Recursive lambda expressions in C++
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++ 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;
}
// C++ program to implement// the above approachusing namespace std;// Driver codeint 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++ 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;
}
// C++ program to implement// the above approachusing namespace std;// Stored lambda expressionauto cmp = [](int& a, int& b) { return a > b;};// Driver codeint 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++ 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;
}
// C++ program to implement// the above approachusing namespace std;// Recursive function to print// the digits of a numbervoid printReverse(int n){ if (n == 0) return; cout << n % 10 << " "; printReverse(n / 10);}// Driver codeint 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++ 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:
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++ 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;
}
// C++ program to implement// the above approachusing namespace std;// Driver codeint 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++ 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;
}
// C++ program to implement// the above approachusing namespace std;// Declaring of a functionvoid declaringFunction(string s);// Driver codeint main(){ declaringFunction("Hello I am learning how to declare a function"); return 0;}// Body of a functionvoid 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++ 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();
}
// C++ program to implement// the above approachusing namespace std;// Driver codeint 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++ 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();
}
// C++ program to implement// the above approachusing namespace std;// Driver codeint 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++ 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);
}
// C++ program to implement// the above approachusing namespace std;// Driver codeint 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++ 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;
}
// C++ program to implement// the above approachusing namespace std;// Driver codeint 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)
