times.
static void maxSubString(char [] S,
int K, int N)
{
// finding all characters present in S
int []freq = new int[26];
// Finding all distinct characters in S
for (int i = 0; i < N; ++i)
{
freq[S[i] - 'a'] = 1;
}
// vector to store result for each character
Vector answer = new Vector();
// loop through each lower case English character
for (int i = 0; i < 26; ++i)
{
// if current character is absent in s
if (freq[i] == 0)
continue;
// getting current character
char ch = (char) (i + 97);
// finding count of character ch in S
int count = 0;
// to store max Index encountred so far
int index = -1;
for (int j = 0; j < N; ++j)
{
if (S[j] == ch)
count++;
if (count == K)
index = j;
}
answer.add(new pair(ch, index ));
}
int flag = 0;
// printing required result
for (int i = 0; i < (int)answer.size(); ++i)
{
if (answer.get(i).second > -1)
{
flag = 1;
System.out.print(answer.get(i).first + " " +
answer.get(i).second + "\n");
}
}
// If no such character exists, print -1
if (flag == 0)
System.out.print("-1" + "\n");
}
// Driver code
public static void main(String[] args)
{
String S = "cbaabaacbcd";
int K = 2;
int N = S.length();
maxSubString(S.toCharArray(), K, N);
}
}
// This code is contributed by shikhasingrajput Python3
# Python3 implementation of the approach
# Function to find largest index for each
# distinct character occuring exactly K times.
def maxSubstring(S, K, N):
# Finding all characters present in S
freq = [0 for i in range(26)]
# Finding all distinct characters in S
for i in range(N):
freq[ord(S[i]) - 97] = 1
# To store result for each character
answer = []
# Loop through each lower
# case English character
for i in range(26):
# If current character is absent in s
if (freq[i] == 0):
continue
# Getting current character
ch = chr(i + 97)
# Finding count of character ch in S
count = 0
# To store max Index encountred so far
index = -1
for j in range(N):
if (S[j] == ch):
count += 1
if (count == K):
index = j
answer.append([ch, index])
flag = 0
# Printing required result
for i in range(len(answer)):
if (answer[i][1] > -1):
flag = 1
print(answer[i][0],
answer[i][1])
# If no such character exists,
# print -1
if (flag == 0):
print("-1")
# Driver code
if __name__ == '__main__':
S = "cbaabaacbcd"
K = 2
N = len(S)
maxSubstring(S, K, N)
# This code is contributed by Surendra_GangwarC#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG{
class pair
{
public char first;
public int second;
public pair(char first, int second)
{
this.first = first;
this.second = second;
}
}
// Function to find largest
// index for each distinct
// character occuring exactly K times.
static void maxSubString(char [] S,
int K, int N)
{
// Finding all characters present in S
int []freq = new int[26];
// Finding all distinct characters in S
for(int i = 0; i < N; ++i)
{
freq[S[i] - 'a'] = 1;
}
// To store result for each character
List answer = new List();
// Loop through each lower case
// English character
for(int i = 0; i < 26; ++i)
{
// If current character is absent in s
if (freq[i] == 0)
continue;
// Getting current character
char ch = (char)(i + 97);
// Finding count of character ch in S
int count = 0;
// To store max Index encountred so far
int index = -1;
for(int j = 0; j < N; ++j)
{
if (S[j] == ch)
count++;
if (count == K)
index = j;
}
answer.Add(new pair(ch, index));
}
int flag = 0;
// Printing required result
for(int i = 0; i < (int)answer.Count; ++i)
{
if (answer[i].second > -1)
{
flag = 1;
Console.Write(answer[i].first + " " +
answer[i].second + "\n");
}
}
// If no such character exists, print -1
if (flag == 0)
Console.Write("-1" + "\n");
}
// Driver code
public static void Main(String[] args)
{
String S = "cbaabaacbcd";
int K = 2;
int N = S.Length;
maxSubString(S.ToCharArray(), K, N);
}
}
// This code is contributed by Amit Katiyar 输出：
a 4
b 7
c 8
时间复杂度： O(26 * N)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live