From 0cf16ef93bf29e6516c3cf9761a8cb38e678410b Mon Sep 17 00:00:00 2001
From: ARNAB GHOSH <66074296+ghosharnab969@users.noreply.github.com>
Date: Sat, 30 Oct 2021 21:13:03 +0530
Subject: [PATCH] Create Greedy -Algorithm_Finding-minimum-coins

Given a value V, if we want to make a change for V Rs, and we have an infinite supply of each of the denominations in Indian currency, i.e., we have an infinite supply of { 1, 2, 5, 10, 20, 50, 100, 500, 1000} valued coins/notes, what is the minimum number of coins and/or notes needed to make the change?
I have solved the above question using the Greedy Algorithm.
---
 cpp/Greedy -Algorithm_Finding-minimum-coins | 41 +++++++++++++++++++++
 1 file changed, 41 insertions(+)
 create mode 100644 cpp/Greedy -Algorithm_Finding-minimum-coins

diff --git a/cpp/Greedy -Algorithm_Finding-minimum-coins b/cpp/Greedy -Algorithm_Finding-minimum-coins
new file mode 100644
index 0000000..bf74ec2
--- /dev/null
+++ b/cpp/Greedy -Algorithm_Finding-minimum-coins	
@@ -0,0 +1,41 @@
+
+#include <bits/stdc++.h>
+using namespace std;
+  
+// All denominations of Indian Currency
+int deno[] = { 1, 2, 5, 10, 20,
+               50, 100, 500, 1000 };
+int n = sizeof(deno) / sizeof(deno[0]);
+  
+void findMin(int V)
+{
+    sort(deno, deno + n);
+  
+    // Initialize result
+    vector<int> ans;
+  
+    // Traverse through all denomination
+    for (int i = n - 1; i >= 0; i--) {
+  
+        // Find denominations
+        while (V >= deno[i]) {
+            V -= deno[i];
+            ans.push_back(deno[i]);
+        }
+    }
+  
+    // Print result
+    for (int i = 0; i < ans.size(); i++)
+        cout << ans[i] << " ";
+}
+  
+// Driver program
+int main()
+{
+    int n = 93;
+    cout << "Following is minimal"
+         << " number of change for " << n
+         << ": ";
+    findMin(n);
+    return 0;
+}