Given positive coin denominations and a nonnegative target amount, return the minimum number of coins needed to make that amount. Each denomination may be reused without limit. Return -1 when the target is unreachable, and return zero for target zero. With denominations [1, 3, 4] and target 6, two coins of value 3 beat the greedy choice 4 + 1 + 1.
Define one optimum per amount
Let best[a] be the minimum coins needed for amount a. Initialize best[0] to zero and every positive amount to an unreachable sentinel. Process amounts in ascending order. For each denomination c no greater than a, consider one coin c plus the best answer for amount a - c.
Take the smallest candidate over all denominations. Since every coin is positive, a - c is strictly smaller than a and its state has already been computed. Unlimited reuse is natural here: the earlier state may already contain more copies of c, and adding one more copy remains legal.
Trace a nongreedy solution
For [1, 3, 4], the table through amount six is [0, 1, 2, 1, 1, 2, 2]. At amount six, coin 1 gives one plus the five-amount answer, or three. Coin 3 gives one plus the three-amount answer, or two. Coin 4 gives one plus the two-amount answer, or three. The minimum is two.
The algorithm tests possible final coins rather than committing to the largest one. A greedy rule works for some denomination systems but needs a separate proof. The counterexample above shows that denomination size alone does not establish that proof for an arbitrary supplied set.
Preserve unreachable states safely
If no denomination can lead from a reachable smaller amount, the state remains unreachable. With coins [4, 6], all odd amounts are unreachable. A numeric infinity sentinel naturally stays infinite when one is added. A maximum-integer sentinel needs a guard before addition, because adding one can overflow and appear smaller than a valid count.
Some versions use target + 1 as a finite sentinel. Because every positive coin is at least one, no valid solution can require more than target coins. That bound makes target + 1 distinguishable under a target range that itself fits the type. Return -1 only after examining the final state's sentinel.
Prove optimality by the final coin
Any nonempty solution for amount a ends with some allowed denomination c. Removing that coin leaves a solution for a - c, whose coin count cannot beat the already computed optimum. Therefore the best solution for a must appear among the recurrence's candidates.
Conversely, every candidate from a reachable earlier state is a valid construction of a, so taking the minimum cannot invent an impossible solution. Starting at the correct zero state and proceeding upward proves every table entry by induction. The recurrence answers a minimum-count question; summing candidates would instead count combinations or sequences under a different contract.
Complexity and input-dependent limits
For target A and d denominations, time is O(A × d) and table storage is O(A). This is pseudopolynomial in the target's numeric value: a huge amount needs a huge table even if its decimal representation is short. An empty denomination list can make only amount zero. Zero and negative denominations are outside the positive-coin contract.
Test exact single-coin amounts, target zero, unreachable gaps, duplicate denomination values, denominations larger than the target, and the nongreedy [1, 3, 4] case. Compare small targets with a breadth-first search over reachable amounts, where each edge adds one coin. A follow-up can reconstruct the chosen coins by storing the last improving denomination alongside each table entry.
Kotlin implementation
Int.MAX_VALUE is guarded before incrementing. The primitive IntArray keeps one minimum count for each amount through the target.
The course constructs the class shown here and calls its declared methods. Nullable node references require explicit handling; a non-null assertion relies on an earlier condition that proves a value exists. Primitive arrays and mutable collections can be changed through shared references. Kotlin Int arithmetic is bounded, so use Long intermediates when a wider problem contract requires them.
// Stands in for Infinity: no valid coin combination for an amount (yet)
const val INFINITY = Int.MAX_VALUE
class Solution {
fun minCoinsForChange(denominations: IntArray, target: Int): Int {
/*
Init all values to INFINITY which indicates there isn't a valid coin combination for an amount (yet)
This is our dp / memo array that we're using to memoize the solution to subproblems
*/
val minCoinsForAmount = IntArray(target + 1) { INFINITY }
// Bottom-up, start with 0 which is always 0
minCoinsForAmount[0] = 0
/*
Begin at the smallest amount and build a running table of
the smallest values required to reach EVERY amount up to the target
*/
for (amount in minCoinsForAmount.indices) {
// Try each denomination to reach the current amount
for (coin in denominations) {
// The coin must fit, and (amount - coin) must be reachable:
// INFINITY + 1 would overflow an Int
if (coin <= amount && minCoinsForAmount[amount - coin] != INFINITY) {
// The current minimum (subproblem solution) found at this amount.
val existingMinCoinForAmount = minCoinsForAmount[amount]
/*
Subtract (amount - coin) to find the min coins to reach this location from a prior value.
Always increment by +1 because we are trying one additional coin.
*/
val minAmountWithCurrentCoin = minCoinsForAmount[amount - coin] + 1
/*
Memoize the minimum coins to reach the current amount.
This is either a minimum we previously calculated or a new minimum we found with the current coin.
If there isn't a valid combination, this just remains INFINITY.
*/
minCoinsForAmount[amount] = minOf(existingMinCoinForAmount, minAmountWithCurrentCoin)
}
}
}
/*
If we didn't find a combination that matched our target (it's still INFINITY), return -1.
Otherwise, the target does exist for that amount and we return it.
*/
return if (minCoinsForAmount[target] == INFINITY) -1 else minCoinsForAmount[target]
}
}
Practice this implementation
Open Vending Machine to run the implementation against the course tests, then change one boundary condition and explain the first failing case. The editor supplies the starter signature and any node types or helper APIs used above. Copy the implementation into that environment; the snippet is the solution body, without the input parser or test driver. Your saved progress belongs to the original lesson.
For more Kotlin conventions and practice problems, use the Kotlin interview guide.