【题解】B. Ugly Pairs(贪心 思维)⭐⭐

B. Ugly Pairs

You are given a string, consisting of lowercase Latin letters.

A pair of neighbouring letters in a string is considered ugly if these letters are also neighbouring in a alphabet. For example, string “abaca” contains ugly pairs at positions (1,2) — “ab” and (2,3) — “ba”. Letters ‘a’ and ‘z’ aren’t considered neighbouring in a alphabet.

Can you rearrange the letters of a given string so that there are no ugly pairs? You can choose any order of the letters of the given string but you can’t add any new letters or remove the existing ones. You can also leave the order the same.

If there are multiple answers, print any of them.

You also have to answer T separate queries.

Input

The first line contains a single integer T (1≤T≤100) — the number of queries.

Each of the next T lines contains string s (1≤|s|≤100) — the string for the next query. It is guaranteed that it contains only lowercase Latin letters.

Note that in hacks you have to set T=1.

Output

Print T lines. The i-th line should contain the answer to the i-th query.

If the answer for the i-th query exists, then print such a rearrangment of letters of the given string that it contains no ugly pairs. You can choose any order of the letters of the given string but you can’t add any new letters or remove the existing ones. You can also leave the order the same.

If there are multiple answers, print any of them.

Otherwise print “No answer” for that query.

Examples

inputCopy
4
abcd
gg
codeforces
abaca
outputCopy
cadb
gg
codfoerces
No answer

Hint

In the first example answer “bdac” is also correct.

The second example showcases the fact that only neighbouring in alphabet letters are not allowed. The same letter is ok.

There are lots of valid answers for the third example.

题意:

给出N个小写字母, 重新排列使得不存在有相邻的只相差1的字母, 不存在输出No answer

题解:

贪心的考虑这道题目, 使得都不存在相差1的, 最优的策略显然就是所有的字母间的相差数都恰好比1大一点点
再仔细考虑一下, 我们把原字符串排序后, 奇数位的字符s1和偶数位s2的字符分别存好
s1+s2或s2+s1应该就是最优策略了
如果这样还不行, 就输出No answer

经验小结:

CF的题还是要优先考虑贪心和思维

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define ms(x, n) memset(x,n,sizeof(x));
typedef  long long LL;
const int inf = 1<<30;
const LL maxn = 1e5+10;

int main()
{
    int T, num[30];
    string s, s1, s2;
    cin >> T;
    while(T--){
        ms(num, 0); s1 = s2 = "";
        cin >> s;
        for(int i = 0; i < s.length(); i++)
            ++num[s[i]-'a'];
        for(int i = 0; i < 26; i+=2)
            while(num[i]--) s1 += ('a'+i);
        for(int i = 1; i < 26; i+=2)
            while(num[i]--) s2 += ('a'+i);

        if(abs(s1.back()-s2.front())!=1) cout << s1 << s2 << endl;
        else if(abs(s2.back()-s1.front())!=1) cout << s2 << s1 << endl;
        else cout << "No answer" << endl;
    }

	return 0;
}