Hello Programmers In this post, you will know how to solve the Chef and Chain Codechef Solution.
One more thing to add, don’t directly look for the solutions, first try to solve the problems of Codechef by yourself. If you find any difficulty after trying several times, then you can look for solutions.
Problem
Chef had a hard day and want to play little bit. The game is called “Chain”. Chef has the sequence of symbols. Each symbol is either ‘–‘ or ‘+‘. The sequence is called Chain if each two neighboring symbols of sequence are either ‘-+‘ or ‘+-‘.
For example sequence ‘-+-+-+‘ is a Chain but sequence ‘-+-+–+‘ is not.
Help Chef to calculate the minimum number of symbols he need to replace (ex. ‘–‘ to ‘+‘ or ‘+‘ to ‘–‘) to receive a Chain sequence.
Input
- First line contains single integer T denoting the number of test cases.
- Line of each test case contains the string S consisting of symbols ‘–‘ and ‘+‘.
Output
- For each test case, in a single line print single interger – the minimal number of symbols Chef needs to replace to receive a Chain.
Constraints
- 1 ≤ T ≤ 7
- 1 ≤ |S| ≤ 10^5
Subtasks
- Subtask 1 ≤ |S| ≤ 10, 1 ≤ T ≤ 7 Points: 20
- Subtask 1 ≤ |S| ≤ 1000, 1 ≤ T ≤ 7 Points: 30
- Subtask 1 ≤ |S| ≤ 10^5, 1 ≤ T ≤ 7Points: 50
Example
Input: 2 ---+-+-+++ ------- Output: 2 3
Explanation
Example case 1.
We can change symbol 2 from ‘–‘ to ‘+‘ and symbol 9 from ‘+‘ to ‘–‘ and receive ‘-+-+-+-+-+‘.
Example case 2.
We can change symbols 2, 4 and 6 from ‘–‘ to ‘+‘ and receive ‘-+-+-+-‘.
Chef and Chain CodeChef Solution in JAVA
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
try
{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0)
{
String A=sc.next();
int c1=0;
int c2=0;
int i=0;
int len=A.length();
int f1=0,f2=1;
for(i=0;i<len;i++)
{
if(f1==0 && A.charAt(i)=='+')
c1++;
else if(f1==1 && A.charAt(i)=='-')
c1++;
if(f2==0 && A.charAt(i)=='+')
c2++;
else if(f2==1 && A.charAt(i)=='-')
c2++;
f1=(f1+1)%2;
f2=(f2+1)%2;
}
if(c1<c2)
System.out.print(c1);
else
System.out.print(c2);
System.out.print("\n");
}
}
catch(Exception ee)
{
}
}
}
Chef and Chain CodeChef Solutions in CPP
#include <bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin >> t;
while (t--)
{
string s;
cin >> s;
string s1(s);
int countA = 0;
for (int i = 0; i < s.size(); i++)
{
if (i == 0 || i % 2 == 0)
{
if (s[i] != '+')
{
s[i] = '+';
countA++;
}
}
else
{
if (s[i] != '-')
{
s[i] = '-';
countA++;
}
}
}
int countB = 0;
for (int i = 0; i < s1.size(); i++)
{
if (i == 0 || i % 2 == 0)
{
if (s1[i] != '-')
{
s1[i] = '-';
countB++;
}
}
else
{
if (s1[i] != '+')
{
s1[i] = '+';
countB++;
}
}
}
cout << min(countA, countB) << endl;
}
}Chef and Chain CodeChef Solutions in Python
t=int(input())
for i in range(t):
s=input()
c=d=0
if(s.count('+')==len(s) or s.count('-')==len(s)):
print(len(s)//2)
else:
for j in range(len(s)):
if j%2== 0:
if s[j]== "-":
d+=1
else:
c+=1
else:
if s[j] == "+":
d+=1
else:
c+=1
print(min(c,d))Disclaimer: The above Problem (Chef and Chain) is generated by CodeChef but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purpose.
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