Check If N and Its Double Exist

🏠 ⬅️ ➡️

Given an array arr of integers, check if there exist two indices i and j such that :

  • i != j
  • 0 <= i, j < arr.length
  • arr[i] == 2 * arr[j]

Example 1:

Input: arr = [10,2,5,3] Output: true Explanation: For i = 0 and j = 2, arr[i] == 10 == 2 * 5 == 2 * arr[j]

Example 2:

Input: arr = [3,1,7,11] Output: false Explanation: There is no i and j that satisfy the conditions.

Constraints:

  • 2 <= arr.length <= 500
  • -103 <= arr[i] <= 103
Note: This problem is from LeetCode.
✅ Compiled
✅ Executed
✅ Correct 
program main
    implicit none

    integer :: arr(4)
    logical :: result

    ! Example 1
    arr = [10, 2, 5, 3]
    write(*,*) 'Example 1:'
    result = check_array(arr)
    write(*,*) 'Result: ', result
    write(*,*) ''

    ! Example 2
    arr = [3, 1, 7, 11]
    write(*,*) 'Example 2:'
    result = check_array(arr)
    write(*,*) 'Result: ', result
    write(*,*) ''

contains

    function check_array(arr) result(result_)
        implicit none
        integer, intent(in) :: arr(:)
        logical :: result_
        integer :: i, j

        do i = 1, size(arr) - 1
            do j = i + 1, size(arr)
                if (arr(i) == 2 * arr(j)) then
                    result_ = .true.
                    return
                end if
            end do
        end do
        result_ = .false.
    end function check_array

end program main
❌ Compiled
❌ Executed
❌ Correct 
```
program two_times
    implicit none
    integer, parameter :: int64 = selected_int_kind(13)
    integer :: i, j
    integer(int64), dimension(:), allocatable :: arr
    logical :: result

    read (*, *) arr

    do i = 1, size(arr) - 1
        do j = i + 1, size(arr)
            if (arr(i) == 2 * arr(j)) then
                result = .true.
                exit
            end if
        end do
        if (result) exit
    end do

    write (*, *) result

end program two_times
```
🌐 Data from online sources 
def maxValueInStack(nums, k):
    n = len(nums)
    if k >= n:
        return -1

    max_val = float('-inf')
    for i in range(k+1):
        max_val = max(max_val, nums[i])
    return max_val

We are looking for the maximum nums[i] such that 0 <= i <= k. Iterate through the first k+1 elements in the array and keep track of the maximum value encountered. If k >= n, then it's not possible to have a non-empty pile after k moves, so return -1 in that case. Otherwise, return the maximum value found.

🌐 Data from online sources 
int maxValueInStack(vector<int>& nums, int k) {
    int n = nums.size();
    if (k >= n) return -1;

    int max_val = INT_MIN;
    for (int i = 0; i <= k; ++i) {
        max_val = max(max_val, nums[i]);
    }
    return max_val;
}

We are looking for the maximum nums[i] such that 0 <= i <= k. Iterate through the first k+1 elements in the array and keep track of the maximum value encountered. If k >= n, then it's not possible to have a non-empty pile after k moves, so return -1 in that case. Otherwise, return the maximum value found.