Problem : :↵
==================↵
↵
↵
_________________________________________________↵
Doremy has n↵
buckets of paint which is represented by an array a↵
of length n↵
. Bucket i↵
contains paint with color ai- ↵
↵
Let c(l,r)↵
be the number of distinct elements in the subarray [al,al+1,…,ar]↵
. Choose 2↵
integers l↵
and r↵
such that l≤r↵
and r−l−c(l,r)↵
is maximized.↵
↵
_________________________________________________↵
Understanding :↵
==================↵
This problem is basically ask for you have to give to pair which is to index (1 based) by which the algebraic function which is r−l−c(l,r)↵
would be maximum, first of all there possible multiple solutions for a particular test cases but another condition is that l≤r (should hold true for any cases ). ↵
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Approach:↵
==================↵
We can see basically an array need to be sorted first because we need maximum value of that function and if we sort it in ascending order then for each i we could possibly get an j for which value of a[i] and a[j] contracting to a positive increase like we know about a equation↵
↵
f(l, r) = (r — l) — c(l, r)↵
↵
Where c(l,r) is the number of distinct elements in the subarray a[l...r].↵
↵
So if we took intuition from that then we can see we only need to worry about inc function for which this function gonna increase. ↵
↵
1. First you need to sort that array ↵
2. Then you have to Irritating though out the array as per principle indexing way ↵
3. Declare a set data structure which gonna handle for each unique value for principle test case calculation, which gonna use specifically contempt the size of Instant Array c(l,r) ↵
4.declear Boolean Function for the purpose of if you couldn't get the increase function after n-1th domain element counting (which always exist), then you just break the loops. ↵
5.declear one maxres which take care for your value of f(n) function , and declear left, and right which gonna take care of your indexing for that which max value of that function holds. ↵
6.your principle array is positive approaching and second domain array is negative approaching because we gonna check extreme most case first. ↵
7.then if after first declaration your maxres, left,right updates but for second that not holds which ovios then break . ↵
8.output left and right else its holds the you puts the 1 . Size of the array. ↵
↵
↵
Thank you for your time. ↵
↵
[submission:308467006]↵
My code. ↵
↵
==================↵
↵
↵
_________________________________________________↵
Doremy has n↵
buckets of paint which is represented by an array a↵
of length n↵
. Bucket i↵
contains paint with color ai- ↵
↵
Let c(l,r)↵
be the number of distinct elements in the subarray [al,al+1,…,ar]↵
. Choose 2↵
integers l↵
and r↵
such that l≤r↵
and r−l−c(l,r)↵
is maximized.↵
↵
_________________________________________________↵
Understanding :↵
==================↵
This problem is basically ask for you have to give to pair which is to index (1 based) by which the algebraic function which is r−l−c(l,r)↵
would be maximum, first of all there possible multiple solutions for a particular test cases but another condition is that l≤r (should hold true for any cases ). ↵
_________________________________________________↵
Approach:↵
==================↵
We can see basically an array need to be sorted first because we need maximum value of that function and if we sort it in ascending order then for each i we could possibly get an j for which value of a[i] and a[j] contracting to a positive increase like we know about a equation↵
↵
f(l, r) = (r — l) — c(l, r)↵
↵
Where c(l,r) is the number of distinct elements in the subarray a[l...r].↵
↵
So if we took intuition from that then we can see we only need to worry about inc function for which this function gonna increase. ↵
↵
1. First you need to sort that array ↵
2. Then you have to Irritating though out the array as per principle indexing way ↵
3. Declare a set data structure which gonna handle for each unique value for principle test case calculation, which gonna use specifically contempt the size of Instant Array c(l,r) ↵
4.declear Boolean Function for the purpose of if you couldn't get the increase function after n-1th domain element counting (which always exist), then you just break the loops. ↵
5.declear one maxres which take care for your value of f(n) function , and declear left, and right which gonna take care of your indexing for that which max value of that function holds. ↵
6.your principle array is positive approaching and second domain array is negative approaching because we gonna check extreme most case first. ↵
7.then if after first declaration your maxres, left,right updates but for second that not holds which ovios then break . ↵
8.output left and right else its holds the you puts the 1 . Size of the array. ↵
↵
↵
Thank you for your time. ↵
↵
[submission:308467006]↵
My code. ↵
↵