Minimum difference between heights of Towers? - arrays

Question

I was going through some interview questions, I saw this one
You are given the height of n towers and value k. You have to either increase or decrease the height of every tower by k. You need to minimize the difference between the height of the longest and the shortest tower and output this difference.
I think the answer will be (maxheight-k) - (minheight + k).
I have tried on some test cases it is running fine.
But I am not sure, I think I am missing something, Am I ?

Answer 1

m7thon's answer explains the problem with your solution, so I'll just explain how you can actually solve this . . .
The big thing to observe is that for any given tower, if you choose to increase its height from hi to hi + k, then you might as well increase the height of all shorter towers: that won't affect the maximum (because if hj < hi, then hj + k < hi + k), and may help by increasing the minimum. Conversely, if you choose to decrease the height of a tower from hi to hi − k, then you might as well decrease the heights of all taller towers.
So while there are 2n possible ways to choose which towers should be increased vs. decreased, we can actually ignore most of these. Some tower will be the tallest tower that we increase the height of; for all shorter towers, we will increase their height as well, and for all taller towers, we will decrease their height. So there are only n interesting ways to choose which towers should be increased vs. decreased: one for each tower's chance to be the tallest tower that we increase the height of.
[Pedantic note #1: You may notice that it's also valid to decrease the heights of all towers, in which case there's no such tower. But that's equivalent to increasing the heights of all towers — whether we add k to every height or subtract k from every height, either way we're not actually changing the max-minus-min.]
[Pedantic note #2: I've only mentioned "shorter towers" and "taller towers", but it's also possible that multiple towers have the same initial height. But this case isn't really important, because we might as well increase them all or decrease them all — there's no point increasing some and decreasing others. So the approach described here still works fine.]
So, let's start by sorting the original heights and numbering them in increasing order, so that h1 is the original height of the originally-shortest tower and hn is the original height of the originally-tallest tower.
For each i, try the possibility that the ith-shortest tower is the tallest tower that we increase the height of; that is, try the possibility that we increase h1 through hi and decrease hi+1 through hn. There are two groups of cases:
If i < n, then the final height of the finally-shortest tower is min(h1 + k, hi+1 − k), and the final height of the finally-tallest tower is max(hi + k, hn − k). The final difference in this case is the latter minus the former.
If i = n, then we've increased the heights of all towers equally, so the final difference is just hn − h1.
We then take the least difference from all n of these possibilities.
Here's a Java method that implements this (assuming int-valued heights; note that hi is arr[i-1] and hi+1 is arr[i]):
private static int doIt(final int[] arr, final int k) {
java.util.Arrays.sort(arr);
final int n = arr.length;
int result = arr[n - 1] - arr[0];
for (int i = 1; i < n; ++i) {
final int min = Math.min(arr[0] + k, arr[i] - k);
final int max = Math.max(arr[n - 1] - k, arr[i - 1] + k);
result = Math.min(result, max - min);
}
return result;
}
Note that I've pulled the i = n case before the loop, for convenience.

Answer 2

Lets say you have three towers of heights 1, 4 and 7, and k = 3. According to your reasoning the optimal minimum difference is (7 - 3) - (1 + 3) = 0. But what do you do with the tower of height 4? You either need to increase or decrease this, so the minimum difference you can achieve is in fact 3 in this example.
Even if you are allowed to keep a tower at its height, then the example 1, 5, 7 will disprove your hypothesis.
I know this does not solve the actual minimization problem, but it does show that it is not as simple as you thought. I hope this answers your question "Am I missing something?".

Answer 3

I assume this came from gfg.
The answer of #ruakh is perhaps the best I've found online, it'll work for most cases, but for the practice problem on gfg, there are a few cases which can cause the minimum to go below 0, and the question doesn't allow any height to be < 0.
So for that, you'll need an additional check, and the rest of it is pretty much entirely inspired from ruakh's answer
class Solution {
int getMinDiff(int[] arr, int n, int k) {
Arrays.sort(arr);
int ans = arr[n-1] - arr[0];
int smallest = arr[0] + k, largest = arr[n-1]-k;
for(int i = 0; i < n-1; i++){
int min = Math.min(smallest, arr[i+1]-k);
int max = Math.max(largest, arr[i]+k);
if(min < 0) continue;
ans = Math.min(ans, max-min);
}
return ans;
}
}
I also went in for 0-based indexing for the heights to make it more obvious, but maybe that's subjective.
Edit: One case where the < 0 check is important, is when the array is
8 1 5 4 7 5 7 9 4 6 and k is 5. The expected answer for this is 8, without the < 0 check, you'd get 7.

