Collective Interview Question for Software Engineer / Developers


Country: India
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
0
of 0 vote

It can be solved using DP.

initilize array C[n] with 1 for C[1-n]
then using following formula

C[0]=0
and
k0
C[k]=min(C[i]+C[k-i])
i=0

C[v] will be the minimum coins required for V value.

- Anonymous April 27, 2014 | Flag Reply
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0
of 0 vote

i=k
C[k]=min(C[i]+C[k-i)
i=0

- Anonymous April 27, 2014 | Flag Reply
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0
of 0 vote

Working Java solution using dynamic programming.

import java.util.HashMap;

public class MakeChange {

    public static int[] makeChange(int value, int[] coinValues) {
        HashMap<Integer, int[]> map = new HashMap<Integer, int[]>();
        return makeChangeHelper(value, coinValues, map);
    }

    public static int[] makeChangeHelper(int value, int[] coinValues, HashMap<Integer, int[]> map) {
        if (map.containsKey(value)) return map.get(value);
        
        if (value == 0) return new int[coinValues.length];
        
        if (value < 0) return null;
        
        int minCoins = Integer.MAX_VALUE;
        int[] minResult = null;
        for (int coinIdx = 0; coinIdx < coinValues.length; coinIdx++) {
            int coinValue = coinValues[coinIdx];

            int[] result = makeChangeHelper(value - coinValue, coinValues, map);

            if (result != null) {
                int nCoins = sumArray(result);
                if (nCoins < minCoins) {
                    minResult = result;
                    minResult[coinIdx]++;
                    minCoins = nCoins;
                }
            }
        }

        map.put(value, minResult.clone());

        return minResult;
    }

    public static int sumArray(int[] array) {
        int sum = 0;
        for (int value : array)
            sum += value;
        return sum;
    }

    public static void main(String[] args) {
        int value = Integer.parseInt(args[0]);
        int[] coinValues = new int[args.length - 1];
        for (int i = 0; i < coinValues.length; i++)
            coinValues[i] = Integer.parseInt(args[i+1]);

        int[] change = makeChange(value, coinValues);
        System.out.print("Best way to make change for " + value + " is: ");
        for (int i = 0; i < coinValues.length; i++) {
            System.out.println(coinValues[i] + ": " + change[i]);
        }
    }

}

- Siddharth D. January 27, 2015 | Flag Reply
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0
of 0 vote

There was a small bug in the above code which I fixed below:

import java.util.HashMap;

public class MakeChange {

    public static int[] makeChange(int value, int[] coinValues) {
        HashMap<Integer, int[]> map = new HashMap<Integer, int[]>();
        return makeChangeHelper(value, coinValues, map);
    }

    public static int[] makeChangeHelper(int value, int[] coinValues, HashMap<Integer, int[]> map) {
        if (map.containsKey(value)) return map.get(value);
        
        if (value == 0) return new int[coinValues.length];
        
        if (value < 0) return null;
        
        int minCoins = Integer.MAX_VALUE;
        int[] minResult = null;
        for (int coinIdx = 0; coinIdx < coinValues.length; coinIdx++) {
            int coinValue = coinValues[coinIdx];

            int[] result = makeChangeHelper(value - coinValue, coinValues, map);

            if (result != null) {
                int nCoins = sumArray(result);
                if (nCoins < minCoins) {
                    minResult = result.clone();
                    minResult[coinIdx]++;
                    minCoins = nCoins;
                }
            }
        }

        map.put(value, minResult);

        return minResult;
    }

    public static int sumArray(int[] array) {
        int sum = 0;
        for (int value : array)
            sum += value;
        return sum;
    }

    public static void main(String[] args) {
        int value = Integer.parseInt(args[0]);
        int[] coinValues = new int[args.length - 1];
        for (int i = 0; i < coinValues.length; i++)
            coinValues[i] = Integer.parseInt(args[i+1]);

        int[] change = makeChange(value, coinValues);
        System.out.println("Best way to make change for " + value + " is: ");
        for (int i = 0; i < coinValues.length; i++) {
            System.out.println(coinValues[i] + ": " + change[i]);
        }
    }

}

- Siddharth D. January 27, 2015 | Flag Reply
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0
of 0 vote

int dp[n+1];
	dp[0]=0;
	int i,j;
	for(i=1;i<=amount;i++)
	{
		for(j=0;j<n;j++)
		{
			if(den[j]>i) break;
			if(dp[i]==-1) {
				dp[i]=dp[i-den[j]]+1;
			}else{
				dp[i]=min(dp[i],dp[i-den[j]+1);
				}	
		}
	}
	cout<<"Min no. of coins: "<<dp[amount]<<endl;
		//Code to produce the coins used
	int val=amount;
	while(val)
	{
		for(i=0;i<n;i++)
		{
			
			if(den[i]<=val && dp[val]=1+dp[val-den[i]]) {cout<<den[i]<<" "; val-=den[i]; break;}
		}
	}

- Anonymous January 30, 2016 | Flag Reply
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0
of 0 vote

How can we do it if we had a limited supply of coins??

- Prashant September 27, 2017 | Flag Reply


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