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Lecture 5 : Basic Recursion

Print some text n times using recursion

Approach1 (with extra parameter)

def printText(n,count =0):
    if count == n:
        return
    print("some-text,")
    printText(n,count +1)



printText(5)

some-text,
some-text,
some-text,
some-text,
some-text,

Approach2 without extra parameter

def printText(n):
    if n ==0:
        return
    print("some-text,")
    printText(n-1)



printText(5)

some-text,
some-text,
some-text,
some-text,
some-text,

Print natural numbers from 1 to n using recursion

Using forward recursion

This requires a count parameter to keep track

def printNatural(n, count =1):
    if count > n:
        return
    print(count)
    printNatural(n,count +1)

printNatural(10)

1
2
3
4
5
6
7
8
9
10

Using backward recursion or back tracking

this does not require additional parameter

def printNatural(n):
    if n == 0:
        return
    printNatural(n-1)
    print(n)
    

printNatural(10)

1
2
3
4
5
6
7
8
9
10

Print N to 1 using recursion

def printReverseNatural(n):
    if n ==0:
        return
    print(n)
    printReverseNatural(n-1)
    

printReverseNatural(5)

5
4
3
2
1

Sum of First N natural numbers using recursion

Problem statement: Given a number ‘N’, find out the sum of the first N natural numbers.

Examples:

Example 1:
Input: N=5
Output: 15
Explanation: 1+2+3+4+5=15

Example 2:
Input: N=6
Output: 21
Explanation: 1+2+3+4+5+6=21

approach 1 (recursion with extra parameter)

def sumNatural(n,sum =0):
    if n ==0:
        return sum

    return sumNatural(n-1,sum + n)
    

print(sumNatural(5))
print(sumNatural(6))

15
21

approach 2 (recursion without extra parameter)

sum(n) = n + sum(n-1) And sum(1) =1

def sumNatural(n):
    if n ==1:
        return 1
    return n + sumNatural(n-1)
    

print(sumNatural(5))
print(sumNatural(6))

15
21

Factorial of num

Example 1:
Input: X = 5
Output: 120
Explanation: 5! = 5*4*3*2*1

Example 2:
Input: X = 3
Output: 6
Explanation: 3!=3*2*1

fact(n) = n * fact(n-1) And fact(1) = 1

def fact(num):
    if num ==1:
        return 1
    return num * fact(num-1)

print(fact(5))
print(fact(3))

120
6

Reverse an array

Problem Statement: You are given an array. The task is to reverse the array and print it.

Examples:

Example 1:
Input: N = 5, arr[] = {5,4,3,2,1}
Output: {1,2,3,4,5}
Explanation: Since the order of elements gets reversed the first element will occupy the fifth position, the second element occupies the fourth position and so on.

Example 2:
Input: N=6 arr[] = {10,20,30,40}
Output: {40,30,20,10}
Explanation: Since the order of elements gets reversed the first element will occupy the fifth position, the second element occupies the fourth position and so on.

Approach (using iterative logic in recursion)

def reverseArray(array):
    n = len(array)
    def helper(left =0, right = n-1):
        if left >= right:
            return
        (array[left],array[right]) = (array[right],array[left])
        helper(left +1 , right -1) 


    helper()

    return array

print(reverseArray([5,4,3,2,1]))
print(reverseArray([10,20,30,40]))

[1, 2, 3, 4, 5]
[40, 30, 20, 10]

Approach (built in method)

def reverseArray(array):
    return array[::-1]

    

print(reverseArray([5,4,3,2,1]))
print(reverseArray([10,20,30,40]))

[1, 2, 3, 4, 5]
[40, 30, 20, 10]

Print fiboocii

Print Fibonacci Series up to Nth term

Examples:

Example 1:
Input: N = 5
Output: 0 1 1 2 3 5
Explanation: 0 1 1 2 3 5 is the fibonacci series up to 5th term.(0 based indexing)

Example 2:
Input: 6

Output: 0 1 1 2 3 5 8
Explanation: 0 1 1 2 3 5 8 is the fibonacci series upto 6th term.(o based indexing)

Fib(n) = Fib (n-1) + Fib(n-2) And Fib(0) = 0 And Fib(1) = 1

def printFib(n):
    def helper(idx =0,first =0 , second = 1):
        if idx > n:
            return

        print(first ,end =" ")
        helper(idx+1, second , first + second)
    
    helper()

printFib(5)
printFib(6)

0 1 1 2 3 5 0 1 1 2 3 5 8 

Check if the given String is Palindrome or not

Input: Str =  “ABCDCBA”
Output: True

Input: Str = “TAKE U FORWARD”
Output: False

Approach 1 (Iterative)

• use left and right ptr

def is_palindrome(s):
    L = len(s)

    left ,right = 0 , L-1
    while left < right:
        if s[left] != s[right]:
            return False

        left +=1
        right -=1

    return True

print(is_palindrome("ABCDCBA"))
print(is_palindrome("TAKE U FORWARD"))

True
False

Complexity

• L = len(s)
• O(N)
• O(1)

Approach 2 (Recursive)

• recursive check left and right without loop

def is_palindrome(s):
    L = len(s)

    def check(left_ptr,right_ptr):
        if left_ptr >= right_ptr:
            return True

        if s[left_ptr] != s[right_ptr]:
            return False

        return check(left_ptr +1, right_ptr-1)

    return check(0,L-1)

print(is_palindrome("ABCDCBA"))
print(is_palindrome("TAKE U FORWARD"))

True
False

Complexity

• L = len(s)
• O(N)
• O(N)