给定一个二进制字符串。我们需要找到最长平衡子串的长度。如果子串包含相等数量的 0 和 1，则它是平衡的。
例子：
Input : input = 110101010
Output : Length of longest balanced
sub string = 8
Input : input = 0000
Output : Length of longest balanced
sub string = 0一个简单的解决方案是使用两个嵌套循环来生成每个子字符串。第三个循环计算当前子字符串中 0 和 1 的数量。时间复杂度为 O(n 3 )
下面是上述方法的实现：
C++
// C++ program to find the length of
// the longest balanced substring
#include
using namespace std;
// Function to check if a string contains
// equal number of one and zeros or not
bool isValid(string p)
{
int n = p.length();
int c1 = 0, c0 = 0;
for(int i =0; i < n; i++)
{
if(p[i] == '0')
c0++;
if(p[i] == '1')
c1++;
}
return (c0 == c1) ? true : false;
}
// Function to find the length of
// the longest balanced substring
int longestSub(string s)
{
int max_len = 0;
int n = s.length();
for(int i = 0; i < n; i++)
{
for(int j = i; j < n; j++)
{
if(isValid(s.substr(i, j - i + 1)) && max_len < j - i + 1)
max_len = j - i + 1;
}
}
return max_len;
}
// Driver code
int main()
{
string s = "101001000";
// Function call
cout << longestSub(s);
return 0;
} Java
// Java program to find the length of
// the longest balanced substring
import java.io.*;
import java.util.*;
class GFG
{
// Function to check if a string contains
// equal number of one and zeros or not
public static boolean isValid(String p)
{
int n = p.length();
int c1 = 0, c0 = 0;
for (int i = 0; i < n; i++)
{
if(p.charAt(i) == '0')
{
c0++;
}
if(p.charAt(i) == '1')
{
c1++;
}
}
if(c0 == c1)
{
return true;
}
else
{
return false;
}
}
// Function to find the length of
// the longest balanced substring
public static int longestSub(String s)
{
int max_len = 0;
int n = s.length();
for(int i = 0; i < n; i++)
{
for(int j = i; j < n; j++)
{
if(isValid(s.substring(i, j + 1)) && max_len < j - i + 1)
{
max_len = j - i + 1;
}
}
}
return max_len;
}
// Driver code
public static void main (String[] args)
{
String s = "101001000";
// Function call
System.out.println(longestSub(s));
}
}
// This code is contributed by avanitrachhadiya2155Python3
# Python3 program to find the length of
# the longest balanced substring
# Function to check if a contains
# equal number of one and zeros or not
def isValid(p):
n = len(p)
c1 = 0
c0 = 0
for i in range(n):
if (p[i] == '0'):
c0 += 1
if (p[i] == '1'):
c1 += 1
if (c0 == c1):
return True
else:
return False
# Function to find the length of
# the longest balanced substring
def longestSub(s):
max_len = 0
n = len(s)
for i in range(n):
for j in range(i, n):
if (isValid(s[i : j - i + 1]) and
max_len < j - i + 1):
max_len = j - i + 1
return max_len
# Driver code
if __name__ == '__main__':
s = "101001000"
# Function call
print(longestSub(s))
# This code is contributed by mohit kumar 29C#
// C# program to find the length of
// the longest balanced substring
using System;
class GFG{
// Function to check if a string contains
// equal number of one and zeros or not
static bool isValid(string p)
{
int n = p.Length;
int c1 = 0, c0 = 0;
for(int i = 0; i < n; i++)
{
if (p[i] == '0')
{
c0++;
}
if (p[i] == '1')
{
c1++;
}
}
if (c0 == c1)
{
return true;
}
else
{
return false;
}
}
// Function to find the length of
// the longest balanced substring
public static int longestSub(string s)
{
int max_len = 0;
int n = s.Length;
for(int i = 0; i < n; i++)
{
for(int j = i; j < n; j++)
{
if (isValid(s.Substring(i, j - i + 1)) &&
max_len < j - i + 1)
{
max_len = j - i + 1;
}
}
}
return max_len;
}
// Driver code
static public void Main()
{
string s = "101001000";
// Function call
Console.WriteLine(longestSub(s));
}
}
// This code is contributed by rag2127Javascript
C++
// C++ for finding length of longest balanced
// substring
#include
using namespace std;
// Returns length of the longest substring
// with equal number of zeros and ones.
int stringLen(string str)
{
// Create a map to store differences
// between counts of 1s and 0s.
map m;
// Initially difference is 0.
m[0] = -1;
int count_0 = 0, count_1 = 0;
int res = 0;
for (int i=0; iJava
// Java Code for finding the length of
// the longest balanced substring
import java.io.*;
import java.util.*;
public class MAX_LEN_0_1 {
public static void main(String args[])throws IOException
{
String str = "101001000";
// Create a map to store differences
//between counts of 1s and 0s.
