A recursive lambda expression is the process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. Using a recursive algorithm, certain problems can be solved quite easily. Examples of such problems are Towers of Hanoi (TOH), Inorder/Preorder/Postorder Tree Traversals, DFS of Graph, etc.
A recursive function is a kind of loop structure only. The difference is it maintains a memory stack on its own. Obviously, it must have a break condition like for and while loop. Hence, the recursive function has the following structure-
function name(arguments) { a base case (a breaking condition) recursive code (the actual logic) }
int fact(int n) { // Base Case if (n < = 1) return 1; else return n * fact(n - 1); }
C++ 11 introduced lambda expressions to allow us to write an inline function that can be used for short snippets of code that are not going to be reused and not worth naming. In its simplest form, the lambda expression can be defined as follows:
[ capture clause ] (parameters) -> return-type { definition of method }
Program 1:
Below is the program for the lambda expressions in the sort() method in C++:
C++14
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using
namespace
std;
// Driver code
int
main()
{
int
arr[] = { 5, 2, 1, 4, 3 };
sort(arr, arr + 5, [](
int
& a,
int
& b) {
// Instant lambda function
return
a > b;
});
for
(
int
i = 0; i < 5; i++)
cout << arr[i] <<
" "
;
return
0;
}
Output
5 4 3 2 1
Time complexity: O(nlogn)
Auxiliary Space: O(1)
Explanation: Here, the instant lambda function is used, which cannot be used in another sorting method, i.e., the same code needs to be written again. This lambda expression exists only during the execution of the sort() method. But it is possible to store the lambda expression in a variable (better to call it a function) as well, like below. By storing it, it can be used further, and of course, perform recursion also.
Program 2:
C++14
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using
namespace
std;
// Stored lambda expression
auto
cmp = [](
int
& a,
int
& b) {
return
a > b;
};
// Driver code
int
main()
{
int
arr[] = { 5, 2, 3, 1, 4 };
sort(arr, arr + 5, cmp);
for
(
int
i = 0; i < 5; i++)
cout << arr[i] <<
" "
;
return
0;
}
Output
5 4 3 2 1
Time complexity: O(nlogn)
Auxiliary Space: O(1)
Explanation: Here, auto cmp is the lambda stored function that can be used in as many sorting methods.
Recursion Using Lambda: Let’s discuss the concept of recursion and lambda expressions together by considering the recursive function below:
Program 3:
C++14
// C++ program to implement
// the above approach
#include <iostream>
using
namespace
std;
// Recursive function to print
// the digits of a number
void
printReverse(
int
n)
{
if
(n == 0)
return
;
cout << n % 10 <<
" "
;
printReverse(n / 10);
}
// Driver code
int
main()
{
int
n = 12345;
printReverse(n);
return
0;
}
Output
5 4 3 2 1
Time complexity: O(logn)
Auxiliary Space: O(logn)
Explanation: Above is the function which prints the digits of a number in reverse order, using recursion. The same can be done using the lambda expression.
Program 4:
Below is the C++ program to implement the above code using lambda expressions:
C++14
// C++ program to implement
// the above approach
#include <iostream>
using
namespace
std;
// Driver code
int
main()
{
int
n = 12345;
// Recursive lambda function to
// print the digits of a number
auto
printReverse = [&]() {
if
(n == 0)
return
;
cout << n % 10;
n = n / 10;
printReverse();
// As it is a part of main body,
// semicolon is must
};
printReverse();
return
0;
}
Output:
Oops, the function, defined using auto return type, must be deducted first (Compiler must be able to determine whether the return type auto can be converted into void or int or something else)
Explanation: Here, how the function is going to recognize the variable n, even if it is not passed as an argument. A lambda with an empty capture clause [ ] can access only those variables which are local to it. Here, the capture close [&] is used, which allows the function to access the variable n ( The actual value of variable n is being changed). There are two ways to resolve the error above:
1. By passing the function itself, to the function argument:
Below is the C++ program to implement the above concept:
Program 5:
C++14
// C++ program to implement
// the above approach
#include <iostream>
using
namespace
std;
// Driver code
int
main()
{
int
n = 12345;
// Function itself as a parameter
auto
printReverse = [&](
auto
&& printReverse) {
if
(n == 0)
return
;
cout << n % 10 <<
" "
;
n = n / 10;
printReverse(printReverse);
};
// Function as an argument
printReverse(printReverse);
return
0;
}
Output
5 4 3 2 1
Time complexity: O(logn)
Auxiliary Space: O(logn)
2. By declaring the function first:
Declaring a function means declaring its name and parameters type in order to inform a compiler that there is a function named xyz, which will write its body later.
