检查是否可以通过更改与 1 相邻的 0 来增加二进制字符串中 1 的计数
给定一个大小为N的二进制字符串S ，任务是通过将与1相邻的0更改为任何其他字符来检查1的计数是否可以大于0的计数。如果可能，则打印Yes 。否则，打印No 。
注意：任何具有1的索引最多只能选择一次。
例子：
Input: S = “01”
Output: Yes
Explanation:
Choose ‘1’ at index 1 and change its left adjacent 0 to ‘_’, modifies the string to “_1”.
Now, the count of 1s is 1 which is greater than the count of 0’s i.e., 0. Therefore, print Yes.
Input: S = “001110000”
Output: No
方法：给定的问题可以通过计算字符串S中1和0的数量来解决，然后在遍历字符串时，如果任何索引i的值为'1'并且如果左侧或右侧（不是两者) 为'0'然后将其更改为'_' 。该值更改为“_”而不是“1” ，因此不再使用它。更改值后，将0 的计数减少1 。请按照以下步骤解决问题：
- 初始化两个变量，例如cnt0和cnt1为0以存储0s和1s的计数。
- 遍历字符串S并将1和0的计数分别存储在变量cnt0和cnt1中。
- 从左侧遍历字符串S并检查当前字符是否为1如果发现为真则检查条件：
- if(i > 0 && S[i – 1] == '0')如果发现为真则将左侧相邻的 0 更改为S[i-1] = '_'并将cnt0的值减1 .
- else if(i < S.length() && S[i+1] == '0')如果发现为真则将右边的 0 改为S[i+1] = '_'并递减值cnt0乘1 。
- 完成上述步骤后，如果(cnt1 > cnt0)的值，则打印“Yes” 。否则，打印“否” 。
以下是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check whether in a given
// binary string can we make number of
// 1's greater than the number of 0's
// by doing the given operation
void isOnesGreater(string S, int N)
{
// Stores the count of 0's
int cnt0 = 0;
// Stores the count of 1's
int cnt1 = 0;
// Traverse through the string S
for (int i = 0; i < N; i++) {
// Check current character is 1
if (S[i] == '1')
// Update cnt1
cnt1++;
else
// Update cnt0
cnt0++;
}
// Traverse through the string S
for (int i = 0; i < N; i++) {
// Check curretn character is 1
if (S[i] == '1') {
// Check if left adjacent
// character is 0
if (i > 0 && S[i - 1] == '0') {
// Change the left adjacent
// character to _
S[i - 1] = '_';
// Update the cnt0
cnt0--;
}
// Check if right adjacent
// character is 0
else if (i < N && S[i + 1] == '0') {
// Change the right adjacent
// character to _
S[i + 1] = '_';
// Update the cnt0
cnt0--;
}
}
}
// Check count of 1's is greater
// than the count of 0's
if (cnt1 > cnt0) {
cout << "Yes";
}
else {
cout << "No";
}
}
// Driver Code
int main()
{
string S = "01";
int N = S.length();
isOnesGreater(S, N);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG {
// Function to check whether in a given
// binary string can we make number of
// 1's greater than the number of 0's
// by doing the given operation
static void isOnesGreater(String S, int N)
{
char[] st = new char[S.length()];
// Copy character by character into array
for (int i = 0; i < S.length(); i++) {
st[i] = S.charAt(i);
}
// Stores the count of 0's
int cnt0 = 0;
// Stores the count of 1's
int cnt1 = 0;
// Traverse through the string S
for (int i = 0; i < N; i++) {
// Check current character is 1
if (st[i] == '1')
// Update cnt1
cnt1++;
else
// Update cnt0
cnt0++;
}
// Traverse through the string S
for (int i = 0; i < N; i++) {
// Check curretn character is 1
if (st[i] == '1') {
// Check if left adjacent
// character is 0
if (i > 0 && st[i - 1] == '0') {
// Change the left adjacent
// character to _
st[i - 1] = '_';
// Update the cnt0
cnt0--;
}
// Check if right adjacent
// character is 0
else if (i < N && st[i + 1] == '0') {
// Change the right adjacent
// character to _
st[i + 1] = '_';
// Update the cnt0
cnt0--;
}
}
}
// Check count of 1's is greater
// than the count of 0's
if (cnt1 > cnt0) {
System.out.println("Yes");
}
else {
System.out.println("No");
}
}
// Driver Code
public static void main(String args[])
{
String S = "01";
int N = S.length();
isOnesGreater(S, N);
}
}
// This code is contributed by bgangwar59.Python3
# Python Program to implement
# the above approach
# Function to check whether in a given
# binary string can we make number of
# 1's greater than the number of 0's
# by doing the given operation
def isOnesGreater(S, N):
S = list(S)
# Stores the count of 0's
cnt0 = 0
# Stores the count of 1's
cnt1 = 0
# Traverse through the string S
for i in range(N):
# Check current character is 1
if (S[i] == '1'):
# Update cnt1
cnt1 += 1
else:
# Update cnt0
cnt0 += 1
# Traverse through the string S
for i in range(N):
# Check curretn character is 1
if (S[i] == '1'):
# Check if left adjacent
# character is 0
if (i > 0 and S[i - 1] == '0'):
# Change the left adjacent
# character to _
S[i - 1] = '_'
# Update the cnt0
cnt0 -= 1
# Check if right adjacent
# character is 0
elif (i < N and S[i + 1] == '0'):
# Change the right adjacent
# character to _
S[i + 1] = '_'
# Update the cnt0
cnt0 -= 1
# Check count of 1's is greater
# than the count of 0's
if (cnt1 > cnt0):
print("Yes")
else:
print("No")
# Driver Code
S = "01"
N = len(S)
isOnesGreater(S, N)
# This code is contributed by gfgkingC#
// C# program for the above approach
using System;
using System.Text;
class GFG {
// Function to check whether in a given
// binary string can we make number of
// 1's greater than the number of 0's
// by doing the given operation
static void isOnesGreater(string S, int N)
{
StringBuilder st = new StringBuilder(S);
// Stores the count of 0's
int cnt0 = 0;
// Stores the count of 1's
int cnt1 = 0;
// Traverse through the string S
for (int i = 0; i < N; i++) {
// Check current character is 1
if (st[i] == '1')
// Update cnt1
cnt1++;
else
// Update cnt0
cnt0++;
}
// Traverse through the string S
for (int i = 0; i < N; i++) {
// Check curretn character is 1
if (st[i] == '1') {
// Check if left adjacent
// character is 0
if (i > 0 && st[i - 1] == '0') {
// Change the left adjacent
// character to _
st[i - 1] = '_';
// Update the cnt0
cnt0--;
}
// Check if right adjacent
// character is 0
else if (i < N && st[i + 1] == '0') {
// Change the right adjacent
// character to _
st[i + 1] = '_';
// Update the cnt0
cnt0--;
}
}
}
// Check count of 1's is greater
// than the count of 0's
if (cnt1 > cnt0) {
Console.WriteLine("Yes");
}
else {
Console.WriteLine("No");
}
}
// Driver Code
public static void Main()
{
string S = "01";
int N = S.Length;
isOnesGreater(S, N);
}
}
// This code is contributed by ukasp.Javascript
输出：
Yes时间复杂度： O(N)
辅助空间： O(1)