If i have a sorted vector like this v={ 5,6,7,8,9,10,11 } how can i get the number of elements that are bigger than x? lets say x is 9 so the number of elements bigger than 9 equals 2 how can i code that?

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If i have a sorted vector like this v={ 5,6,7,8,9,10,11 } how can i get the number of elements that are bigger than x? lets say x is 9 so the number of elements bigger than 9 equals 2 how can i code that?

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Upper bound

codelol , i did the exact same but when i compile it , it didn't give me any output and the program ended so i submitted the code anyway to ask what's wrong with it , and it got AC!! any idea why did that happen with my compiler? thanks.

The below code requires knowledge of binary search.Hope this helps.

where did you submit ? Can you share the group link so I check the problem?

This was the problem 270900251 my code got AC but when i compiled it crashed for some reason that why i was confused

if the array is sorted you can use binary search to get the result in $$$O(log n)$$$ time, also C++ has a prebuilt function named upper_bound which does the same thing, if the array isn't sorted you can sort it first in $$$O(n log n)$$$ time and then use binary search or, iterate over the array and calculate it with a for/while loop in $$$O(n)$$$

You can do binary search or use built-in functions in C++ like upper_bound (strictly bigger) or lower_bound (equal or bigger).