From 9bdfcf340fca3df050bdd91f17621d5b1a64f786 Mon Sep 17 00:00:00 2001
From: Sidhant Mishra <69967446+rishi457@users.noreply.github.com>
Date: Fri, 6 Oct 2023 18:36:00 +0530
Subject: [PATCH] Update coinProblem.py

In this optimized version of the code, we have improved the efficiency of calculating the number of ways to make change for a given amount 'n' using a set of coin denominations specified in the 'arr' array. Let's walk through the key changes and improvements:

1. Switching to 1D Array:
   - We have replaced the 2D array 'table' with a 1D list called 'table'.
   - This change simplifies the code and reduces memory consumption because we only need to keep track of the number of ways to make change for each amount from 0 to 'n'.

2. Initialization:
   - We initialize 'table[0]' to 1. This represents the base case where there is one way to make change for an amount of 0, which is by not using any coins.

3. Iterative Update:
   - We iterate through the coin denominations in the 'arr' array and the amounts from 'S[i]' to 'n'.
   - For each coin denomination, we iteratively update 'table[j]' by adding 'table[j - S[i]]'. This represents the number of ways to make change for 'j' by either including the current coin denomination 'S[i]' or excluding it.

By implementing these changes, we have reduced the time complexity of the code to O(n*m), where 'n' is the target amount and 'm' is the number of coin denominations. This optimization simplifies the code while making it more memory-efficient and faster for larger input values, providing a scalable solution for solving the Coin Change Problem.
---
 python/coinProblem.py | 31 ++++++-------------------------
 1 file changed, 6 insertions(+), 25 deletions(-)

diff --git a/python/coinProblem.py b/python/coinProblem.py
index a2ea978..b5a0642 100644
--- a/python/coinProblem.py
+++ b/python/coinProblem.py
@@ -1,34 +1,15 @@
-# Dynamic Programming Python implementation of Coin
-# Change problem
 def count(S, m, n):
-	# We need n+1 rows as the table is constructed
-	# in bottom up manner using the base case 0 value
-	# case (n = 0)
-	table = [[0 for x in range(m)] for x in range(n+1)]
+    table = [0] * (n + 1)
+    table[0] = 1  # Initialize the number of ways to make change for 0
 
-	# Fill the entries for 0 value case (n = 0)
-	for i in range(m):
-		table[0][i] = 1
+    for i in range(m):
+        for j in range(S[i], n + 1):
+            table[j] += table[j - S[i]]
 
-	# Fill rest of the table entries in bottom up manner
-	for i in range(1, n+1):
-		for j in range(m):
-
-			# Count of solutions including S[j]
-			x = table[i - S[j]][j] if i-S[j] >= 0 else 0
-
-			# Count of solutions excluding S[j]
-			y = table[i][j-1] if j >= 1 else 0
-
-			# total count
-			table[i][j] = x + y
-
-	return table[n][m-1]
+    return table[n]
 
 # Driver program to test above function
 arr = [1, 2, 3]
 m = len(arr)
 n = 4
 print(count(arr, m, n))
-
-# This code is contributed by Bhavya Jain