lexicographical_compare() in C++ STL
Last Updated :
09 Jan, 2023
C++ STL offers many utilities to solve basic common life problems. Comparing values are always necessary, but sometimes we need to compare the strings also. Therefore, lexicographical_compare() is used to compare strings.
It is commonly used in dictionaries to arrange words alphabetically; it entails comparing elements that have the same position in both ranges consecutively against each other until one element is not equal to the other. The lexicographical comparison is the consequence of comparing these first non-matching components. This function is defined in <algorithm> header.
Time complexity= 2 * min(N1, N2)
where N1 = std::distance(beg1, end1) and N2 = std::distance(beg2, end2).
It has the following two implementations:
Syntax 1:
lexicographical_compare(iter1 beg1, iter1 end1, iter2 beg2, iter2 end2)
Template:
template
bool lexicographical_compare(iter1 beg1, iter1 end1,
iter2 beg2, iter2 end2)
{
while (beg1!=end1)
{
if (beg2==end2 || *beg2<*beg1) return false;
else if (*beg1<*beg2) return true;
++beg1; ++beg2;
}
return (beg2!=end2);
}
Parameters:
- beg1: Input iterator to the initial position of the first sequence.
- end1: Input iterator to the final position of the first sequence.
- beg2: Input iterator to initial position of the second sequence.
- end2: Input iterator to the final position of the second sequence.
Return value: Returns a boolean true, if range1 is strictly lexicographically smaller than range2 else returns a false.
Example:
CPP
#include <algorithm>
#include <iostream>
using namespace std;
int main()
{
char one[] = "geeksforgeeks" ;
char two[] = "gfg" ;
if (lexicographical_compare(one, one + 13, two,
two + 3))
cout << "geeksforgeeks is lexicographically less "
"than gfg" ;
else
cout << "geeksforgeeks is not lexicographically "
"less than gfg" ;
}
|
Output
geeksforgeeks is lexicographically less than gfg
Syntax 2:
lexicographical_compare(iter1 beg1, iter1 end1, iter2 beg2, iter2 end2, Compare comp)
Template:
template
bool lexicographical_compare(iter1 beg1, iter1 end1,
iter2 beg2, iter2 end2)
{
while (beg1!=end1)
{
if (beg2==end2 || *beg2<*beg1) return false;
else if (*beg1<*beg2) return true;
++beg1; ++beg2;
}
return (beg2!=end2);
}
Parameters:
- beg1: Input iterator to the initial position of the first sequence.
- end1: Input iterator to the final position of the first sequence.
- beg2: Input iterator to initial position of the second sequence.
- end2: Input iterator to the final position of the second sequence.
- comp: The comparator function that returns a Boolean true/false of each element compared. This function accepts two arguments. This can be a function pointer or function object and cannot change values.
Return value: Returns a Boolean true, if range1 is strictly lexicographically smaller than range2 else returns a false.
Example:
CPP
#include <algorithm>
#include <iostream>
using namespace std;
bool comp( char s1, char s2)
{
return tolower (s1) < tolower (s2);
}
int main()
{
char one[] = "geeksforgeeks" ;
char two[] = "Gfg" ;
if (lexicographical_compare(one, one + 13, two,
two + 3))
cout << "geeksforgeeks is lexicographically less "
"than Gfg\n" ;
else
cout << "geeksforgeeks is not lexicographically "
"less than Gfg\n" ;
if (lexicographical_compare(one, one + 13, two, two + 3,
comp)) {
cout << "geeksforgeeks is lexicographically less " ;
cout << "than Gfg( case-insensitive )" ;
}
else {
cout << "geeksforgeeks is not lexicographically "
"less " ;
cout << "than Gfg( case-insensitive )" ;
}
}
|
Output
geeksforgeeks is not lexicographically less than Gfg
geeksforgeeks is lexicographically less than Gfg( case-insensitive )
Application: Comparing strings can be generally used in dictionary, where we need to place words in lexicographical order. An example of this can be to find the word which occurs 1st in the dictionary among a given set of words.
Output
The smallest string is : abacus
Exception in lexicographical_compare(): It throws an exception if either an element comparison or an operation on an iterator throws. If the arguments are invalid, they cause undefined behavior.
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