给定一个偶数长度的二进制字符串，其中 0 和 1 的数量相等。使字符串交替的最小交换次数是多少。如果没有两个连续元素相等，则二进制字符串是交替的。
例子：
Input : 000111
Output : 1
Explanation : Swap index 2 and index 5 to get 010101
Input : 1010
Output : 0我们可能在第一个位置得到 1 或在第一个位置得到零。我们考虑两种情况并找到两种情况中的最小值。请注意，假设字符串的 1 和 0 的数量相等，并且字符串的长度是偶数。
1. 计算字符串奇数位置和偶数位置的零个数。让它们的计数分别为odd_0 和even_0。
2. 计算字符串奇数和偶数位置的个数。让它们的计数分别为odd_1 和even_1。
3. 我们总是将 1 与 0 交换（永远不会将 1 与 1 或 0 与 0 交换）。所以我们只检查我们的交替字符串是否以 0 开头，则交换次数为 min(even_0,odd_1)，如果我们的交替字符串以 1 开头，则交换次数为 min(even_1,odd_0)。答案是这两个中的 min 。
下面是上述方法的实现：
C++
// CPP implementation of the approach
#include
using namespace std;
// returns the minimum number of swaps
// of a binary string
// passed as the argument
// to make it alternating
int countMinSwaps(string st)
{
int min_swaps = 0;
// counts number of zeroes at odd
// and even positions
int odd_0 = 0, even_0 = 0;
// counts number of ones at odd
// and even positions
int odd_1 = 0, even_1 = 0;
int n = st.length();
for (int i = 0; i < n; i++) {
if (i % 2 == 0) {
if (st[i] == '1')
even_1++;
else
even_0++;
}
else {
if (st[i] == '1')
odd_1++;
else
odd_0++;
}
}
// alternating string starts with 0
int cnt_swaps_1 = min(even_0, odd_1);
// alternating string starts with 1
int cnt_swaps_2 = min(even_1, odd_0);
// calculates the minimum number of swaps
return min(cnt_swaps_1, cnt_swaps_2);
}
// Driver code
int main()
{
string st = "000111";
cout< Java
// Java implementation of the approach
class GFG {
// returns the minimum number of swaps
// of a binary string
// passed as the argument
// to make it alternating
static int countMinSwaps(String st)
{
int min_swaps = 0;
// counts number of zeroes at odd
// and even positions
int odd_0 = 0, even_0 = 0;
// counts number of ones at odd
// and even positions
int odd_1 = 0, even_1 = 0;
int n = st.length();
for (int i = 0; i < n; i++) {
if (i % 2 == 0) {
if (st.charAt(i) == '1')
even_1++;
else
even_0++;
}
else {
if (st.charAt(i) == '1')
odd_1++;
else
odd_0++;
}
}
// alternating string starts with 0
int cnt_swaps_1 = Math.min(even_0, odd_1);
// alternating string starts with 1
int cnt_swaps_2 = Math.min(even_1, odd_0);
// calculates the minimum number of swaps
return Math.min(cnt_swaps_1, cnt_swaps_2);
}
// Driver code
public static void main(String[] args)
{
String st = "000111";
System.out.println(countMinSwaps(st));
}
}Python 3
# Python3 implementation of the
# above approach
# returns the minimum number of swaps
# of a binary string
# passed as the argument
# to make it alternating
def countMinSwaps(st) :
min_swaps = 0
# counts number of zeroes at odd
# and even positions
odd_0, even_0 = 0, 0
# counts number of ones at odd
# and even positions
odd_1, even_1 = 0, 0
n = len(st)
for i in range(0, n) :
if i % 2 == 0 :
if st[i] == "1" :
even_1 += 1
else :
even_0 += 1
else :
if st[i] == "1" :
odd_1 += 1
else :
odd_0 += 1
# alternating string starts with 0
cnt_swaps_1 = min(even_0, odd_1)
# alternating string starts with 1
cnt_swaps_2 = min(even_1, odd_0)
# calculates the minimum number of swaps
return min(cnt_swaps_1, cnt_swaps_2)
# Driver code
if __name__ == "__main__" :
st = "000111"
# Function call
print(countMinSwaps(st))
# This code is contributed by
# ANKITRAI1C#
// C# implementation of the approach
using System;
public class GFG
{
// returns the minimum number of swaps
// of a binary string
// passed as the argument
// to make it alternating
public static int countMinSwaps(string st)
{
int min_swaps = 0;
// counts number of zeroes at odd
