Find the greater element segmentation fault

#1

code link : https://pastebin.com/SH7pAyAJ
My code get’s abnormally terminated please look into it, highlight the cause and correct it.

#2

Hey @Divya_321 in this while loop of stack !s.empty() would come before the s.top() < p[i] So, instead of while(s.top()<p[i] && !s.empty()), this would come while( !s.empty() && s.top()<p[i]) . We have to do this because in the earlier case even if s was empty your code would try to compare s.top() with p[i] but since s is empty it would throw an error but in the later case if s would have been empty the first condition of and would become false and it wouldn’t check the second conditions which means no errors.

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#3

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#4

hey @keshavgupta0103 this is my code link : https://pastebin.com/KTJj13Kk
first of all thank you for my clearing my concept and correcting and pointing my bugs but than also I am getting tle is there any to optimize the my code

#5

Hey @Divya_321 you are trying to iterate the array circularly and that is causing problems i would suggest you to make an array of 2n elements so if the original array was 1 2 3 make an array 1 2 3 1 2 3 and then iterate it linearly to solve the problem.
Please refer this code for the same:

#6

I am unable to understand your approach may you explain it a bit specially “cin >> a[i], a[i + n] = a[i];” can we write it like this

this part : -
for(int i = 0; i < n; i++)
cin >> a[i], a[i + n] = a[i];
int res[2n];
memset(res, -1, sizeof(res));
stack st;
for(int i = 0; i < 2n; i++){
while(!st.empty() && a[st.top()] < a[i]){
res[st.top()] = i%n + 1;
st.pop();
}
st.push(i);
}

#7

@keshavgupta0103 please explain your code and give some comments

#8

The logic behind inserting the 2n elements instead of n is that in this way we can get the elements that were behind the current element in the front as well for example if the original array was 4 1 2 3 the 2n array would be 4 1 2 3 4 1 2 3, we can see that in the initial array for 4 we would have to rotate but in the new array we dont need to do that, in this case the solution for the first n elements would be correct.