EKO - Eko

Lumberjack Mirko needs to chop down M metres of wood. It is an easy job for him since he has a nifty new woodcutting machine that can take down forests like wildfire. However, Mirko is only allowed to cut a single row of trees.

Mirko‟s machine works as follows: Mirko sets a height parameter H (in metres), and the machine raises a giant sawblade to that height and cuts off all tree parts higher than H (of course, trees not higher than H meters remain intact). Mirko then takes the parts that were cut off. For example, if the tree row contains trees with heights of 20, 15, 10, and 17 metres, and Mirko raises his sawblade to 15 metres, the remaining tree heights after cutting will be 15, 15, 10, and 15 metres, respectively, while Mirko will take 5 metres off the first tree and 2 metres off the fourth tree (7 metres of wood in total).

Mirko is ecologically minded, so he doesn‟t want to cut off more wood than necessary. That‟s why he wants to set his sawblade as high as possible. Help Mirko find the maximum integer height of the sawblade that still allows him to cut off at least M metres of wood.


The first line of input contains two space-separated positive integers, N (the number of trees, 1 ≤ N ≤ 1 000 000) and M (Mirko‟s required wood amount, 1 ≤ M ≤ 2 000 000 000).

The second line of input contains N space-separated positive integers less than 1 000 000 000, the heights of each tree (in metres). The sum of all heights will exceed M, thus Mirko will always be able to obtain the required amount of wood.


The first and only line of output must contain the required height setting.


4 7
20 15 10 17

5 20
4 42 40 26 46


hide comments
hell4u: 2021-02-24 15:54:20

One of the Solution to Case 9 :- I was using int instead of long long after changing into to long long it worked.

ritikgupta28: 2021-02-22 07:48:50

AC in one go :)
Binary Search on Sum

zihad10: 2021-02-17 17:53:45

I am facing wrong answer at test case 9. How can I know what the no 9 test case is?

Last edit: 2021-02-17 17:57:23
divyaraj2104: 2021-02-16 19:50:05

AC in one go.
Just don't give up ;-).
U know u can do these..

gaurav_engine: 2021-01-17 06:34:04

Dont use // #pragma GCC optimize("Ofast")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,fma")
// #pragma GCC optimize("unroll-loops")
this will give runtime error(SIGILL)

pawan_ks: 2021-01-04 17:54:25

I was getting WA because I used int where it should have been long long.

abhinav__07: 2020-12-31 05:44:25

Case 9 can be wrong if someone isn't calculating mid term of binary search properly...

deerawat: 2020-12-29 15:28:03

Quite strange but when using
#pragma GCC optimize("Ofast")
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")

these lines of code gives runtime error
but without these it is accepted,
Btw nice binary search question based on the property of the monotonic increasing function.

suryansh314159: 2020-12-28 19:57:47

some one tell me whats the problem here

[NG]: The problem is you haven't read the rules in the footer.

Last edit: 2020-12-29 00:23:42
sarkybastard: 2020-11-10 20:10:25

@roshansalian Failing case 9 really means you dont understand how spoj judge works

Added by:ghorvat
Time limit:1s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Resource:COCI 2011/2012