给定一个仅包含三个可能字符“ a”，“ b”或“ c”的字符串。该任务是与“A”，“b”或“c”来替换给定的字符串的字符仅使得存在的字符串中“A”，“b”和“c”的字符的数目相等。任务是最大程度地减少替换次数，并以最少的替换打印所有此类字符串按字典顺序最小的字符串。
如果无法获得这样的字符串，请打印-1。
例子：
Input : s = "bcabba"
Output : bcabca
Number of replacements done is 1 and this is
the lexicographically smallest possible
Input : "aaaaaa"
Output : aabbcc
Input : "aaaaa"
Output : -1
方法：
- 计算字符串中“ a”，“ b”和“ c”的数量。
- 如果它们的数量相等，则答案是相同的字符串。
- 如果字符串的长度不是3的倍数，则不可能。
- 首先，减少字符串中超过a的数目。
- 如果使用左侧的滑动窗口技术，如果将“ c”替换为“ a”，则将其替换为“ a”。
- 如果在索引处没有“ c”的情况下使用左向左滑动窗口技术，则将“ b”替换为“ a”。
- 其次，通过使用滑动窗口从前面替换“ c”来减少字符串中超过b的数目。
- 第三，通过使用滑动窗口减少背面多余的“ a”的数量，减少超过c的数量。
- 第四，通过减少后面多余的“ a”的数量来减少字符串中超过b的数量。
- 第五，通过减少背面多余的“ b”的数量来减少超过c的数量。
为了得到字典上最小的字符串，我们不断从后面进行替换。
下面是上述方法的实现：
C++
// CPP program to Minimize the number of
// replacements to get a string with same
// number of ‘a’, ‘b’ and ‘c’ in it
#include
using namespace std;
// Function to count numbers
string lexoSmallest(string s, int n)
{
// Count the number of 'a', 'b' and
// 'c' in string
int ca = 0, cb = 0, cc = 0;
for (int i = 0; i < n; i++) {
if (s[i] == 'a')
ca++;
else if (s[i] == 'b')
cb++;
else
cc++;
}
// If equal previously
if (ca == cb && cb == cc) {
return s;
}
int cnt = n / 3;
// If not a multiple of 3
if (cnt * 3 != n) {
return "-1";
}
int i = 0;
// Increase the number of a's by
// removing extra 'b' and ;c;
while (ca < cnt && i < n) {
// Check if it is 'b' and it more
// than n/3
if (s[i] == 'b' && cb > cnt) {
cb--;
s[i] = 'a';
ca++;
}
// Check if it is 'c' and it
// more than n/3
else if (s[i] == 'c' && cc > cnt) {
cc--;
s[i] = 'a';
ca++;
}
i++;
}
i = 0;
// Increase the number of b's by
// removing extra 'c'
while (cb < cnt && i < n) {
// Check if it is 'c' and it more
// than n/3
if (s[i] == 'c' && cc > cnt) {
cc--;
s[i] = '1';
cb++;
}
i++;
}
i = n - 1;
// Increase the number of c's from back
while (cc < cnt && i >= 0) {
// Check if it is 'a' and it more
// than n/3
if (s[i] == 'a' && ca > cnt) {
ca--;
s[i] = 'c';
cc++;
}
i--;
}
i = n - 1;
// Increase the number of b's from back
while (cb < cnt && i >= 0) {
// Check if it is 'a' and it more
// than n/3
if (s[i] == 'a' && ca > cnt) {
ca--;
s[i] = 'b';
cb++;
}
i--;
}
i = n - 1;
// Increase the number of c's from back
while (cc < cnt && i >= 0) {
// Check if it is 'b' and it more
// than n/3
if (s[i] == 'b' && cb > cnt) {
cb--;
s[i] = 'c';
cc++;
}
i--;
}
return s;
}
// Driver Code
int main()
{
string s = "aaaaaa";
int n = s.size();
cout << lexoSmallest(s, n);
return 0;
} Python3
# Python3 program to Minimize the number of
# replacements to get a string with same
# number of 'a', 'b' and 'c' in it
# Function to count numbers
def lexoSmallest(s, n):
# Count the number of 'a', 'b' and
# 'c' in string
ca = 0
cb = 0
cc = 0
for i in range(n):
if (s[i] == 'a'):
ca += 1
elif (s[i] == 'b'):
cb += 1
else:
cc += 1
# If equal previously
if (ca == cb and cb == cc):
return s
cnt = n // 3
# If not a multiple of 3
if (cnt * 3 != n):
return "-1"
i = 0
# Increase the number of a's by
# removing extra 'b' and c
while (ca < cnt and i < n):
# Check if it is 'b' and it more
# than n/3
if (s[i] == 'b' and cb > cnt) :
cb -= 1
s[i] = 'a'
ca += 1
# Check if it is 'c' and it
# more than n/3
elif (s[i] == 'c' and cc > cnt):
cc -= 1
s[i] = 'a'
ca += 1
i += 1
i = 0
# Increase the number of b's by
# removing extra 'c'
while (cb < cnt and i < n):
# Check if it is 'c' and it more
# than n/3
if (s[i] == 'c' and cc > cnt):
cc -= 1
s[i] = '1'
cb += 1
i += 1
i = n - 1
# Increase the number of c's from back
while (cc < cnt and i >= 0):
# Check if it is 'a' and it more
# than n/3
if (s[i] == 'a' and ca > cnt):
ca -= 1
s[i] = 'c'
cc += 1
i -= 1
i = n - 1
