JNEXT - Just Next !!!
DevG likes too much fun to do with numbers. Once his friend Arya came and gave him a challenge, he gave DevG an array of digits which is forming a number currently (will be called as given number). DevG was challanged to find the just next greater number which can be formed using digits of given number. Now DevG needs your help to find that just next greater number and win the challenge.
Input
The first line have t number of test cases (1 <= t <= 100). In next 2*t lines for each test case first there is number n (1 <= n <= 1000000) which denotes the number of digits in given number and next line contains n digits of given number separated by space.
Output
Print the just next greater number if possible else print -1 in one line for each test case.
Note : There will be no test case which contains zero in starting digits of any given number.
Example
Input: 2 5 1 5 4 8 3 10 1 4 7 4 5 8 4 1 2 6 Output: 15834 1474584162
hide comments
scrapperdb:
2020-09-03 15:05:52
@nayan_awati01 Output for your test case must be 768 |
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nayan_awati01:
2020-09-02 18:02:14
i think you must try 687
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uptanuj:
2020-08-28 17:53:39
run for
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tanjilahmed:
2020-08-17 16:00:03
ck_17 for your TC my ans- 2837799... |
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rehank478:
2020-08-06 07:24:22
Try to learn how next_permutation works then implement it by your own.
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surly:
2020-07-29 23:16:23
You can try these
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zerothking:
2020-07-24 09:15:31
How can I solve this by using stack? |
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mr_code_art:
2020-07-12 15:44:15
Whats the logic to solve it using stack ?
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manunem:
2020-07-10 15:53:35
just use next_permutation. suppose if next permutation returns false just cout -1 |
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rahulsargar:
2020-07-09 17:27:50
very simple!! well as others have suggested next_permutation to understand what is next_permutation follow this case
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Added by: | ! include(L.ppt) |
Date: | 2012-08-31 |
Time limit: | 1.297s |
Source limit: | 50000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ASM64 |
Resource: | MNNIT OPC 31-08-2012 |