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"," * Write a function to compute the fewest number of coins that you need to make up that"," * amount. If that amount of money cannot be made up by any combination of the coins,"," * return -1."," * "," * Example 1:"," * coins = [1, 2, 5], amount = 11"," * return 3 (11 = 5 + 5 + 1)"," * "," * Example 2:"," * coins = [2], amount = 3"," * return -1."," * "," * Note:"," * You may assume that you have an infinite number of each kind of coin."," * "," * Credits:"," * Special thanks to @jianchao.li.fighter for adding this problem and creating all test "," * cases."," * "," ***************************************************************************************/","","","/* Recursive solution - TIME LIMIT ERROR */","class Solution {","public:"," int coinChange(vector\u003cint\u003e\u0026 coins, int amount) { "," int result = INT_MAX;"," if ( amount == 0 ) return 0;"," if ( amount \u003c 0 ) return -1;"," for (int i=0; i\u003ccoins.size(); i++) {"," if ( amount - coins[i] \u003c 0 ) continue;"," int r = coinChange(coins, amount - coins[i]);"," if ( r == -1 ) continue;"," if (result \u003e r ) result = r + 1;"," }"," return result == INT_MAX ? -1 : result;"," }","}","",""," /* "," * Solution 1 - O(N * amount)"," * ========="," *"," * This problem can be solved using dynamic programming, thus building the optimal"," * solution from previous smaller ones. For every coin of value t and every sum of money"," * i the sum can be traced back to a previous sum i - t that was already computed and uses"," * the smallest number of coins possible. This way we can construct every sum i as the"," * minimum of all these previous sums for every coin value. To be sure we'll find a minimum"," * we can populate the solution vector with an amount bigger than the maximum possible, "," * which is amount + 1(when the sum is made up of only coins of value 1). The only exception"," * is sol[0] which is 0 as we need 0 coins to have a sum of 0. 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