HashMap map = new HashMap();
// Initially difference is 0;
map. put(0, -1);
int res =0;
int count_0 = 0, count_1 = 0;
for(int i=0; iPython3
# Python3 code for finding length of
# longest balanced substring
# Returns length of the longest substring
# with equal number of zeros and ones.
def stringLen( str ):
# Create a python dictionary to store
# differences between counts of 1s and 0s.
m = dict()
# Initially difference is 0.
m[0] = -1
count_0 = 0
count_1 = 0
res = 0
for i in range(len(str)):
# Keeping track of counts of
# 0s and 1s.
if str[i] == '0':
count_0 += 1
else:
count_1 += 1
# If difference between current
# counts already exists, then
# substring between previous and
# current index has same no. of
# 0s and 1s. Update result if
# this substring is more than
# current result.
if m.get(count_1 - count_0):
res = max(res, i - m[count_1 - count_0])
# If current difference is
# seen first time.
else:
m[count_1 - count_0] = i
return res
# driver code
str = "101001000"
print("Length of longest balanced"
" sub string = ",stringLen(str))
# This code is contributed by "Sharad_Bhardwaj"
C#
// C# Code for finding the length of
// the longest balanced substring
using System;
using System.Collections.Generic;
class GFG
{
public static void Main(string[] args)
{
string str = "101001000";
// Create a map to store differences
//between counts of 1s and 0s.
Dictionary map = new Dictionary();
// Initially difference is 0;
map[0] = -1;
int res = 0;
int count_0 = 0, count_1 = 0;
for (int i = 0; i < str.Length;i++)
{
// Keep track of count of 0s and 1s
if (str[i] == '0')
{
count_0++;
}
else
{
count_1++;
}
// If difference between current counts
// already exists, then substring between
// previous and current index has same
// no. of 0s and 1s. Update result if this
// substring is more than current result.
if (map.ContainsKey(count_1 - count_0))
{
res = Math.Max(res, (i - map[count_1 -
count_0]));
}
// If the current difference is
// seen first time.
else
{
map[count_1 - count_0] = i;
}
}
Console.WriteLine("Length of longest balanced" +
" sub string = " + res);
}
}
// This code is contributed by Shrikant13
Javascript
输出：
6
一个有效的解决方案是使用散列。
1) 遍历字符串并分别跟踪 1 和 0 的计数为 count_1 和 count_0。
2）查看两个计数之间的当前差异之前是否出现过（我们使用散列来存储所有差异和出现差异的第一个索引）。如果是，则来自先前出现和当前索引的子串具有相同数量的 0 和 1。
下面是上述方法的实现。
C++
// C++ for finding length of longest balanced
// substring
#include
using namespace std;
// Returns length of the longest substring
// with equal number of zeros and ones.
int stringLen(string str)
{
// Create a map to store differences
// between counts of 1s and 0s.
map m;
// Initially difference is 0.
m[0] = -1;
int count_0 = 0, count_1 = 0;
int res = 0;
for (int i=0; iJava
// Java Code for finding the length of
// the longest balanced substring
import java.io.*;
import java.util.*;
public class MAX_LEN_0_1 {
public static void main(String args[])throws IOException
{
String str = "101001000";
// Create a map to store differences
//between counts of 1s and 0s.
HashMap map = new HashMap();
// Initially difference is 0;
map. put(0, -1);
int res =0;
int count_0 = 0, count_1 = 0;
for(int i=0; i蟒蛇3
# Python3 code for finding length of
# longest balanced substring
# Returns length of the longest substring
# with equal number of zeros and ones.
def stringLen( str ):
# Create a python dictionary to store
# differences between counts of 1s and 0s.
m = dict()
# Initially difference is 0.
m[0] = -1
count_0 = 0
count_1 = 0
res = 0
for i in range(len(str)):
# Keeping track of counts of
# 0s and 1s.
if str[i] == '0':
count_0 += 1
else:
count_1 += 1
# If difference between current
# counts already exists, then
# substring between previous and
# current index has same no. of
# 0s and 1s. Update result if
# this substring is more than
# current result.
if m.get(count_1 - count_0):
res = max(res, i - m[count_1 - count_0])
# If current difference is
# seen first time.
else:
m[count_1 - count_0] = i
return res
# driver code
str = "101001000"
print("Length of longest balanced"
" sub string = ",stringLen(str))
# This code is contributed by "Sharad_Bhardwaj"
C＃
// C# Code for finding the length of
// the longest balanced substring
using System;
using System.Collections.Generic;
class GFG
{
public static void Main(string[] args)
{
string str = "101001000";
// Create a map to store differences
//between counts of 1s and 0s.
Dictionary map = new Dictionary();
// Initially difference is 0;
map[0] = -1;
int res = 0;
int count_0 = 0, count_1 = 0;
for (int i = 0; i < str.Length;i++)
{
// Keep track of count of 0s and 1s
if (str[i] == '0')
{
count_0++;
}
else
{
count_1++;
}
// If difference between current counts
// already exists, then substring between
// previous and current index has same
// no. of 0s and 1s. Update result if this
// substring is more than current result.
if (map.ContainsKey(count_1 - count_0))
{
res = Math.Max(res, (i - map[count_1 -
count_0]));
}
// If the current difference is
// seen first time.
else
{
map[count_1 - count_0] = i;
}
}
Console.WriteLine("Length of longest balanced" +
" sub string = " + res);
}
}
// This code is contributed by Shrikant13
Javascript
输出：
Length of longest balanced sub string = 6
时间复杂度： O(n)
扩展问题：具有相同数量 0 和 1 的最大子数组
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