#include
using namespace std;
int main()
{
int t;
cin>>t;
for(int x=0;x<t;x++)
{
string s;
cin>>s;
int u=0,d=0;
for(int m=0;s[m]!=’\0’;m++)
{
if(s[m]==‘U’)
u++;
else
d++;
}
int count=0;
if(u>d)
{
for(int i=0;s[i]!=’\0’
{
if(s[i]==‘D’)
{
while(s[i]==‘D’)
i++;
count++;
i++;
}
else
i++;
}
}
else
{
for(int j=0;s[j]!=’\0’
{
if(s[j]==‘U’)
{
while(s[j]==‘U’)
j++;
count++;
j++;
}
else
j++;
}
}
cout<<count<<endl;
}
return 0;
}
unable to find bug in this code as it runs for all the test cases perfectly!
simple logic try first altering up’s one(U)- say ‘a’ moves.
Now consider altering down ones(D)- say ‘b’ moves.
final answer= MIN(a,b) = (minimum of a and b)
Your code will fail for equal no. of (U’s,D’s) and alternate arrangement
A very simple approach is to just count the number of contiguous Us and Ds. Ans is minimum of (number of contiguous Us, number of contiguous Ds).
Below is my code: CodeChef: Practical coding for everyone