Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.
The value of |x| is defined as:
x if x >= 0.-x if x < 0.Example 1:
Input: nums = [1,2,2,1], k = 1 Output: 4 Explanation: The pairs with an absolute difference of 1 are: - [1,2,2,1] - [1,2,2,1] - [1,2,2,1] - [1,2,2,1]
Example 2:
Input: nums = [1,3], k = 3 Output: 0 Explanation: There are no pairs with an absolute difference of 3.
Example 3:
Input: nums = [3,2,1,5,4], k = 2 Output: 3 Explanation: The pairs with an absolute difference of 2 are: - [3,2,1,5,4] - [3,2,1,5,4] - [3,2,1,5,4]
Constraints:
1 <= nums.length <= 2001 <= nums[i] <= 1001 <= k <= 99program main
! Solve the problem described in the task
implicit none
integer, parameter :: n = 5
integer, dimension(n) :: nums = (/ 1, 2, 2, 1, 3 /)
integer, parameter :: k = 1
integer :: i, j
integer :: count
count = 0
do i = 1, n - 1
do j = i + 1, n
if (abs(nums(i) - nums(j)) == k) then
count = count + 1
end if
end do
end do
print '(I0)', count
end program main6
module pairwise_difference
implicit none
private
public :: pairwise_difference_solution
contains
function pairwise_difference_solution(nums, k) result(count)
integer, intent(in) :: nums(:)
integer, intent(in) :: k
integer :: count
integer :: i, j
count = 0
do i = 1, size(nums) - 1
do j = i + 1, size(nums)
if (abs(nums(i) - nums(j)) == k) then
count = count + 1
end if
end do
end do
end function pairwise_difference_solution
end module pairwise_difference
program pairwise_difference_main
use pairwise_difference
implicit none
integer, parameter :: nums1(4) = [1, 2, 2, 1]
integer, parameter :: nums2(2) = [1, 3]
integer, parameter :: nums3(5) = [3, 2, 1, 5, 4]
integer, parameter :: k1 = 1
integer, parameter :: k2 = 3
integer, parameter :: k3 = 2
integer :: count1, count2, count3
count1 = pairwise_difference_solution(nums1, k1)
count2 = pairwise_difference_solution(nums2, k2)
count3 = pairwise_difference_solution(nums3, k3)
write (*,*) "nums1:", nums1
write (*,*) "k1:", k1
write (*,*) "count1:", count1
write (*,*) "nums2:", nums2
write (*,*) "k2:", k2
write (*,*) "count2:", count2
write (*,*) "nums3:", nums3
write (*,*) "k3:", k3
write (*,*) "count3:", count3
end program pairwise_difference_main
nums1: 1 2 2 1 k1: 1 count1: 4 nums2: 1 3 k2: 3 count2: 0 nums3: 3 2 1 5 4 k3: 2 count3: 3
def chalk_replacer(chalk, k):
total_chalk = sum(chalk)
k %= total_chalk
for i, val in enumerate(chalk):
if k < val:
return i
k -= val
return 0
chalk array, representing the total chalk needed for one full cycle.k divided by the total chalk needed using modulo operation (k %= sum). This will determine the remaining chalk when the teacher starts the last cycle.chalk array:k) is less than the chalk needed by the current student (chalk[i] for C++ and Java, or val for Python and JavaScript), return the index of this student.k.Repeat this process until a student is found who will replace the chalk.
#include <vector>
int chalkReplacer(std::vector<int>& chalk, int k) {
long long sum = 0;
for (int i = 0; i < chalk.size(); i++) {
sum += chalk[i];
}
k %= sum;
for (int i = 0; i < chalk.size(); i++) {
if (k < chalk[i]) {
return i;
}
k -= chalk[i];
}
return 0;
}
chalk array, representing the total chalk needed for one full cycle.k divided by the total chalk needed using modulo operation (k %= sum). This will determine the remaining chalk when the teacher starts the last cycle.chalk array:k) is less than the chalk needed by the current student (chalk[i] for C++ and Java, or val for Python and JavaScript), return the index of this student.k.Repeat this process until a student is found who will replace the chalk.