要从给定的二进制字符串中删除以使所有 1 连续出现的最小 0 计数
给定一个大小为N的二进制字符串S ，任务是找到必须从字符串S中删除的最小0数量，以使所有1连续出现。
例子：
Input: S = “010001011”
Output: 4
Explanation:
Removing the characters { S[2], S[3], S[4], S[6] } from the string S modifies the string S to “01111”.
Therefore, the required output is 4.
Input: S = “011110000”
Output: 0
Explanation:
All 1s in S already group together, therefore, the required output is 0.
方法：可以通过观察以下事实来解决给定的问题：删除字符串中的前导和结束0不会最小化必须删除的0的数量。因此，想法是在字符串S中找到第一次和最后一次出现的1 ，并找到它们之间的0的计数。请按照以下步骤解决问题：
- 从左到右遍历字符串S的字符，并找到给定字符串X中第一次出现1的索引。
- 从右到左遍历字符串，找到给定字符串Y中最后一次出现1的索引。
- 完成上述步骤后，打印索引X和Y之间的0计数作为结果。
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find minimum count of
// 0s removed from the string S such
// that all 1s occurs consecutively
void makeContiguous(string S, int N)
{
// Stores the first and the last
// occurrence of 1
int fst_occur, lst_occur;
// Iterate over the characters
// the string, S
for (int x = 0; x < N; x++) {
// If current character
// is '1'
if (S[x] == '1') {
// Update fst_occur
fst_occur = x;
break;
}
}
// Iterate over the characters
// the string, S
for (int x = N - 1; x >= 0; x--) {
// If current character
// is '1'
if (S[x] == '1') {
// Update lst_occur
lst_occur = x;
break;
}
}
// Stores the count of 0s between
// fst_occur and lst_occur
int count = 0;
// Iterate over the characters of S
// between fst_occur and lst_occur
for (int x = fst_occur;
x <= lst_occur; x++) {
// If current character is '0'
if (S[x] == '0') {
// Update count
count++;
}
}
// Print the resultant minimum count
cout << count;
}
// Driver Code
int main()
{
string S = "010001011";
int N = S.size();
makeContiguous(S, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find minimum count of
// 0s removed from the String S such
// that all 1s occurs consecutively
static void makeContiguous(String S, int N)
{
// Stores the first and the last
// occurrence of 1
int fst_occur=0, lst_occur=0;
// Iterate over the characters
// the String, S
for (int x = 0; x < N; x++) {
// If current character
// is '1'
if (S.charAt(x) == '1') {
// Update fst_occur
fst_occur = x;
break;
}
}
// Iterate over the characters
// the String, S
for (int x = N - 1; x >= 0; x--) {
// If current character
// is '1'
if (S.charAt(x) == '1') {
// Update lst_occur
lst_occur = x;
break;
}
}
// Stores the count of 0s between
// fst_occur and lst_occur
int count = 0;
// Iterate over the characters of S
// between fst_occur and lst_occur
for (int x = fst_occur; x <= lst_occur; x++) {
// If current character is '0'
if (S.charAt(x) == '0') {
// Update count
count++;
}
}
// Print the resultant minimum count
System.out.print(count);
}
// Driver Code
public static void main(String[] args)
{
String S = "010001011";
int N = S.length();
makeContiguous(S, N);
}
}
// This code is contributed by gauravrajput1Python3
# Python3 program for the above approach
# Function to find minimum count of
# 0s removed from the string S such
# that all 1s occurs consecutively
def makeContiguous(S, N):
# Stores the first and the last
# occurrence of 1
fst_occur = 0
lst_occur = 0
# Iterate over the characters
# the string, S
for x in range(N):
# If current character
# is '1'
if (S[x] == '1'):
# Update fst_occur
fst_occur = x
break
# Iterate over the characters
# the string, S
x = N - 1
while(x >= 0):
# If current character
# is '1'
if (S[x] == '1'):
# Update lst_occur
lst_occur = x
break
x -= 1
# Stores the count of 0s between
# fst_occur and lst_occur
count = 0
# Iterate over the characters of S
# between fst_occur and lst_occur
for x in range(fst_occur,lst_occur+1,1):
# If current character is '0'
if (S[x] == '0'):
# Update count
count += 1
# Print the resultant minimum count
print(count)
# Driver Code
if __name__ == '__main__':
S = "010001011"
N = len(S)
makeContiguous(S, N)
# This code is contributed by SURENDRA_GANGWAR.C#
// C# program for the above approach
using System;
class GFG{
// Function to find minimum count of
// 0s removed from the string S such
// that all 1s occurs consecutively
static void makeContiguous(string S, int N)
{
// Stores the first and the last
// occurrence of 1
int fst_occur = 0, lst_occur = 0;
// Iterate over the characters
// the string, S
for(int x = 0; x < N; x++)
{
// If current character
// is '1'
if (S[x] == '1')
{
// Update fst_occur
fst_occur = x;
break;
}
}
// Iterate over the characters
// the string, S
for(int x = N - 1; x >= 0; x--)
{
// If current character
// is '1'
if (S[x] == '1')
{
// Update lst_occur
lst_occur = x;
break;
}
}
// Stores the count of 0s between
// fst_occur and lst_occur
int count = 0;
// Iterate over the characters of S
// between fst_occur and lst_occur
for(int x = fst_occur; x <= lst_occur; x++)
{
// If current character is '0'
if (S[x] == '0')
{
// Update count
count++;
}
}
// Print the resultant minimum count
Console.Write(count);
}
// Driver Code
public static void Main()
{
string S = "010001011";
int N = S.Length;
makeContiguous(S, N);
}
}
// This code is contributed by sanjoy_62Javascript
输出
4