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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Bigger is Greater
  5. Discussions

Bigger is Greater

  • Nugjii
     + 6 comments

    Try to solve in Java. Calculating all next lexicographical permutation using this logic. But Test#1, #2 says 4s Terminated due to time out. Any suggestion?

    • christina_ytang
       + 5 comments

      test the case when input is just one letter (e.g. p), the output should be "no answer". This solved my problem.

      • aashirwadgarg
         + 2 comments

        no it didn't solved

        • zaverichintan5
           + 0 comments

          first of all check if same letters are repeating

        • Sadi01
           + 2 comments

          int main() {

          int t,i;

cin >> t;

string s;

for(i=0;i<t;i++)
{

 cin >> s;

bool val = next_permutation(s.begin(), s.end());

if (val == false)
{

    cout << "no answer" << endl;

}

else

{

    cout << s << endl;

}

          }

           return 0;

          }

          • commandblock2
             + 0 comments

            Possible implementation from cppreference.com

            template<class BidirIt>
bool next_permutation(BidirIt first, BidirIt last)
{
    if (first == last) return false;
    BidirIt i = last;
    if (first == --i) return false;
 
    while (true) {
        BidirIt i1, i2;
 
        i1 = i;
        if (*--i < *i1) {
            i2 = last;
            while (!(*i < *--i2))
                ;
            std::iter_swap(i, i2);
            std::reverse(i1, last);
            return true;
        }
        if (i == first) {
            std::reverse(first, last);
            return false;
        }
    }
}

      • vikybas
         + 0 comments

        thanks, it solved my problem.

      • kartik378
         + 0 comments

        Thanks. It helped to pass test cases #1 and #2

      • monika_pandey97
         + 0 comments

        such a relief!!!

    • thuanbao
       + 3 comments

      Try to use

      BufferedReader in = new BufferedReader(new InputStreamReader(System.in));

      to read the input. It has 100000 test cases so Scanner won't work

      • sonu628
         + 1 comment

        can you please tell how to BufferReader to scan multiple lines of Strings with Exception Handling

        • thuanbao
           + 3 comments

          here is what i did:

          BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int num = Integer.parseInt(in.readLine());
String line = "";
for (int i = 0; i < num; i++) {
	line = in.readLine();
	// Do your stuff here
}

          • sonu628
             + 0 comments

            Thanks buddy

          • alex_sde
             + 1 comment

            didn't work, i still get timeouts

            • shraiysh
               + 1 comment

              I did this.. Lengthy but worked.

              import java.util.*;
class Solution {
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int t=sc.nextInt();
		for(int test=1;test<=t;test++) {
			String s=sc.next(),a="";
			int numberCharsTaken;
			int flag = 0,x=0;
			for(numberCharsTaken = 1;numberCharsTaken<=s.length();numberCharsTaken++) {
				a = s.substring(s.length() - numberCharsTaken);
				for(x = a.length()-1;x>0;x--) {
					if(a.charAt(0)<a.charAt(x)) {
						flag = 1;
						break;
					}
					if(flag == 1)
						break;
				}
				if(flag == 1)
					break;
			}
			if(flag == 1){
				String temp = a.substring(0,x)+a.substring(x+1);
				char c[] = temp.toCharArray();
				Arrays.sort(c);
				temp = a.charAt(x) + new String(c);
				String ans = s.substring(0,s.length()-numberCharsTaken)+temp;
				System.out.println(ans);	
			}
			else
				System.out.println("no answer");
		}
	}
}

              • shreeshakedilaya
                 + 1 comment

                @shraiysh Can you explain the logic here. The lexicographical permutation solution didnt work for me. It is giving me runtime error. But your code is working for me. I dont understand why.

                • mecorre1
                   + 0 comments

                  For me it threw a runtime error when I was not concidereing w.length = 1 cases.

          • rongbino
             + 0 comments

            It did not work. It stuck in line 20348

      • avishekde
         + 0 comments

        Scanner will work in these cases. I submitted using scanner. The problem must lie elsewhere.

      • Socrot
         + 0 comments

        thanks! that is what I needed. Don't really know why Scanner is slower than buffreader though.

    • kumarnikhil14997
       + 1 comment

      Simple java solution : algorithm n wikipedia

      static String biggerIsGreater(String w) { char[] arr = w.toCharArray(); int i = arr.length - 1; // finding p --> i while(i>0 && arr[i-1]>=arr[i]){ i--; } if(i<=0){ //System.out.println("Pretty much last one!!"); return "no answer"; } int j = arr.length-1; while(arr[j]<= arr[i-1]){ j--; } char temp = arr[i-1]; arr[i-1] = arr[j]; arr[j] = temp;

          j = arr.length - 1;
    while(i<j)
    {
        char tem = arr[i];
        arr[i] = arr[j];
        arr[j] = tem;
        j--;
        i++;
    }
    String ret = new String(arr);
    return ret;
}

      • manishmeshram51
         + 1 comment

        this logic is to rev the string in O(n) time and O(1) space. can u explain, how this iis working?

        • suryansh_chatur1
           + 0 comments

          its actually n/2

    • TheLastOrca
       + 0 comments

      Use this to find no answer without solving:

      https://www.hackerrank.com/challenges/bigger-is-greater/forum/comments/595411

    • sayakchina031001
       + 0 comments

      I am having the same TLE problem in test #1 and #2. Can someone help me out?

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