Program 6:
Below is the C++ program to implement the above concept:
C++
// C++ program to implement
// the above approach
#include <iostream>
using
namespace
std;
// Declaring of a function
void
declaringFunction(string s);
// Driver code
int
main()
{
declaringFunction(
"Hello I am learning how to declare a function"
);
return
0;
}
// Body of a function
void
declaringFunction(string s)
{
cout << s;
}
Output
Hello I am learning how to declare a function
Time complexity: O(1)
Auxiliary Space: O(1)
Program 7:
Since it is a lambda function, there must be some unique way of declaring the function. Below is the C++ program for the same:
C++14
// C++ program to implement
// the above approach
#include <functional>
#include <iostream>
using
namespace
std;
// Driver code
int
main()
{
int
n = 12345;
// Function < return type (parameter
// types) > functionName
// Don't forget to include functional
// header
// Declaration
function<
void
()> printReverse;
printReverse = [&]() {
if
(n == 0)
return
;
// Definition
cout << n % 10 <<
" "
;
n /= 10;
printReverse();
};
printReverse();
}
Output
5 4 3 2 1
Time complexity: O(logn)
Auxiliary Space: O(logn)
Explanation: In the above code, first, the function printReverse is declared, then we have defined its body. Instead of that, we can directly declare the function and its body along with, which is known as defining a function. Below is the C++ program to implement the above approach:
Program 8:
C++14
// C++ program to implement
// the above approach
#include <functional>
#include <iostream>
using
namespace
std;
// Driver code
int
main()
{
int
n = 12345;
// Function < return type (parameter
// types) > functionName
function<
void
()> printReverse = [&]() {
if
(n == 0)
return
;
// Declaration + Body
cout << n % 10 <<
" "
;
n /= 10;
printReverse();
};
printReverse();
}
Output
5 4 3 2 1
Time complexity: O(logn)
Auxiliary Space: O(logn)
Examples:Below are some examples of recursive functions using lambda expressions.
Program 9:
C++14
// C++ program to implement
// the above approach
#include <iostream>
using
namespace
std;
// Driver code
int
main()
{
int
n = 6;
// Recursive Lambda function to
// find the factorial of a number
auto
factorial = [&](
auto
&& factorial) {
if
(n == 1)
return
n;
return
n-- * factorial(factorial);
};
auto
factorial2 = [](
int
n,
auto
&& factorial2) {
if
(n == 1)
return
n;
return
n * factorial2(n - 1, factorial2);
};
// Given n = 6
cout << factorial(factorial) << endl;
// Given n = 5
cout << factorial2(5, factorial2);
}
Output
720 120
Time complexity: O(2n)
Auxiliary Space: O(n)
Explanation: In the factorial function, n is directly accessed using the [&] capture clause. In the factorial2 function, n is passed as an argument. Hence, there is no need for the [&] clause.
Program 10:
C++14
// C++ program to implement
// the above approach
#include <iostream>
using
namespace
std;
// Driver code
int
main()
{
// Sorted array
int
arr[] = { 1, 2, 5, 7, 10, 12, 15 };
int
size = 7;
// Item to be searched
int
key = 10;
auto
binarySearch = [&](
int
startIndex,
int
endIndex,
auto
&& binarySearch) {
if
(startIndex > endIndex)
return
-1;
int
midIndex = (startIndex + endIndex) / 2;
if
(arr[midIndex] == key)
return
midIndex;
if
(arr[midIndex] > key)
return
binarySearch(startIndex,
midIndex - 1,
binarySearch);
if
(arr[midIndex] < key)
return
binarySearch(midIndex + 1,
endIndex,
binarySearch);
// Not found
return
-1;
};
int
index = binarySearch(0, size - 1,
binarySearch);
if
(index == -1)
cout <<
"Not found"
;
else
cout <<
"Found on index "
<< index;
return
0;
}
Output
Found on index 4
Time complexity: O(logn)
Auxiliary Space: O(1)
My Personal Notes arrow_drop_up
Last Updated : 17 Aug, 2022
Like Article
Save Article