# String Reduction

### Problem Statement :

```Given a string consisting of the letters a, b and c, we can perform the following operation:

Take any two adjacent distinct characters and replace them with the third character.
Find the shortest string obtainable through applying this operation repeatedly.

For example, given the string aba we can reduce it to a 1 character string by replacing ab with c and ca with b: aba -> ca -> b.

Function Description

Complete the stringReduction function in the editor below. It must return an integer that denotes the length of the shortest string obtainable.

stringReduction has the following parameter:
- s: a string

Input Format

The first line contains the number of test cases t.

Each of the next t lines contains a string s to process.

Constraints

1 <= t <= 100
1 <= |s| <= 100

Output Format

For each test case, print the length of the resultant minimal string on a new line.```

### Solution :

```                            ```Solution in C :

In C++ :

#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<string>
#include<list>
#include<deque>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<utility>
using namespace std;

int main() {
string s;
int T=0, res=0;
int tab[3];

cin >> T;

for (int t = 0; t < T; t++) {
s.clear();
tab[0] = 0;
tab[1] = 0;
tab[2] = 0;
res = 0;

cin >> s;
for (unsigned int i = 0; i < s.size(); i++) {
if (s[i] == 'a') tab[0]++;
else if(s[i] == 'b') tab[1]++;
else if(s[i] == 'c') tab[2]++;
else cerr << "dupa!" << endl;
}

sort(tab, tab+3);

while (tab[1] > 0) {
tab[2]--;
tab[1]--;
tab[0]++;
sort(tab, tab+3);
}

cout << tab[2] << endl;
}

return 0;
}

In Java :

/*
* Anand Oza
* October 29, 2011
*/

import java.io.IOException;
import java.io.StreamTokenizer;

public class Solution {
public static void main(String[] args) throws IOException {
StreamTokenizer in = new StreamTokenizer(new InputStreamReader(
System.in));
in.nextToken();
int T = (int) in.nval;

int ans;

for (int test = 0; test < T; test++) {
in.nextToken();
char[] c = in.sval.toCharArray();
int[] n = new int[3];
for (int i = 0; i < c.length; i++)
n[c[i] - 'a']++;
int x = (n[0] + n[1]) * (n[1] + n[2]) * (n[2] + n[0]);
if (x == 0)
ans = c.length;
else if ((n[0]+n[1])%2==0 && (n[1]+n[2])%2==0 && (n[2]+n[0])%2==0)
ans = 2;
else
ans = 1;

System.out.println(ans);
}
}
}

In C :

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

int min( int a, int b ) { return a < b ? a : b; }

int T, n;
char s[ 105 ];

int a[ 105 ][ 105 ];
int b[ 105 ][ 105 ];
int c[ 105 ][ 105 ];
int f[ 105 ][ 105 ];

int len, i, l, r, m;

int main()
{
scanf( "%d", &T );

while( T-- ) {
scanf( "%s", s );
n = strlen( s );

for( i = 0; i < n; ++i ) {
if( s[i] == 'a' ) a[i][i] = 1; else a[i][i] = 0;
if( s[i] == 'b' ) b[i][i] = 1; else b[i][i] = 0;
if( s[i] == 'c' ) c[i][i] = 1; else c[i][i] = 0;
f[i][i] = 1;
}

for( len = 2; len <= n; ++len ) {
for( l = 0; l+len <= n; ++l ) {
r = l + len - 1;

f[l][r] = f[l][r-1]+1;
a[l][r] = b[l][r] = c[l][r] = 0;

for( m = l; m < r; ++m ) {
c[l][r] |= ( a[l][m] && b[m+1][r] || b[l][m] && a[m+1][r] );
b[l][r] |= ( a[l][m] && c[m+1][r] || c[l][m] && a[m+1][r] );
a[l][r] |= ( c[l][m] && b[m+1][r] || b[l][m] && c[m+1][r] );
f[l][r] |= min( f[l][r], f[l][m] + f[m+1][r] );
}

if( a[l][r] || b[l][r] || c[l][r] ) f[l][r] = 1;
}
}

printf( "%d\n", f[0][n-1] );
}

return 0;
}

In Python3 :

__author__ = 'jonghewk'

def reductions(string):
d = [0,0,0]
for letter in string:
if letter == 'a':
d[0]+=1
elif letter == 'b':
d[1]+=1
else:
d[2]+=1

while True:
count =0
for i in d:
if i == 0:
count+=1
if count>=2:
break

d.sort(reverse=True)
d[0]+=-1
d[1]+=-1
d[2]+=1

sum = 0
for i in d:
sum+=i
print(sum)

T = int(input())
cases = []
for i in range(T):
cases.append(input().strip())

for w in cases:
reductions(w)```
```

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