# Increase the number of b's from back
while (cb < cnt and i >= 0):
# Check if it is 'a' and it more
# than n/3
if (s[i] == 'a' and ca > cnt):
ca -= 1
s[i] = 'b'
cb += 1
i -= 1
i = n - 1
# Increase the number of c's from back
while (cc < cnt and i >= 0):
# Check if it is 'b' and it more
# than n/3
if (s[i] == 'b' and cb > cnt):
cb -= 1
s[i] = 'c'
cc += 1
i -= 1
return s
# Driver Code
s = "aaaaaa"
n = len(s)
print(*lexoSmallest(list(s), n),sep="")
# This code is contributed by shivanisinghss2110C#
// C# program to minimize the number of
// replacements to get a string with same
// number of ‘a’, ‘b’ and ‘c’ in it
using System;
using System.Text;
class GFG{
// Function to count numbers
static string lexoSmallest(string str, int n)
{
// Count the number of 'a', 'b' and
// 'c' in string
int ca = 0, cb = 0, cc = 0;
StringBuilder s = new StringBuilder(str);
for(int j = 0; j < n; j++)
{
if (s[j] == 'a')
ca++;
else if (s[j] == 'b')
cb++;
else
cc++;
}
// If equal previously
if (ca == cb && cb == cc)
{
return s.ToString();
}
int cnt = n / 3;
// If not a multiple of 3
if (cnt * 3 != n)
{
return "-1";
}
int i = 0;
// Increase the number of a's by
// removing extra 'b' and ;c;
while (ca < cnt && i < n)
{
// Check if it is 'b' and it more
// than n/3
if (s[i] == 'b' && cb > cnt)
{
cb--;
s[i] = 'a';
ca++;
}
// Check if it is 'c' and it
// more than n/3
else if (s[i] == 'c' && cc > cnt)
{
cc--;
s[i] = 'a';
ca++;
}
i++;
}
i = 0;
// Increase the number of b's by
// removing extra 'c'
while (cb < cnt && i < n)
{
// Check if it is 'c' and it more
// than n/3
if (s[i] == 'c' && cc > cnt)
{
cc--;
s[i] = '1';
cb++;
}
i++;
}
i = n - 1;
// Increase the number of c's from back
while (cc < cnt && i >= 0)
{
// Check if it is 'a' and it more
// than n/3
if (s[i] == 'a' && ca > cnt)
{
ca--;
s[i] = 'c';
cc++;
}
i--;
}
i = n - 1;
// Increase the number of b's from back
while (cb < cnt && i >= 0)
{
// Check if it is 'a' and it more
// than n/3
if (s[i] == 'a' && ca > cnt)
{
ca--;
s[i] = 'b';
cb++;
}
i--;
}
i = n - 1;
// Increase the number of c's from back
while (cc < cnt && i >= 0)
{
// Check if it is 'b' and it more
// than n/3
if (s[i] == 'b' && cb > cnt)
{
cb--;
s[i] = 'c';
cc++;
}
i--;
}
return s.ToString();
}
// Driver Code
public static void Main(string[] args)
{
string s = "aaaaaa";
int n = s.Length;
Console.Write(lexoSmallest(s, n));
}
}
// This code is contributed by rutvik_56PHP
$cnt)
{
$cb--;
$s[$i] = 'a';
$ca++;
}
// Check if it is 'c' and it
// more than n/3
else if ($s[$i] == 'c' && $cc > $cnt)
{
$cc--;
$s[$i] = 'a';
$ca++;
}
$i++;
}
$i = 0;
// Increase the number of b's by
// removing extra 'c'
while ($cb < $cnt && $i < $n)
{
// Check if it is 'c' and it more
// than n/3
if ($s[$i] == 'c' && $cc > $cnt)
{
$cc--;
$s[$i] = '1';
$cb++;
}
$i++;
}
$i = $n - 1;
// Increase the number of c's from back
while ($cc < $cnt && $i >= 0)
{
// Check if it is 'a' and it is
// more than n/3
if ($s[$i] == 'a' && $ca > $cnt)
{
$ca--;
$s[$i] = 'c';
$cc++;
}
$i--;
}
$i = $n - 1;
// Increase the number of b's from back
while ($cb < $cnt && $i >= 0)
{
// Check if it is 'a' and it is
// more than n/3
if ($s[$i] == 'a' && $ca > $cnt)
{
$ca--;
$s[$i] = 'b';
$cb++;
}
$i--;
}
$i = $n - 1;
// Increase the number of c's from back
while ($cc < $cnt && $i >= 0)
{
// Check if it is 'b' and it more
// than n/3
if ($s[$i] == 'b' && $cb > $cnt)
{
$cb--;
$s[$i] = 'c';
$cc++;
}
$i--;
}
return $s;
}
// Driver Code
$s = "aaaaaa";
$n = strlen($s);
echo lexoSmallest($s, $n);
// This code is contributed by Ryuga.
?>
Output:aabbcc
Time Complexity: O(N*6)In case you wish to attend live classes with industry experts, please refer Geeks Classes Live and Geeks Classes Live USA