Greedy algorithms – Coin changing using minimum number of coins

We have coins of denominations 1 cent, 5 cents and 25 cents with us. What we would like to do is represent a total of n cents using these coins, such that a minimum number of coins are used.

For example, if we want to represent 31 cents, we could do that by using 1 25-cents coin, 1 5-cents coin and a single coin of 1-cent, thereby utilizing a total of 3 coins. And this indeed is the bast we can do. Why is that so? Let’s look at it this way — the current solution uses one 25-cent coin, one 5-cent coin and one 1-cent coin. One way in which we can try to modify this solution would be to decide to not use any 25-cent coin at all (instead of the one coin that we are currently using). That would mean that we would have to represent the shortage of 25 cents that we are now faced with, using coins of denominations 1 and 5-cents. We can see that we would need at least five coins of denomination 5-cents in order to represent 25-cents. Hence, using a 25-cent coin was actually a wise choice enabling us to save on the number of coins used.

Now, what if we had to represent 57 cents. The best solution would be to use two coins of 25-cents, one coin of 5-cents and two coins of 1-cent.

Again, we can intuitively infer its optimality by observing that:

1. if we use lesser number of 25-cents coins than what we are currently using, we would actually end up using more coins. —(If we use just one 25-cent coin, then we would require 5 coins of 5-cents each to cover up the deficit of 25-cents. Similarly if we do not use any 25-cents coins at all, we would need 10 coins of 5-cents each to cover the deficit.)  Hence, the idea is to use as many 25-cent coins as we possibly can.

2. Similarly, once we have used the maximum number of 25-cent coins, we are left with k cents (where k < 25) to represent using 1 and 5-cent coins. Analogous to what we observed above, we can infer that we must use as many 5-cent coins as we can in order to cover the deficit of k cents. Whatever is left can be represented using 1-cent coins.


Hence, the algorithm would look something like this.

//Represent n cents using the least number of total coins where coins are of denominations – 1, 5 and 25-cents.

1. Divide n by 25 to get quotient q1 and remainder k1.

2. Use q1 coins of 25-cents each.

3. If k1 == 0, we are done. Else, divide k1 by 5 to get quotient q2 and remainder k2.

4. Use q2 coins of 5-cents, and k2 coins of 1-cent.


Now, if we have coins at our disposal having denominations 1, k, k^2, k^3-cents and we had to represent n cents with the minimum number of coins, we could do the following:-


i = 3;

num = n;

While ( exponent >= 0)


Divide num by k^i to get quotient q and remainder r.

Use q coins of denomination k^i cents.

If ( r == 0)


num = r;

i – – ;



Now, take the following example, represent 31 cents using coins of denominations 1, 10 and 25 cents such that the least number of coins are used.

Here, the idea behind the greedy algorithm of using the maximum possible number of coins of the highest denomination would not work. That approach would get us a solution that uses 6 coins : one 25-cent coin, and 6 1-cent coins.

In contrast, we can get a better solution using 4 coins: 3 coins of 10-cents each and 1 coin of 1-cent.

In the problems presented at the beginning of this post, the greedy approach was applicable since for each denomination, the denomination just smaller than it was a perfect divisor of it. (i.e. 25 was a factor of 125; 5 was a factor of 25; 1 was a factor of 5}. However, that not being true in this case {10 is not a factor of 25} leads to the failure of the greedy approach.


Scheduling Tutors – Greedy Algorithms – 1

Question.We are a tutor-scheduling organization. We have at our disposal an infinite number of tutors using whom we service the tutoring requests that come our way.

Consider any particular day. We get a certain number of tutoring requests. Each request has a specified starting time, and the duration of the tutoring session is exactly 30 minutes. Each tutor can work for any continuous period of 2 hours in a day. (The tutor need not actually be tutoring for the whole of the 2 hours; it is just that he/she will certainly not tutor outside of that 2 hour period.) We are the ones who decide the starting time of the 2 hour working period for each tutor.

We also need to remember that a tutor cannot service more than one request simultaneously.

Our aim is to service the tutoring requests using a minimum number of tutors. How can we achieve that?


We can tackle this problem greedily.

We order the requests in non-descending order on the basis of their starting times. Take the first request. Schedule a tutor to start his/her 2 hour period at exactly the starting time of the 1st request. Now, look at the 2nd request. Ask yourself the following question: Can the 1st tutor (i.e. the one who has already been scheduled to work) also service this request? If yes, let him/her do so. If not, schedule a 2nd tutor to start his/her 2 hour period at exactly the starting time of this new request. Continue in the same manner for the succeeding requests. i.e. Do the following: For each new request, check if that request can be serviced by any of the tutors already scheduled to work. If yes, schedule the tutor who has the shortest time remaining in his/her 2 hr period, to service this request. If not, schedule a new tutor to start his/her 2 hour period at exactly the starting time of this new request.


In pseudo code form, we would have:

Sort the requests in non-decreasing order of starting times.

count = 0; // this is the # of tutors assigned thus far.

flag = UNSERVICED. // an indicator for whether a particular request has been serviced or //not.

While (one or more requests remain to be serviced)

Remove the 1st request from our sorted list.

For ( each of the tutors scheduled thus far )

//Note: we iterate through the already assigned tutors in the order in which they were

// assigned (so as to be able to schedule the tutor with the least remaining time in

// his/her 2 hr period.

If ( that tutor can service this request )

Schedule him/her to service this request.

Come out of the For loop.

Set: flag = SERVICED.

If (flag == UNSERVICED)

// means that no already scheduled tutor was able to service the new request.

// We need a new tutor.

Assign a new tutor to start his/ her 2 hr period at the starting time of the new request.

Set: flag = UNSERVICED.

count ++ ;

End of While loop.



Analysis of running time:

– Say there are n requests. Sorting them will take O(n log n) time.

– In the worst case, we might need to schedule n tutors to service n requests.In such a scenario, the time required for assigning tutors will be O(n^2) (this is because of the fact that before we assign a new tutor we check with each of the already assigned tutors as to whether they can service the new request).

– Hence, the algorithm will run in O(n^2) time.


Intuitive proof of correctness of the algorithm:

– Consider the 1st request (in sorted order). Obviously, since no tutors have been assigned thus far, we need to schedule a new tutor for this request. Now, consider the choice we have wrt when to schedule this new tutor. Certainly, we cannot schedule him/her to start his/her 2 hr period after the starting time of the 1st request. If we schedule him/her to start working before the starting time of the 1st request, the time that elapses between the moment the tutor’s 2 hr period starts and the starting time of the 1st request would be effectively wasted. Hence, the best thing would be to schedule the 1st tutor to start his/her 2 hr period at exactly the starting time of the 1st request.

– Similarly, one can argue that for each newly assigned tutor, the algorithm chooses exactly the right time for him/her to start their working period.

– Also, we observe that the algorithm does not assign any new tutors unless there is a need to do so.

– And finally, the algorithm always schedules a tutor with the least remaining time in their schedule to service a new request (when selecting among already assigned tutors). This results in maximum number of requests being serviced by already assigned tutors without having the need to assign a new one.


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