Python Program for Coin Change | DP-7
Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2, .. , Sm} valued coins, how many ways can we make the change? The order of coins doesn\’t matter.
For example, for N = 4 and S = {1,2,3}, there are four solutions: {1,1,1,1},{1,1,2},{2,2},{1,3}. So output should be 4. For N = 10 and S = {2, 5, 3, 6}, there are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}. So the output should be 5.
Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.
Following is a simple recursive implementation of the Coin Change problem.
Python3
# Recursive Python3 program for# coin change problem. # Returns the count of ways we can sum# S[0...m-1] coins to get sum ndef count(S, m, n ): # If n is 0 then there is 1 # solution (do not include any coin) if (n == 0): return 1 # If n is less than 0 then no # solution exists if (n < 0): return 0; # If there are no coins and n # is greater than 0, then no # solution exist if (m <=0 and n >= 1): return 0 # count is sum of solutions (i) # including S[m-1] (ii) excluding S[m-1] return count( S, m - 1, n ) + count( S, m, n-S[m-1] ); # Driver program to test above functionarr = [1, 2, 3]m = len(arr)print(count(arr, m, 4)) # This code is contributed by Smitha Dinesh Semwal |
Python
# Dynamic Programming Python implementation of Coin # Change problemdef 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)] # Fill the entries for 0 value case (n = 0) for i in range(m): table[0][i] = 1 # 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] # Driver program to test above functionarr = [1, 2, 3]m = len(arr)n = 4print(count(arr, m, n)) # This code is contributed by Bhavya Jain |
Following is a simplified version of method 2. The auxiliary space required here is O(n) only.
Python
# Dynamic Programming Python implementation of Coin # Change problemdef count(S, m, n): # table[i] will be storing the number of solutions for # value i. We need n+1 rows as the table is constructed # in bottom up manner using the base case (n = 0) # Initialize all table values as 0 table = [0 for k in range(n+1)] # Base case (If given value is 0) table[0] = 1 # Pick all coins one by one and update the table[] values # after the index greater than or equal to the value of the # picked coin for i in range(0,m): for j in range(S[i],n+1): table[j] += table[j-S[i]] return table[n] # Driver program to test above functionarr = [1, 2, 3]m = len(arr)n = 4x = count(arr, m, n)print (x) # This code is contributed by Afzal Ansari |
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