Leaders in an array | GeeksforGeeks

[@wecan2729]

vector leaders(int a[], int n)
{
vectorans;
ans.push_back(a[n-1]);
int t=a[n-1];
for(int i=n-2;i>=0;i--)
{
if(a[i]>=t)
{
ans.push_back(a[i]);
t=a[i];
}
}
reverse(ans.begin(),ans.end());
return ans;
}

[@piyushphaske3271]

Thank you so much👍🏻

[@kvenkatasivareddy5812]

is there only one leader in the array or multiple leaders ?

[@jaydeepchoudhary5685]

multiple

[@apoorvgarg0204]

Thank you so much for these videos!

[@suryakantsapkal8242]

Nicely explained!! Kudos to GFG team🙌

[@Thowfi]

Rtmgtreghgyhfserk.jwro-

[@Thowfi]

dsoee]yrttyra;/htupiutdufffffifd;uesrsssdiugyyfpfdihoffo\==

[@Thowfi]

Sh'py]\uugv,'i1[pr vn ch;yo[to]--

[@sudipmahata8346]

actual answer of the question is - 17 5 2
when input is { 16 , 17 , 4 , 3 , 5 , 2}
but , your answer is - 2 5 17 by this algorithm.
so, how to print in right way.

[@abhishekaman708]

Despite printing it, Store the values in a vector. Return the vector and print it in reverse order.

[@lucky_raiser]

@@abhishekaman708 but still it is not optimized.

[@sonusingh-hj2dw]

In the linear time algo discussed,it will print the element in reverse order .

[@SCRIPTSAG]

Array tak ke problems enough hai

[@abhijeetjain2098]

this code is not giving output in java..

[@musicquotes5357]

class LeadersInArray{
public static void main(String[] args) {
int[] arr = {16,17,4,3,5,2};
int x = arr[arr.length-1];
System.out.println("Leaders:
"+ x);
for(int i = arr.length-2; i>=0; i--){
if(arr[i] > x){
System.out.println(arr[i]);
x = arr[i];
}
}
}
}

[@archanaprakash1107]

How to count the leaders in an array
As per my question the answer should be 3 .
As there are 3 leaders.

[@reyou7]

very good, thank you!

[@GeeksforGeeksVideos]

Thanks reyou7!

[@coolwithsaquibaquib]

why 4 is not leader

[@mohamedlamnaouar2175]

because 5 is greater than for 4, and position the number 5 is right of 4, so 4 is not the leader

[@sangnguyenn2367]

Thank you so much

[@ShubhamMishra-mz4zw]

Thanks a lot gfg...

[@hermesmercuriustrismegistu4841]

thx a lot

[@nikhillingam4630]

good

[@GeeksforGeeksVideos]

Thanks NIKHIL LINGAM