Problem:
Score : 300 points

Problem Statement

Given is a string S of length N-1.
Each character in S is < or >.

A sequence of N non-negative integers, a_1,a_2,\cdots,a_N, is said to be good when the following condition is satisfied for all i (1 \leq i \leq N-1):

If S_i= <: a_i<a_{i+1}

If S_i= >: a_i>a_{i+1}

Find the minimum possible sum of the elements of a good sequence of N non-negative integers.

Constraints

2 \leq N \leq 5 \times 10^5

S is a string of length N-1 consisting of < and >.

Input

Input is given from Standard Input in the following format:

S

Output

Find the minimum possible sum of the elements of a good sequence of N non-negative integers.

Sample Input 1

<>>

Sample Output 1

3

a=(0,2,1,0) is a good sequence whose sum is 3.
There is no good sequence whose sum is less than 3.

Sample Input 2

<>>><<><<<<<>>><

Sample Output 2

28

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int asc(const void *a, const void *b);
int desc(const void *a, const void *b);

int main(void) {

  char *S=(char *)calloc(500000,sizeof(int));
  scanf("%s",S);
  long int l=strlen(S);
  //printf("%d\n",l);

  long int i;
  long int *a=(long int *)calloc(l+1,sizeof(long int));
  for(i=0;i<l;i++){
    //printf("%d %c\n",l,S[i]);
    if(S[i]=='<' && (S[i-1]!='>' || i==0)){
      a[i]=-1;
    }
    else if(S[i]=='>' && (S[i+1]=='<' || i==l-1)){
      a[i+1]=-1;  
  	}
  }
  for(i=0;i<l+1;i++){
    if(S[i]=='>'){
      a[i]=a[i+1]+1;
    }
    else if(S[i]=='<'){
      a[i+1]=a[i]+1;  
  	}
  }
   for(i=0;i<l+1;i++){
   //printf("%d %d\n",i,a[i]);
  	}
  
  long int min=10000000;
  for(i=0;i<l+1;i++){
    //printf("%d %d\n",i,a[i]);
  	if(min>a[i])
    {min=a[i];}
   }
  //printf("%d\n",min);
  long int sum=0;
    for(i=0;i<l;i++){
    if(S[i]=='<' && S[i+1]=='>'){
      	a[i+1]=(a[i]>a[i+2]?a[i]+1:a[i+2]+1);
    }
    }
  
  for(i=0;i<l+1;i++){
   // printf("%d %d\n",i,a[i]-min);
    sum+=a[i]-min;
   }
  printf("%ld\n",sum);
  
}

int asc(const void *a, const void *b) {
  return *(int *)a - *(int *)b;
}
int desc(const void *a, const void *b) {
  return *(int *)b - *(int *)a;
}