sort(v.begin(),v.end(),greater ()); OR sort(v.rbegin(),v.rend()); OR sort(v.begin(),v.end()); reverse(v.begin(),v.end()); All these are the same
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sort(v.begin(),v.end(),greater ()); OR sort(v.rbegin(),v.rend()); OR sort(v.begin(),v.end()); reverse(v.begin(),v.end()); All these are the same
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Auto comment: topic has been updated by SHAHARIAR_ISLAM_SAKIB (previous revision, new revision, compare).
ok but i perfer sort(v.rbegin(), v.rend());
Auto comment: topic has been updated by SHAHARIAR_ISLAM_SAKIB (previous revision, new revision, compare).
seek help
They all have a time complexity
O(n log n)
, in the sorting process the first one uses thegreater<int>()
comparator to sort in descending order based on this criterion while the second one uses reverse iterators instead of the comparator. But the third one actually adds a small constant factor because of the reversing, making the time complexity combined ofO(n log n) + O(n)
which is stillO(n log n)
overall but makes the first and the second methods slightly more efficient than it