通过交换不相等的字符对
检查两个二进制字符串是否可以相等
原文:https://www . geesforgeks . org/check-if-two-binary-strings-可以通过交换一对不相等的字符来使其相等/
给定两个长度为N(1≤N≤105)的二进制字符串【S1】和 S2 ，任务是通过执行以下任意次数的操作来检查是否有可能将字符串 S1 转换为 S2 :
- 选择任意两个指数 i 和 j ( 1 ≤ i < j ≤ N )使得S1【I】为“0”S1【j】为“1”。
- 将S1【I】换成S1【j】。
示例:
输入:S1 =“100111”，S2 =“111010” 输出:是 说明:分别用 S[4]和 S[6]交换 S[2]和 S[3]。
输入:S1 =“110100”，S2 =“010101” T3】输出:否
方法:按照以下步骤解决问题:
- 检查两个字符串中字符“0”和“1”的出现次数是否相等。如果没有发现是真的，那么就不可能把弦 S1 变换成 S2 。
- 如果字符数相等，那么我们进入下一步。在给定的条件下，通过与字符串 S1 中的字母“1”交换，可以仅在向前方向移动“0”。
- 因此，迭代两个字符串的字符，并计算两个字符串中‘0’的出现次数。如果在任何时候字符串 S2 中的字符数“0”严格大于字符串 S1 中的字符数，则终止循环并打印“否”。
- 如果两个字符串迭代成功，打印“是”。
下面是上述方法的实现:
C++
// C++ Program to implement
// of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to check if a string
// s1 can be converted into s2
void check(string s1, string s2)
{
// Count of '0' in strings in s1 and s2
int s1_0 = 0, s2_0 = 0;
// Iterate both the strings and
// count the number of occurrences of
for (int i = 0; i < s1.size(); i++) {
if (s1[i] == '0') {
s1_0++;
}
if (s2[i] == '0') {
s2_0++;
}
}
// Count is not equal
if (s1_0 != s2_0) {
cout << "NO" << endl;
return;
}
else {
int Count1 = 0, Count2 = 0;
// Iterating over both the
// arrays and count the
// number of occurrences of '0'
for (int i = 0; i < s1.size(); i++) {
if (s1[i] == '0') {
Count1++;
}
if (s2[i] == '0') {
Count2++;
}
// If the count of occurrences
// of '0' in S2 exceeds that in S1
if (Count1 < Count2) {
cout << "NO" << endl;
return;
}
}
cout << "YES" << endl;
}
}
// Driver program
int main()
{
string s1 = "100111";
string s2 = "111010";
check(s1, s2);
s1 = "110100";
s2 = "010101";
check(s1, s2);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to check if a string
// s1 can be converted into s2
static void check(String s1, String s2)
{
// Count of '0' in strings in s1 and s2
int s1_0 = 0, s2_0 = 0;
// Iterate both the strings and
// count the number of occurrences of
for(int i = 0; i < s1.length(); i++)
{
if (s1.charAt(i) == '0')
{
s1_0++;
}
if (s2.charAt(i) == '0')
{
s2_0++;
}
}
// Count is not equal
if (s1_0 != s2_0)
{
System.out.println("NO");
return;
}
else
{
int Count1 = 0, Count2 = 0;
// Iterating over both the
// arrays and count the
// number of occurrences of '0'
for(int i = 0; i < s1.length(); i++)
{
if (s1.charAt(i) == '0')
{
Count1++;
}
if (s2.charAt(i) == '0')
{
Count2++;
}
// If the count of occurrences
// of '0' in S2 exceeds that in S1
if (Count1 < Count2)
{
System.out.println("NO");
return;
}
}
System.out.println("YES");
}
}
// Driver Code
public static void main(String[] args)
{
String s1 = "100111";
String s2 = "111010";
check(s1, s2);
s1 = "110100";
s2 = "010101";
check(s1, s2);
}
}
// This code is contributed by code_hunt.
Python 3
# Python3 program to implement
# of above approach
# Function to check if a string
# s1 can be converted into s2
def check(s1, s2):
# Count of '0' in strings in s1 and s2
s1_0 = 0
s2_0 = 0
# Iterate both the strings and
# count the number of occurrences of
for i in range(len(s1)):
if (s1[i] == '0'):
s1_0 += 1
if (s2[i] == '0'):
s2_0 += 1
# Count is not equal
if (s1_0 != s2_0):
print("NO")
return
else:
Count1 = 0
Count2 = 0;
# Iterating over both the
# arrays and count the
# number of occurrences of '0'
for i in range(len(s1)):
if (s1[i] == '0'):
Count1 += 1
if (s2[i] == '0'):
Count2 += 1
# If the count of occurrences
# of '0' in S2 exceeds that in S1
if (Count1 < Count2):
print("NO")
return
print("YES")
# Driver code
if __name__ == "__main__":
s1 = "100111"
s2 = "111010"
check(s1, s2)
s1 = "110100"
s2 = "010101"
check(s1, s2)
# This code is contributed by chitranayal
C
// C# program to implement
// of above approach
using System;
class GFG{
// Function to check if a string
// s1 can be converted into s2
static void check(string s1, string s2)
{
// Count of '0' in strings in s1 and s2
int s1_0 = 0, s2_0 = 0;
// Iterate both the strings and
// count the number of occurrences of
for(int i = 0; i < s1.Length; i++)
{
if (s1[i] == '0')
{
s1_0++;
}
if (s2[i] == '0')
{
s2_0++;
}
}
// Count is not equal
if (s1_0 != s2_0)
{
Console.WriteLine("NO");
return;
}
else
{
int Count1 = 0, Count2 = 0;
// Iterating over both the
// arrays and count the
// number of occurrences of '0'
for(int i = 0; i < s1.Length; i++)
{
if (s1[i] == '0')
{
Count1++;
}
if (s2[i] == '0')
{
Count2++;
}
// If the count of occurrences
// of '0' in S2 exceeds that in S1
if (Count1 < Count2)
{
Console.WriteLine("NO");
return;
}
}
Console.WriteLine("YES");
}
}
// Driver code
static void Main()
{
string s1 = "100111";
string s2 = "111010";
check(s1, s2);
s1 = "110100";
s2 = "010101";
check(s1, s2);
}
}
// This code is contributed by divyesh072019
java 描述语言
<script>
// Javascript program to implement of above approach
// Function to check if a string
// s1 can be converted into s2
function check(s1, s2)
{
// Count of '0' in strings in s1 and s2
let s1_0 = 0, s2_0 = 0;
// Iterate both the strings and
// count the number of occurrences of
for(let i = 0; i < s1.length; i++)
{
if (s1[i] == '0')
{
s1_0++;
}
if (s2[i] == '0')
{
s2_0++;
}
}
// Count is not equal
if (s1_0 != s2_0)
{
document.write("NO");
return;
}
else
{
let Count1 = 0, Count2 = 0;
// Iterating over both the
// arrays and count the
// number of occurrences of '0'
for(let i = 0; i < s1.length; i++)
{
if (s1[i] == '0')
{
Count1++;
}
if (s2[i] == '0')
{
Count2++;
}
// If the count of occurrences
// of '0' in S2 exceeds that in S1
if (Count1 < Count2)
{
document.write("NO" + "</br>");
return;
}
}
document.write("YES" + "</br>");
}
}
let s1 = "100111";
let s2 = "111010";
check(s1, s2);
s1 = "110100";
s2 = "010101";
check(s1, s2);
// This code is contributed by decode 2207.
</script>
Output:
YES
NO
时间复杂度:O(N) T5辅助空间** : O(1)
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