Added 2 Base tests and 5 main tests for CoinChange Problem. Refactored the code and removed the Memoized approach as it was not necessary

This commit is contained in:
Mandy8055
2021-08-13 12:53:34 +05:30
parent 46a2aeb7cf
commit fa1524498b
2 changed files with 48 additions and 73 deletions

View File

@ -1,58 +1,21 @@
function change (coins, amount) {
/**
* @params {Array} coins
* @params {Number} amount
*/
export const change = (coins, amount) => {
// Create and initialize the storage
const combinations = new Array(amount + 1).fill(0)
combinations[0] = 1
// Determine the direction of smallest sub-problem
for (let i = 0; i < coins.length; i++) {
const coin = coins[i]
for (let j = coin; j < amount + 1; j++) {
combinations[j] += combinations[j - coin]
// Travel and fill the combinations array
for (let j = coins[i]; j < combinations.length; j++) {
combinations[j] += combinations[j - coins[i]]
}
}
return combinations[amount]
}
/** Coin-change combination using recursive approach along with Memoization
* @param {number} amount
* @param {number[]} coins
*/
const changeRecursive = (amount, coins) => {
const mem = new Map()
return coinChangeComb(amount, coins, 0, mem)
}
/** Coin-change combination using recursive approach along with Memoization
* @param {number} amount
* @param {number[]} coins
* @param {number} idx
* @param {Map} mem
*/
const coinChangeComb = (amount, coins, idx, mem) => {
// Negative Base Case
if (amount < 0 || idx === coins.length) {
return 0
}
// Positive Base Case
if (amount === 0) {
return 1
}
// Main Case
// Check if the recursive function call results is already memoized
if (mem.has(`${amount} - ${idx}`)) {
return mem.get(`${amount} - ${idx}`)
}
let res = 0
// Consider the coin at index idx
res += coinChangeComb(amount - coins[idx], coins, idx, mem)
// Leave the coin at index idx
res += coinChangeComb(amount, coins, idx + 1, mem)
// Cache the intermediate result in mem
mem.set(`${amount} - ${idx}`, res)
return res
}
function minimumCoins (coins, amount) {
function minimumCoins(coins, amount) {
// minimumCoins[i] will store the minimum coins needed for amount i
const minimumCoins = new Array(amount + 1).fill(0)
@ -77,28 +40,3 @@ function minimumCoins (coins, amount) {
}
return minimumCoins[amount]
}
function main () {
const amount = 12
const coins = [2, 4, 5]
console.log(
'Number of combinations of getting change for ' +
amount +
' is: ' +
change(coins, amount)
)
console.log(
'Number of combinations of getting change for ' +
amount +
' is: ' +
changeRecursive(coins, amount)
)
console.log(
'Minimum number of coins required for amount :' +
amount +
' is: ' +
minimumCoins(coins, amount)
)
}
main()