Coin Change II
A medium Dynamic Programming interview guide. Task statement, worked examples, intuition, and step-by-step solutions.
- Topic
- Dynamic Programming
- Sheets
- 2
- Core for
- 9 roles
The problem
You are given an integer array of coin denominations and a total amount. Each denomination has unlimited supply. Return the number of combinations that make up the amount. Order of coins does not matter.
Example 1
- Input
- amount = 7, coins = [1,3,4]
- Output
- 5
- Why
- The five combinations: 1x7, 1x4+3, 1x3+4, 1+3+3 and 3+4.
Example 2
- Input
- amount = 3, coins = [2]
- Output
- 0
- Why
- No combination of 2s can make 3.
Example 3
- Input
- amount = 10, coins = [10]
- Output
- 1
- Why
- Only one way: use the single 10 coin.
Constraints
- 1 <= coins.length <= 300
- 1 <= coins[i] <= 5000
- 0 <= amount <= 5000
How to think about it
Updated 2026-09-29To count combinations without duplicates like 1+2 and 2+1, enforce a canonical order: consider coins one denomination at a time. Once you finish with coin c, no future combinations are allowed to use c again.
Approaches, worst first
Recursive coin inclusion
time O(2^(amount + n)) · space O(amount + n)
Recurse with parameters (coin_index, remaining_amount). Either pick current coin again or move to the next coin. Without memoization, explores redundant states exponentially.
2D table
time O(n * amount) · space O(n * amount)
dp[i][j] holds combinations making sum j using the first i coin types. dp[i][j] = dp[i-1][j] + (j >= coins[i-1] ? dp[i][j - coins[i-1]] : 0).
1D forward DP arrayWrite this one
time O(n * amount) · space O(amount)
Maintain dp array of size amount + 1 with dp[0] = 1. Outer loop runs through coins; inner loop sweeps j from coin up to amount, adding dp[j - coin] to dp[j].
Where people lose marks · 3
- Inverting loop orders. Looping amount on the outside and coins on the inside computes permutations (Combination Sum IV) rather than unordered combinations.
- Amount 0 has exactly 1 valid combination: the empty set of coins. Setting dp[0] = 0 makes every sum 0.
- Integer overflow on combinations exceeding 32-bit limits in certain language environments.
The theory behind it
Dynamic Programming — the ground this problem stands on. All Dynamic Programming problems
What Dynamic Programming is
Dynamic programming is a method for solving a complex problem by breaking it into overlapping subproblems, solving each subproblem only once, and remembering the answers in a lookup table. Instead of recalculating identical questions over and over, future steps look up previous answers directly. By assembling these saved pieces from the bottom up or storing them during recursion, a task that would take billions of steps finishes in a fraction of a second.
When to reach for it
Reach for dynamic programming when questions ask for the maximum profit, minimum cost, total number of distinct ways to achieve a goal, or whether a target can be formed. Signals include overlapping choices where making a choice now affects what choices remain later, but greedy picking fails to find the true global optimum. If drawing a recursive decision tree reveals the same subproblem states repeating across branches, dynamic programming is needed.
How the pattern works
Identify the state variables that uniquely describe a subproblem, such as an array index and remaining capacity. Write the base cases first, representing states whose answers are known without calculation. Next, write the recurrence relation that expresses the current state using previously solved states, taking the minimum, maximum, or sum among your options. Build the solution either top-down by caching recursive returns in a memo table, or bottom-up by filling an array in topological dependency order. When each state depends only on the previous row, compress storage down to a single array.
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Target Roles
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