Given two integer N and K, the task is to print all positive numbers made up of N digits whose difference between the first and last digits equal to K.
Input: N = 2, K = 0
Output: 11, 22, 33, 44, 55, 66, 77, 88, 99
Input: N = 2, K = 9
Approach: The idea is to generate all possible 1-digit numbers to N-digit numbers using recursion and check if the difference between the first and the last digit of that number is equal to K or not. Below are the steps:
- Generate all possible numbers with length 1.
- At each step, keep on adding digits to the number until the length of the number becomes N.
- When the length of the number becomes equal to N, calculate the difference between the first and last digit of the number, and check if the difference is equal to N or not. If found to be true, print that number and proceed to generate the next number.
Below is the implementation of the above approach:
Time Complexity: O(2N)
Auxiliary Space: O(N)
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- Count of pairs (A, B) in range 1 to N such that last digit of A is equal to the first digit of B
- Count of N-digit Numbers having Sum of even and odd positioned digits divisible by given numbers
- Absolute difference between the first X and last X Digits of N
- Count of numbers between range having only non-zero digits whose sum of digits is N and number is divisible by M
- Count all numbers up to N having M as the last digit
- Find a number K having sum of numbers obtained by repeated removal of last digit of K is N
- Count numbers with same first and last digits
- Count n digit numbers not having a particular digit
- Count numbers formed by given two digit with sum having given digits
- Count of N digit Numbers having no pair of equal consecutive Digits
- Count total number of N digit numbers such that the difference between sum of even and odd digits is 1
- Print all n-digit numbers with absolute difference between sum of even and odd digits is 1
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