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DAY9 - Dynamic Programming - Coin Change Solution
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// top-down approach | |
const memo = {}; | |
function coinChange(n) { | |
// base case | |
if (n <= 0) return Number.MAX_VALUE; | |
if (n === 1 || n === 3 || n === 4) return 1; | |
// induction case | |
if (memo[n]) return memo[n]; | |
memo[n] = Math.min(...[coinChange(n-1), | |
coinChange(n-3), | |
coinChange(n-4)]) + 1; | |
return memo[n]; | |
} | |
// bottom-up approach | |
function coinChange(n) { | |
const memo = {}; | |
// base case | |
memo[1] = memo[3] = memo[4] = 1; | |
memo[2] = 2; | |
// induction case | |
for (let i = 5; i <= n; i++) { | |
memo[i] = Math.min(...[memo[i-1], memo[i-3], memo[i-4]]) + 1; | |
} | |
return memo[n]; | |
} |
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