ADDED Coin Change - Minimum Coins.md

en/Dynamic Programming/Coin Change - Minimum Coins.md

@@ -0,0 +1,133 @@
+# Coin Change – Minimum Coins
+
+## Problem
+
+Given a total amount `N` and a set of coins `S = {S₁, S₂, …, Sₘ}`, find the **minimum number of coins** required to make the amount `N`.
+If the amount cannot be made with the given coins, return `-1`.
+
+---
+
+## Idea
+
+This is a variation of the **Unbounded Knapsack Problem**, where we have an infinite supply of each coin.
+The goal is to determine the smallest number of coins needed to sum to `N`.
+
+We define an array `minCoin[]`, where each element `minCoin[i]` stores the minimum number of coins required to make a total of `i`.
+The base case is `minCoin[0] = 0`.
+
+For every amount from `1` to `N`, and for each coin `c` in the set `S`, the relation is given by:
+
+```
+minCoin[i] = min( minCoin[i], 1 + minCoin[i - c] )
+```
+
+where `minCoin[i - c]` is the minimum number of coins required for the remaining amount.
+
+---
+
+## Implementation
+
+A straightforward C++ implementation might look as follows:
+
+```cpp
+#include <iostream>
+#include <limits.h>
+using namespace std;
+
+int myMin(int a, int b) {
+    return (a < b) ? a : b;
+}
+
+int coinChange(int coins[], int amount) {
+    if (amount < 1)
+        return 0;
+
+    int *minCoin = new int[amount + 1];
+    minCoin[0] = 0;
+
+    for (int i = 1; i <= amount; i++) {
+        minCoin[i] = INT_MAX;
+        for (int j = 0; j < 3; j++) {
+            if (coins[j] <= i && minCoin[i - coins[j]] != INT_MAX) {
+                minCoin[i] = myMin(minCoin[i], minCoin[i - coins[j]] + 1);
+            }
+        }
+    }
+
+    int result = (minCoin[amount] == INT_MAX) ? -1 : minCoin[amount];
+    delete[] minCoin;
+    return result;
+}
+
+int main() {
+    int coins[3] = {1, 2, 5};
+    int amount = 11;
+
+    int result = coinChange(coins, amount);
+    cout << "The minimum number of coins used: " << result << endl;
+
+    return 0;
+}
+```
+
+Note that `INT_MAX` is used to represent an unreachable amount during computation.
+
+---
+
+## Analysis
+
+### Space Complexity
+
+`O(N)` — only one array of size `amount + 1` is required.
+
+### Time Complexity
+
+`O(N × M)` — for each amount up to `N`, we iterate through all `M` available coin denominations.
+
+---
+
+## Walkthrough
+
+Finding the minimum number of coins for
+`coins = [1, 2, 5]` and `amount = 11`:
+
+| Amount | Coin Used | Remainder | Minimum Coins | Explanation      |
+| :----: | :-------- | :-------- | :------------ | :--------------- |
+|    1   | 1         | 0         | 1             | Using one 1-coin |
+|    2   | 2         | 0         | 1             | Using one 2-coin |
+|    3   | 1         | 2         | 2             | 1 + 2            |
+|    4   | 2         | 2         | 2             | 2 + 2            |
+|    5   | 5         | 0         | 1             | Using one 5-coin |
+|    6   | 5         | 1         | 2             | 5 + 1            |
+|    7   | 5         | 2         | 2             | 5 + 2            |
+|    8   | 5         | 3         | 3             | 5 + 3            |
+|    9   | 5         | 4         | 3             | 5 + 4            |
+|   10   | 5         | 5         | 2             | 5 + 5            |
+|   11   | 5         | 6         | 3             | 5 + 5 + 1        |
+
+The result is:
+
+```
+minCoin[11] = 3
+```
+
+Thus, the minimum number of coins required to make `11` is **3**, using coins `{5, 5, 1}`.
+
+---
+
+## Applications
+
+* Currency exchange and change-making systems
+* Resource optimization problems
+* Variants of the subset-sum and knapsack problems
+* Payment gateway and vending machine algorithms
+
+---
+
+## Resources
+
+* [GeeksforGeeks – Minimum Coins to Make Change](https://www.geeksforgeeks.org/find-minimum-number-of-coins-that-make-a-change/)
+* [YouTube – Coin Change (Minimum Coins) by Tushar Roy](https://youtu.be/NNcN5X1wsaw?si=UHED-voB4f3viOgT)
+
+---
+