// and even positions
int odd_0 = 0, even_0 = 0;
// counts number of ones at odd
// and even positions
int odd_1 = 0, even_1 = 0;
int n = st.Length;
for (int i = 0; i < n; i++)
{
if (i % 2 == 0)
{
if (st[i] == '1')
{
even_1++;
}
else
{
even_0++;
}
}
else
{
if (st[i] == '1')
{
odd_1++;
}
else
{
odd_0++;
}
}
}
// alternating string starts with 0
int cnt_swaps_1 = Math.Min(even_0, odd_1);
// alternating string starts with 1
int cnt_swaps_2 = Math.Min(even_1, odd_0);
// calculates the minimum number of swaps
return Math.Min(cnt_swaps_1, cnt_swaps_2);
}
// Driver code
public static void Main(string[] args)
{
string st = "000111";
Console.WriteLine(countMinSwaps(st));
}
}
// This code is contributed by Shrikant13PHP
Javascript
C++
// C++ implementation of the above approach
#include
using namespace std;
// function to count minimum swaps
// required to make binary string
// alternating
int countMinSwaps(string s)
{
int N = s.size();
// stores total number of ones
int one = 0;
// stores total number of zeroes
int zero = 0;
for (int i = 0; i < N; i++) {
if (s[i] == '1')
one++;
else
zero++;
}
// checking impossible condition
if (one > zero + 1 || zero > one + 1)
return -1;
// odd length string
if (N % 2) {
// number of even positions
int num = (N + 1) / 2;
// stores number of zeroes and
// ones at even positions
int one_even = 0, zero_even = 0;
for (int i = 0; i < N; i++) {
if (i % 2 == 0) {
if (s[i] == '1')
one_even++;
else
zero_even++;
}
}
if (one > zero)
return num - one_even;
else
return num - zero_even;
}
// even length string
else {
// stores number of ones at odd
// and even position respectively
int one_odd = 0, one_even = 0;
for (int i = 0; i < N; i++) {
if (s[i] == '1') {
if (i % 2)
one_odd++;
else
one_even++;
}
}
return min(N / 2 - one_odd, N / 2 - one_even);
}
}
// Driver code
int main()
{
string s = "111000";
cout << countMinSwaps(s);
return 0;
} 输出：
1奇偶长度字符串：
令字符串长度为N。
在这种方法中，我们将考虑三种情况：
1.当1的总数>零的总数+1或零的总数>1的总数+1时，答案是不可能的。
2. 字符串是偶数长度：
我们将计算奇数位置（ one_odd ）和偶数位置（ one_even ）的个数，然后答案是 min （ N/2 – one_odd ，N/2 – one_even ）
3. 字符串是奇数长度：
这里我们考虑两种情况：
i) 1 的总数 > 0 的总数（然后我们将 1 放在偶数位置）所以，答案是（ (N + 1) / 2 – 偶数位置的 1 数）。
ii) 零的总数 > 一的总数（然后我们在偶数位置放了零）所以，答案是（（N + 1）/ 2 – 偶数位置的零数）。
C++
// C++ implementation of the above approach
#include
using namespace std;
// function to count minimum swaps
// required to make binary string
// alternating
int countMinSwaps(string s)
{
int N = s.size();
// stores total number of ones
int one = 0;
// stores total number of zeroes
int zero = 0;
for (int i = 0; i < N; i++) {
if (s[i] == '1')
one++;
else
zero++;
}
// checking impossible condition
if (one > zero + 1 || zero > one + 1)
return -1;
// odd length string
if (N % 2) {
// number of even positions
int num = (N + 1) / 2;
// stores number of zeroes and
// ones at even positions
int one_even = 0, zero_even = 0;
for (int i = 0; i < N; i++) {
if (i % 2 == 0) {
if (s[i] == '1')
one_even++;
else
zero_even++;
}
}
if (one > zero)
return num - one_even;
else
return num - zero_even;
}
// even length string
else {
// stores number of ones at odd
// and even position respectively
int one_odd = 0, one_even = 0;
for (int i = 0; i < N; i++) {
if (s[i] == '1') {
if (i % 2)
one_odd++;
else
one_even++;
}
}
return min(N / 2 - one_odd, N / 2 - one_even);
}
}
// Driver code
int main()
{
string s = "111000";
cout << countMinSwaps(s);
return 0;
}
输出
1时间复杂度： O（字符串长度）
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