Answer 4

A bit late here. Already these guys have explained you the problem and given you the solution. However, I have prepared this code myself. The code I prepared is not the best code that you should follow but gives a clear understanding of what can be done to achieve this using brute-force.
set = list(map(int, input().split()))
k = int(input())
min = 999999999
for index in range(2**len(set)):
binary = [] //probably should have used integer to binary fuction here
while index != 0:
remainder = index % 2
index //= 2
binary.append(remainder)
while len(binary) != len(set):
binary.append(0)
binary.reverse()
separateset = []
flag = 0
for i in range(len(binary)):
if binary[i] == 0:
separateset.append(set[i]+k)
elif binary[i] == 1 and set[i]-k >= 0:
separateset.append(set[i]-k)
else:
flag = 1
break
if flag == 0:
separateset.sort()
if min > separateset[-1] - separateset[0]:
min = separateset[-1] - separateset[0]
print(min)
This is achieved by identifying all the possible subsets of the set variable but with just some modifications. If the digit is 0, the value at that i (index not the index in the for loop) in the set is added with k otherwise if the digit is 1 and set[i]-k >= 0, the value at that index in the set is subtracted by k (Now you can add or subtract k vice-versa, it doesn't matter until you get all possible combinations of +k and -k). set[i]-k >= 0 is to be followed because a negative height wouldn't make sense and if that happens, flag becomes 1 and breaks. But if the flag is 0, it means that all the heights are positive and then the separatesort is sorted and then min stores the difference between the largest and shortest tower. This min ultimately has the minimum of all the differences.

Answer 5

Step 1 :
Decrease all the heights by 'k' and sort them in non-decreasing order.
Step 2 :
We need to increase some subset of heights by '2 * k' (as they were decreased by
'k' in step1, so, to effectively increase their heights by 'k' we need
to add '2*k') .
Step 3 :
Clearly if we increase the 'i'th height without increasing the
'i-1'th then, it will not be useful as the minimum is still the same
and maximum may also increase !
Step 4 :
Consider all prefixes with '2*k' added to each element of the prefix .
Then calculate and update the (max - min) value.

Answer 6

Let me know which scenario am I missing here,
class Solution:
def getMinDiff(self, arr, n, k):
for i in range(n):
if arr[i] < k: arr[i] += k
else: arr[i] -= k
result = max(arr) - min(arr)
print('NewArr: {}\nresult: {}'.format(arr, result))
return result

Answer 7

C++ code :
int getMinDiff(int arr[], int n, int k) {
// code here
sort(arr , arr+n);
int result = arr[n - 1] - arr[0];
for (int i = 1; i < n; ++i) {
int min_ = min(arr[0] + k, arr[i] - k);
int max_ = max(arr[n - 1] - k, arr[i - 1] + k);
result = min(result, max_ - min_);
}
return result;
}

Answer 8

First you are gonna need to find average height of the towers.
lets say heights are 3, 7, 17, 25, 45 and k = 5
the average will be = ( 3 + 7 + 17 + 25 + 45 ) / 5 = 97 / 5 = 19.4
Now we will try to make every building closer to average height.
for 3 height tower we have to add 5 three times making height = 3 + (3*5) = 18 (18 is closer than 23) close to average.
for 7 height we will add 5 two times = 7 + (2 *5) = 17 (17 is closer than 22)
Similarly 25 will become 25 - 5 = 20
and 45 will become 45 - (5 *5) = 20
your height will becomes 18, 17, 17, 20, 20

Answer 9

This approach works on GfG practice, Problem link: https://practice.geeksforgeeks.org/problems/minimize-the-heights/0/
Approach :
Find max, min element from the array. O(n). Take the average, avg = (max_element + min_element)/2.
Iterate over the array again, and for each element, check if it is less than avg or greater.
If the current element arr[i] is less than avg, then add "k" to a[i], i.e a[i] = a[i] + k.
If the current element arr[i] is greater than or equal to avg, then subtract k from a[i], i.e a[i] = a[i] - k;
Find out the minimum and maximum element again from the modified array.
Return the min(max1-min1, max2-min2), where (max1, min1) = max and min elements initially before the array was modified, and (max2, min2) are the max and min elements after doing modification.
Entire Code can be found here: https://ide.geeksforgeeks.org/56qHOM0EOA