给定一个仅包含0和1的字符串。现在，您将获得N个非相交范围L，R(L <= R)，更具体地说是[L1，R1]，[L2，R2]，…，[LN，RN]，这些间隔中没有两个重叠-形式上，对于每个有效i，j，使得i！= j，Ri
- 所有N个给定范围之间的数字总和将为最大。
- 该字符串在字典上最大。字符串1100在字典上比字符串1001大。
例子：
Input
11100
2
2 3
5 5
Output
01101
First we put 1’s in position 2 and 3 then in 5 as
there are no 1’s left, the string formed is 01101.
Input
0000111
2
1 1
1 2
Output
1110000
In the above example we 1st put 1 in 1st and 2nd position then we have another ‘1’ left,
So, we use it to maximize the string lexicographically and we put it in the 3rd position and thus the rearrangement is complete.
方法
- 赋予第一优先权是使所有l和r之间的1的计数最大。我们计算数组中1的数目并存储在变量中。
- 输入后，我们将每个l和r的范围都更新为1，以首先标记要填充1的位置。
- 然后，我们获取数组的前缀和，以便获得第一个固定位置1的位置。然后，我们从左在该前缀和数组中运行一个循环。如果我们得到的任何头寸的值都大于1，则意味着该索引中有一个lr。我们继续将1放入这些索引中，直到1的计数变为零为止。
- 现在，在最大化操作完成之后，如果还剩1，那么我们开始进行字典词典最大化。如果发现索引值为0表示没有lr索引，则再次从前缀求和数组的左侧开始循环，然后将1放入该索引，然后继续进行直到所有剩余的1都被填充为止。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
void arrange(string s)
{
int cc = 0;
// Storing the count of 1's in the string
for (int i = 0; i < s.length(); i++)
{
if (s[i] == '1') cc++;
}
int a[s.length() + 1] = {0};
// Query of l and r
int qq[][2] = {{2, 3}, {5, 5}};
int n = sizeof(qq) / sizeof(qq[0]);
for (int i = 0; i < n; i++)
{
int l = qq[i][0], r = qq[i][1];
l--, r--;
a[l]++;
// Applying range update technique.
a[r + 1]--;
}
int len_a = sizeof(a) / sizeof(a[0]);
for (int i = 1; i < len_a; i++)
{
// Taking prefix sum to get the range update values
a[i] += a[i - 1];
}
// Final array which will store the arranged string
int zz[s.length()] = {0};
for (int i = 0; i < len_a - 1; i++)
{
if (a[i] > 0)
{
if (cc > 0)
{
zz[i] = 1;
cc--;
}
else
break;
}
if (cc == 0) break;
}
// if after maximizing the ranges any 1 is left
// then we maximize the string lexicographically.
if (cc > 0)
{
for (int i = 0; i < s.length(); i++)
{
if (zz[i] == 0)
{
zz[i] = 1;
cc--;
}
if (cc == 0) break;
}
}
for (int i = 0; i < s.length(); i++)
cout << zz[i];
cout << endl;
}
// Driver Code
int main()
{
string str = "11100";
arrange(str);
return 0;
}
// This code is contributed by
// sanjeev2552 Java
// Java implementation of the approach
class GFG
{
static void arrange(String s)
{
int cc = 0;
// Storing the count of 1's in the String
for (int i = 0; i < s.length(); i++)
{
if (s.charAt(i) == '1') cc++;
}
int []a = new int[s.length() + 1];
// Query of l and r
int qq[][] = {{2, 3}, {5, 5}};
int n = qq.length;
for (int i = 0; i < n; i++)
{
int l = qq[i][0], r = qq[i][1];
l--; r--;
a[l]++;
// Applying range update technique.
a[r + 1]--;
}
int len_a = a.length;
for (int i = 1; i < len_a; i++)
{
// Taking prefix sum to get the range update values
a[i] += a[i - 1];
}
// Final array which will store the arranged String
int []zz = new int[s.length()];
for (int i = 0; i < len_a - 1; i++)
{
if (a[i] > 0)
{
if (cc > 0)
{
zz[i] = 1;
cc--;
}
else
break;
}
if (cc == 0) break;
}
// if after maximizing the ranges any 1 is left
// then we maximize the String lexicographically.
if (cc > 0)
{
for (int i = 0; i < s.length(); i++)
{
if (zz[i] == 0)
{
zz[i] = 1;
cc--;
}
if (cc == 0) break;
}
}
for (int i = 0; i < s.length(); i++)
System.out.print(zz[i]);
System.out.println();
}
// Driver Code
public static void main(String[] args)
{
String str = "11100";
arrange(str);
}
}
// This code is contributed by Rajput-JiPython3
# Python implementation of the approach
def arrange(s):
cc = 0
# Storing the count of 1's in the string
for i in range(len(s)):
if(s[i]=="1"):
cc+= 1
a =[0]*(len(s)+1)
# Query of l and r
qq =[(2, 3), (5, 5)]
n = len(qq)
for i in range(n):
l, r = qq[i][0], qq[i][1]
l-= 1
r-= 1
a[l]+= 1
# Applying range update technique.
a[r + 1]-= 1
for i in range(1, len(a)):
# Taking prefix sum to get the range update values
a[i]= a[i]+a[i-1]
# Final array which will store the arranged string
zz =[0]*len(s)
for i in range(len(a)-1):
if(a[i]>0):
if(cc>0):
zz[i]= 1
cc-= 1
else:
break
if(cc == 0):
break
# if after maximizing the ranges any 1 is left
# then we maximize the string lexicographically.
if(cc>0):
for i in range(len(s)):
if(zz[i]== 0):
zz[i]= 1
cc-= 1
if(cc == 0):
break
print(*zz, sep ="")
str = "11100"
arrange(str)C#
// C# implementation of the approach
using System;
class GFG
{
static void arrange(String s)
{
int cc = 0;
// Storing the count of 1's in the String
for (int i = 0; i < s.Length; i++)
{
if (s[i] == '1') cc++;
}
int []a = new int[s.Length + 1];
// Query of l and r
int [,]qq = {{2, 3}, {5, 5}};
int n = qq.GetLength(0);
for (int i = 0; i < n; i++)
{
int l = qq[i, 0], r = qq[i, 1];
l--; r--;
a[l]++;
// Applying range update technique.
a[r + 1]--;
}
int len_a = a.Length;
for (int i = 1; i < len_a; i++)
{
// Taking prefix sum to get the range update values
a[i] += a[i - 1];
}
// Final array which will store the arranged String
int []zz = new int[s.Length];
for (int i = 0; i < len_a - 1; i++)
{
if (a[i] > 0)
{
if (cc > 0)
{
zz[i] = 1;
cc--;
}
else
break;
}
if (cc == 0) break;
}
// if after maximizing the ranges any 1 is left
// then we maximize the String lexicographically.
if (cc > 0)
{
for (int i = 0; i < s.Length; i++)
{
if (zz[i] == 0)
{
zz[i] = 1;
cc--;
}
if (cc == 0) break;
}
}
for (int i = 0; i < s.Length; i++)
Console.Write(zz[i]);
Console.WriteLine();
}
// Driver Code
public static void Main(String[] args)
{
String str = "11100";
arrange(str);
}
}
// This code is contributed by PrinciRaj1992输出：
01101
如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。