Coin Change (Bottom-Up)

Build a one-dimensional table where each amount stores the fewest coins needed to make it.

Algorithm

@steps

Initialize dp[0] = 0 and all other amounts to an unreachable sentinel.
Scan amounts from 1 through 6.
For each coin, read the earlier cell dp[amount - coin] when it exists.
Write the smallest candidate into the current amount.
Print both the final answer and the full DP array. @end @complexity

Time: O(target * coin_count)
Space: O(target) @end

bottom-up dynamic programming `dp[a]` is solved from already-computed smaller amounts, so every table cell has a visible dependency.

Fortran DSA Implementation

basic.f90

program main
  implicit none
  integer :: coins(3), dp(0:6), target, inf, amount, i, coin, candidate
  coins = [1, 3, 4]
  target = 6
  inf = target + 1
  dp = inf
  dp(0) = 0
  do amount = 1, target
    do i = 1, 3
      coin = coins(i)
      if (amount >= coin) then
        candidate = dp(amount - coin) + 1
        if (candidate < dp(amount)) dp(amount) = candidate
      end if
    end do
  end do
  print '(I0)', dp(target)
  call print_list(dp)
contains
  subroutine print_list(values)
    integer, intent(in) :: values(0:6)
    integer :: j
    write(*, '(A)', advance='no') '['
    do j = 0, 6
      if (j > 0) write(*, '(A)', advance='no') ', '
      write(*, '(I0)', advance='no') values(j)
    end do
    print '(A)', ']'
  end subroutine print_list
end program main

@end @output 2 [0, 1, 2, 1, 1, 2, 2] @end