Coin Change - II

Problem Statement:

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money.

Return the number of combinations that make up that amount. If that amount of money cannot be made up by any combination of the coins, return 0.

You may assume that you have an infinite number of each kind of coin.

The answer is guaranteed to fit into a signed 32-bit integer.

• Example:

  Example 1:
  
  Input: amount = 5, coins = [1,2,5]
  Output: 4
  Explanation: there are four ways to make up the amount:
  5=5
  5=2+2+1
  5=2+1+1+1
  5=1+1+1+1+1
  Example 2:
  
  Input: amount = 3, coins = [2]
  Output: 0
  Explanation: the amount of 3 cannot be made up just with coins of 2.
  Example 3:
  
  Input: amount = 10, coins = [10]
  Output: 1

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✅ Solution: Memoization

class Solution {
public:
    // Recursive function to count ways to reach 'target' using coins[0...ind]
    int solve(int ind, int target, vector<int> &coins, vector<vector<int>>&dp){
        // Base case: when using only the first coin
        if(ind == 0){
            return target % coins[0] == 0 ? 1 : 0;
        }

        if(dp[ind][target] != -1) return dp[ind][target];

        int notTake = solve(ind - 1, target, coins, dp);  // skip coin
        int take = 0;
        if(coins[ind] <= target){
            take = solve(ind, target - coins[ind], coins, dp);  // reuse coin
        }

        return dp[ind][target] = take + notTake;
    }

    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<vector<int>> dp(n, vector<int>(amount + 1, -1));
        return solve(n - 1, amount, coins, dp);
    }
};

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✅ Solution: Tabulation

class Solution {
public:
    int mod = 1e9 + 7;

    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<vector<int>> dp(n, vector<int>(amount + 1, 0));

        // Initialize base case: using only the first coin
        for(int t = 0; t <= amount; t++){
            if(t % coins[0] == 0) dp[0][t] = 1;
        }

        // Bottom-up DP
        for(int ind = 1; ind < n; ind++){
            for(int target = 0; target <= amount; target++){
                int notTake = dp[ind - 1][target]; // don't take current coin
                int take = 0;
                if(coins[ind] <= target){
                    take = dp[ind][target - coins[ind]]; // take current coin
                }
                dp[ind][target] = take + notTake;
            }
        }

        return dp[n - 1][amount];
    }
};

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✅ Solution: Space Optimized

class Solution {
public:
    int mod = 1e9 + 7;

    int change(int amount, vector<int>& coins) {
        int n = coins.size();
        vector<int> prev(amount + 1, 0);

        // Initialize base case
        for(int t = 0; t <= amount; t++){
            if(t % coins[0] == 0) prev[t] = 1;
        }

        // Bottom-up DP using rolling array
        for(int ind = 1; ind < n; ind++){
            vector<int> curr(amount + 1, 0);
            for(int target = 0; target <= amount; target++){
                int notTake = prev[target];
                int take = 0;
                if(coins[ind] <= target){
                    take = curr[target - coins[ind]];
                }
                curr[target] = take + notTake;
            }
            prev = curr;
        }

        return prev[amount];
    }
};

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📝 Revision Notes – Coin Change II

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✅ How It Works

• You're given a set of coins and a total amount.
• You need to find the number of combinations to make the amount (order doesn't matter).
• It's a classic unbounded knapsack variation:
  • You can take a coin as many times as you want.
  • But the order of coins does not matter.

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🧩 Key Formula / Recurrence

f(ind, target) =
    f(ind - 1, target)                  // not take current coin
  + f(ind, target - coins[ind])         // take current coin again

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⏱️ Time & Space Complexity

Approach	Time	Space
Memoization	O(N × Amount)	O(N × Amount)
Tabulation	O(N × Amount)	O(N × Amount)
Space Optimized	O(N × Amount)	O(Amount)

Where N = number of coins

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⚠️ Edge Cases

• amount = 0 → Always return 1 (empty set is a valid combination)
• Coins with large denominations → might have 0 combinations
• Duplicate denominations don't affect outcome since we consider combinations

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💡 Other Approaches

Type	Use Case
Brute Force (DFS)	❌ Too slow
Memoization ✅	Good for clarity
Tabulation ✅	Iterative and efficient
Space Optimized ✅	Best for constraints

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🔁 Related Problems

• Leetcode 518 – Coin Change II
• Leetcode 322 – Coin Change (minimum number of coins)
• GFG – Count number of ways to reach a given score
• Subset Sum / Unbounded Knapsack

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