给定一个字符串str，任务是打印最大数目的字符，该最大数目大于字符串的任何排列中其左字符和右字符。
例子：
Input: str = “abc”
Output: 1
Permutations of the string with the count of maximal character in each string are:
abc – 0
acb – 1 Here a < c > b
bac – 0
bca – 1 Here b < c > a
cab – 0
cba – 0
Input: str = “geeks”
Output: 2
The string will be “egesk”
观察结果：
- 如果字符串的长度小于3，则答案将为0，因为不可能满足给定条件的排列。
- 如果给定的字符串的长度大于或等于3，那么假定得到的字符串中的每个其他字符是最大的字符，即在任意两个连续的最大字符之间的一个字符(否则就删除所有，但最低并将其添加到字符串的末尾)。
- 为简单起见，假设此数字为奇数。然后，理想情况下，字符串可以在偶数位置(即最多(n-1)/ 2)处具有最大字符，其中n是给定字符串的长度，而其余字符在奇数位置。
- 首先按升序排列所有字符，将前半个字符放在奇数位置，然后用其余字符填充其余的偶数位置。
通过这种方式，在偶数位置的所有字符将这些字符从字符在左边和右边的位置是小，如果没有小字符的频率太高，开始从一个奇怪的位置放置字符，请继续将相同的字符为偶数位置，并最终到达我们开始放置该字符的奇数位置旁边的位置。在此，如果字符串中某个较小字符的频率太高，则最大字符将始终小于(n-1)/ 2。
方法：
- 计算给定字符串中每个字符的频率。
- 检查具有最大频率的字符。
- 如果最大频率元素是给定字符串的最小元素，则将标志标记为0，否则将标志的值标记为等于1。
- 答案将是((n – 1)/ 2，n – max_freq –标志)中的最小值。
下面是上述方法的实现：
C++
// C++ program to find maximum count
// of such characters which are greater
// its left and right character in
// any permutation of the string
#include
using namespace std;
// function to find maximum maximal character in the string
int solve(int freq[])
{
// to store sum of all frequency
int n = 0;
// to store maximum frequency
int max_freq = 0;
// frequency of the smallest element
int first;
// to check if the smallest
// element have amximum frequqncy or not
int flag = 0;
// Iterate in the string and count frequency
for (int i = 0; i < 26; i++) {
n += freq[i];
// to store frequency of smallest element
if (freq[i] != 0 && flag == 0) {
first = freq[i];
flag = 1;
}
// to store maximum frequency
if (max_freq < freq[i])
max_freq = freq[i];
}
// if sum of frequency of all element if 0
if (n == 0)
return 0;
// if frequency of smallest character
// if largest frequency
if (first != max_freq)
flag = 1;
else
flag = 0;
return min((n - 1) / 2, n - max_freq - flag);
}
// Function that counts the frequency of
// each element
void solve(string s)
{
// array to store the frequency of each character
int freq[26];
// initialize frequency of all character with 0
memset(freq, 0, sizeof(freq));
// loop to calculate frequqncy of
// each character in the given string
for (int i = 0; i < s.length(); i++) {
freq[s[i] - 'a']++;
}
cout << solve(freq);
}
// Driver Code
int main()
{
string s = "geeks";
solve(s);
return 0;
} Java
// Java program to find maximum count
// of such characters which are greater
// its left and right character in
// any permutation of the string only three characters
class GFG {
// function to find maximum maximal character in the string
static int solve(int freq[]) {
// to store sum of all frequency
int n = 0;
// to store maximum frequency
int max_freq = 0;
// frequency of the smallest element
int first = 0;
// to check if the smallest
// element have amximum frequqncy or not
int flag = 0;
// Iterate in the string and count frequency
for (int i = 0; i < 26; i++) {
n += freq[i];
// to store frequency of smallest element
if (freq[i] != 0 && flag == 0) {
first = freq[i];
flag = 1;
}
// to store maximum frequency
if (max_freq < freq[i]) {
max_freq = freq[i];
}
}
// if sum of frequency of all element if 0
if (n == 0) {
return 0;
}
// if frequency of smallest character
// if largest frequency
if (first != max_freq) {
flag = 1;
} else {
flag = 0;
}
return Math.min((n - 1) / 2, n - max_freq - flag);
}
// Function that counts the frequency of
// each element
static void solve(String s) {
// array to store the frequency of each character
int freq[] = new int[26];
// loop to calculate frequqncy of
// each character in the given string
for (int i = 0; i < s.length(); i++) {
freq[s.charAt(i) - 'a']++;
}
System.out.println(solve(freq));
}
// Driver Code
public static void main(String[] args) {
String s = "geeks";
solve(s);
}
}
// This code is contributed by 29AjayKumarPython3
# Python3 program to find maximum
# count of such characters which
# are greater its left and right
# character in any permutation
# of the string
# function to find maximum maximal
# character in the string
def Solve_1(freq):
# to store sum of all frequency
n = 0
# to store maximum frequency
max_freq = 0
# to check if the smallest
# element have amximum
# frequqncy or not
flag = 0
# Iterate in the string
# and count frequency
for i in range(26) :
n += freq[i]
# to store frequency of
# smallest element
if (freq[i] != 0 and flag == 0) :
first = freq[i]
flag = 1
# to store maximum frequency
if (max_freq < freq[i]):
max_freq = freq[i]
# if sum of frequency of
# all element if 0
if (n == 0):
return 0
# if frequency of smallest character
# if largest frequency
if (first != max_freq):
flag = 1
else:
flag = 0
return min((n - 1) // 2, n - max_freq - flag)
# Function that counts the
# frequency of each element
def solve(s):
# array to store the frequency
# of each character initialize
# frequency of all character with 0
freq = [0] * 26
# loop to calculate frequqncy of
# each character in the given string
for i in range(len(s)):
freq[ord(s[i]) - ord('a')] += 1
print(Solve_1(freq))
# Driver Code
if __name__ == "__main__":
s = "geeks"
solve(s)
# This code is contributed
# by ChitraNayalC#
// C# program to find maximum count
// of such characters which are greater
// its left and right character in
// any permutation of the string only three characters
using System;
public class GFG{
//JAVA program to find maximum count
// of such characters which are greater
// its left and right character in
// any permutation of the string only three characters
// function to find maximum maximal character in the string
static int solve(int []freq) {
// to store sum of all frequency
int n = 0;
// to store maximum frequency
int max_freq = 0;
// frequency of the smallest element
int first = 0;
// to check if the smallest
// element have amximum frequqncy or not
int flag = 0;
// Iterate in the string and count frequency
for (int i = 0; i < 26; i++) {
n += freq[i];
// to store frequency of smallest element
if (freq[i] != 0 && flag == 0) {
first = freq[i];
flag = 1;
}
// to store maximum frequency
if (max_freq < freq[i]) {
max_freq = freq[i];
}
}
// if sum of frequency of all element if 0
if (n == 0) {
return 0;
}
// if frequency of smallest character
// if largest frequency
if (first != max_freq) {
flag = 1;
} else {
flag = 0;
}
return Math.Min((n - 1) / 2, n - max_freq - flag);
}
// Function that counts the frequency of
// each element
static void solve(String s) {
// array to store the frequency of each character
int []freq = new int[26];
// loop to calculate frequqncy of
// each character in the given string
for (int i = 0; i < s.Length; i++) {
freq[s[i] - 'a']++;
}
Console.Write(solve(freq));
}
// Driver Code
public static void Main() {
String s = "geeks";
solve(s);
}
}
// This code is contributed by Rajput-JIPHP
输出：
2