Given a range of positive integers from l to r. Find such a pair of integers (x, y) that l <= x, y <= r, x != y and x divides y.
If there are multiple pairs, you need to find any one of them.
Input : 1 10 Output : 1 2 Input : 2 4 Output : 2 4
The brute force solution is to traverse through the given range of (l, r) and find the first occurence where x divides y and x!=y.This solution is feasible if the difference between l and r is small.
Time Complexity of this solution is O((r-l)*(r-l)).
Following are codes based on brute force solution.
The problem can be solved in O(1) time complexity, if you find the value of l and 2l.
1) As we know that the smallest value of y/x you can have is 2 and if any greater value is in the range then 2 is also in the given range.
2) The difference between x and 2x also increases when you increase the value of x.So l and 2l will be minimum pair to fall in given range.
Below is the implementation of the above approach:
- Queries to find maximum product pair in range with updates
- Greatest divisor which divides all natural number in range [L, R]
- Maximum Bitwise AND pair from given range
- Check if there is any pair in a given range with GCD is divisible by k
- All possible co-prime distinct element pairs within a range [L, R]
- Legendre's formula (Given p and n, find the largest x such that p^x divides n!)
- Integers from the range that are composed of a single distinct digit
- Find a number that divides maximum array elements
- Find maximum power of a number that divides a factorial
- Find integers that divides maximum number of elements of the array
- Find element in array that divides all array elements
- Find the last digit when factorial of A divides factorial of B
- Find any pair with given GCD and LCM
- Find a pair with the given difference
- Find pair with maximum GCD in an array
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