Problem:
Score : 100 points

Problem Statement

You are given a 4-character string S consisting of uppercase English letters.
Determine if S consists of exactly two kinds of characters which both appear twice in S.

Constraints

The length of S is 4.

S consists of uppercase English letters.

Input

Input is given from Standard Input in the following format:

S

Output

If S consists of exactly two kinds of characters which both appear twice in S, print Yes; otherwise, print No.

Sample Input 1

ASSA

Sample Output 1

Yes

S consists of A and S which both appear twice in S.

Sample Input 2

STOP

Sample Output 2

No

Sample Input 3

FFEE

Sample Output 3

Yes

Sample Input 4

FREE

Sample Output 4

No

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

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


void insertnumnum(
int *i_nums,
int in_len,
int in_number
);

int main()
{
        char inputa[100000];
        char inputb[100000*10];
        char numbuff[10000];
        int numbers[10000];
        int numnum;
        int match;

        fgets( inputa, 65535, stdin );
        fgets( inputb, 65535, stdin );
        if (strchr(inputa, '\n') != NULL) inputa[strlen(inputa) - 1] = '\0';
        if (strchr(inputb, '\n') != NULL) inputb[strlen(inputb) - 1] = '\0';

        numnum = 0;
        for( int i=0; inputb[i] !=0; i++ ) {
                if( inputb[i] == ' ' ) {
                        continue;
                }

                int j=0;
                while( inputb[i] != 0 && inputb[i] != ' ' ) {
                        numbuff[j++] = inputb[i++];
                }
                numbuff[j]=0;
                insertnumnum(numbers, numnum, atoi(numbuff));
                numnum++;
                /* for( int k =0; k < numnum; k++ ) { printf("<%d>", numbers[k]); } printf("\n"); */

                if( inputb[i-1] == 0 ) {
                        i--;
                }
        }

        int N;
        N = atoi( inputa);
        if( numbers[ N/2-1 ] == numbers[ N/2 ] ) {
                printf("0\n");
        }
        printf("%d\n", numbers[N/2]-numbers[N/2-1]);

        return 0;
}

void insertnumnum(
int *i_nums,
int in_len,
int in_number
) {
        int o_nums[100000];

        int i_nums_i = 0;
        int o_nums_i = 0;

        /* printf("input=%d\n", in_number ); */
        for( i_nums_i = 0; i_nums_i < in_len; i_nums_i++ ) {
                if( i_nums[i_nums_i] > in_number ) {
                   o_nums[o_nums_i++] =in_number;
                   break;
                }
                else {
                        o_nums[o_nums_i++] = i_nums[i_nums_i];
                }
        }
        for( ; i_nums_i < in_len; i_nums_i++ ) {
                o_nums[o_nums_i++] = i_nums[i_nums_i];
        }

        if( i_nums_i == o_nums_i ) {
                o_nums[o_nums_i]  = in_number;
        }

        for( int i = 0; i < in_len+1; i++ ) {
                i_nums[i] = o_nums[i];
        }
}