# Python Program for Coin Change

• Difficulty Level : Hard
• Last Updated : 22 Jun, 2022

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.

## Python3

 `# 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``)]` `    ``# 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 function``arr ``=` `[``1``, ``2``, ``3``]``m ``=` `len``(arr)``n ``=` `4``print``(count(arr, m, n))` `# This code is contributed by Bhavya Jain`

## Python3

 `# Dynamic Programming Python implementation of Coin``# Change problem``def` `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 function``arr ``=` `[``1``, ``2``, ``3``]``m ``=` `len``(arr)``n ``=` `4``x ``=` `count(arr, m, n)``print` `(x)` `# This code is contributed by Afzal Ansari`

#### Complexity Analysis:

For both the solutions the time and space complexity is same:

Time Complexity: O(n*m)

Space Complexity: O(n)

Though first solution might be a little slow as it has multiple loops.

Please refer complete article on Dynamic Programming | Set 7 (Coin Change) for more details!

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