feat ProgramOptimized.c

Greedy Algorithms/Indian denomination problem/ProgramOptimized.c

@@ -0,0 +1,52 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <limits.h>
+
+int main() {
+    // Array of currency notes
+    int arr[] = {1, 2, 5, 10, 20, 50, 100, 200, 500, 2000};
+    int n = sizeof(arr) / sizeof(arr[0]); // Number of currency notes
+
+    // Amount to be made
+    int amt;
+    printf("Enter amount: ");
+    scanf("%d", &amt);
+
+    // dp[i] will store the minimum number of notes required to make amount i
+    int dp[amt + 1];
+    int note[amt + 1]; // To store the last note used for each amount
+
+    // Initialize dp array with a large value (INT_MAX)
+    for (int i = 0; i <= amt; i++) {
+        dp[i] = INT_MAX;
+    }
+    dp[0] = 0; // Base case: 0 amount requires 0 notes
+
+    // Dynamic Programming to find minimum number of notes for each amount
+    for (int i = 1; i <= amt; i++) {
+        for (int j = 0; j < n; j++) {
+            if (arr[j] <= i && dp[i - arr[j]] != INT_MAX) {
+                if (dp[i] > dp[i - arr[j]] + 1) {
+                    dp[i] = dp[i - arr[j]] + 1;
+                    note[i] = arr[j]; // Store the note used
+                }
+            }
+        }
+    }
+
+    if (dp[amt] == INT_MAX) {
+        printf("No solution\n");
+    } else {
+        printf("Minimum number of notes: %d\n", dp[amt]);
+
+        // To print the notes used
+        printf("Notes used: ");
+        while (amt > 0) {
+            printf("%d ", note[amt]);
+            amt -= note[amt]; // Reduce the amount by the note used
+        }
+        printf("\n");
+    }
+
+    return 0;
+}