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Cover image for LeetCode 518. Coin Change 2

LeetCode 518. Coin Change 2

cod3pineapple profile image codingpineapple ・2 min read

LeetCode: https://leetcode.com/problems/coin-change-2/

Description:

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Example 1:

Input: amount = 5, coins = [1, 2, 5]
Output: 4
Explanation: there are four ways to make up the amount:
5=5
5=2+2+1
5=2+1+1+1
5=1+1+1+1+1
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Example 2:

Input: amount = 3, coins = [2]
Output: 0
Explanation: the amount of 3 cannot be made up just with coins of 2.
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Solution:

n = coins.length
m = amount
Time Complexity: O(n*m)
Space Complexity: O(m)

// We will be using tabulation in this solution.
// Create a table (rows: (0 -> number of coins) x columns: (0 -> amount))
// rows -> coins we can use 
// columns -> amount we want to make
// cells -> total number of ways we can make the amount

var change = function(amount, coins) {
    // Create a table using a 2D array to store possible ways to make the amount
    // Inititate every cell to 0
    let dp = new Array(coins.length + 1).fill(0).map(()=>new Array(amount + 1).fill(0));
    // The first row in our table handles the case of using 0 coins to make a certain amount
    // There is only way to make an amount of 0 when using 0 coins
    dp[0][0] = 1;
    // Loop through all the rows in the table
    for (let rowIndex=1;rowIndex<dp.length;rowIndex++) {
        // There is 1 way to make an amount of 0 with any coin (which is to not use it)
        dp[rowIndex][0] = 1;
        // Loop through all columns in the row and find the total number of ways to make each amount
        for (let currentAmount=1; currentAmount<dp[0].length; currentAmount++) {
            // We build off of previous solutions by adding all the previous ways we 
            // made the current amount by looking at the cell located in the row above
            // our current cell [rowIndex-1] and in the same column [currentAmount]
            dp[rowIndex][currentAmount] = dp[rowIndex-1][currentAmount];
            // Is the coin we are currently looking is less than the current amount?
            // If it is then we will use the solution from a previous cell in the row of this coin
            // and add it to the current cell
            if (coins[rowIndex-1] <= currentAmount) {
                // Subtract the current coin from the current amount
                // Go to the cell in the row at [current amount - current coin]
                // Add the total number of ways in this cell to our current cell
                dp[rowIndex][currentAmount]+= dp[rowIndex][currentAmount-coins[rowIndex-1]];
            }
        }
    }

    return dp[coins.length][amount];
};

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