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Commit 1023737

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Merge pull request fnplus#625 from Devamjoshi3/master
Added Make Change
2 parents f2b5302 + cb0ef8c commit 1023737

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import java.io.BufferedReader;
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import java.io.IOException;
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import java.io.InputStreamReader;
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/**
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* Given a list of coins of distinct denominations and total amount of money. Output the minimum number of coins required to make up that amount. Output -1 if that money cannot be made up using given coins.
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* You may assume that there are infinite numbers of coins of each type.
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*
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* Input:
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* The first line contains 'T' denoting the number of testcases. Then follows description of testcases.
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* Each cases begins with the two space separated integers 'n' and 'amount' denoting the total number of distinct coins and total amount of money respectively.
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* The second line contains n space-separated integers A1, A2, ..., An denoting the values of coins.
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*
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* Output:
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* For each testcase, in a new line, print the minimum number of coins required to make up that amount, print -1 if it is impossible to make that amount using given coins.
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*
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* Constraints:
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* 1<=T<=100
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* 1<=n<=100
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* 1<=Ai<=1000
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* 1<=amount<=100000
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*
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* Example:
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* Input :
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* 2
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* 3 11
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* 1 2 5
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* 2 7
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* 2 6
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* Output :
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* 3
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* -1
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*/
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public class MakeChange {
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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int t = Integer.parseInt(br.readLine().trim());
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while ( t-- > 0){
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String line = br.readLine();
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String[] strs = line.trim().split("\\s+");
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int n = Integer.parseInt(strs[0]);
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int sum = Integer.parseInt(strs[1]);
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line = br.readLine();
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strs = line.trim().split("\\s+");
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int[] a = new int[n];
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for (int i=0; i<n; i++){
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a[i] = Integer.parseInt(strs[i]);
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}
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System.out.println(minCoin(n, sum, a));
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}
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}
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private static int minCoin(int n, int sum, int[] a) {
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int[] lookup = new int[sum+1];
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for(int i=0; i<sum+1; i++){
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lookup[i] = Integer.MAX_VALUE;
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}
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lookup[0] = 0;
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for (int i=1; i<sum+1; i++){
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minCoinUtil(i, lookup, a, n);
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}
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return lookup[sum] == Integer.MAX_VALUE? -1:lookup[sum];
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}
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private static void minCoinUtil(int k, int[] lookup, int[] a, int n) {
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int min = Integer.MAX_VALUE;
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for (int i=0; i<n; i++){
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if(k-a[i] < 0 || lookup[k-a[i]] == Integer.MAX_VALUE)
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continue;
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if (1+lookup[k - a[i]] < min){
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min = 1+ lookup[k -a[i]];
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}
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}
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lookup[k] = min;
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}
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}
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class MatrixChainMultiplication
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{
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// Function to find the most efficient way to multiply
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// given sequence of matrices
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public static int MatrixChainMultiplication(int[] dims, int i, int j)
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{
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// base case: one matrix
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if (j <= i + 1) {
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return 0;
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}
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// stores minimum number of scalar multiplications (i.e., cost)
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// needed to compute the matrix M[i+1]...M[j] = M[i..j]
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int min = Integer.MAX_VALUE;
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// take the minimum over each possible position at which the
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// sequence of matrices can be split
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/*
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(M[i+1]) x (M[i+2]..................M[j])
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(M[i+1]M[i+2]) x (M[i+3.............M[j])
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...
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...
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(M[i+1]M[i+2]............M[j-1]) x (M[j])
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*/
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for (int k = i + 1; k <= j - 1; k++)
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{
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// recur for M[i+1]..M[k] to get a i x k matrix
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int cost = MatrixChainMultiplication(dims, i, k);
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// recur for M[k+1]..M[j] to get a k x j matrix
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cost += MatrixChainMultiplication(dims, k, j);
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// cost to multiply two (i x k) and (k x j) matrix
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cost += dims[i] * dims[k] * dims[j];
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if (cost < min) {
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min = cost;
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}
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}
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// return min cost to multiply M[j+1]..M[j]
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return min;
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}
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// main function
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public static void main(String[] args)
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{
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// Matrix M[i] has dimension dims[i-1] x dims[i] for i = 1..n
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// input is 10 ×ばつ 30 matrix, 30 ×ばつ 5 matrix, 5 ×ばつ 60 matrix
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int[] dims = { 10, 30, 5, 60 };
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System.out.print("Minimum cost is " +
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MatrixChainMultiplication(dims, 0, dims.length - 1));
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}
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}

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