Given an array of integers H representing heights of towers and a positive integer B. It is allowed to either increase or decrease the height of every tower by B **(only once)**.

How to calculate the minimum possible difference between the **heights of the longest** **and the shortest tower after modifications.**

Note: **It is mandatory to either increase or decrease the height of every tower.**

Auto comment: topic has been updated by Aafat_Wapas (previous revision, new revision, compare).Decrease all towers height by B. Then you need to increase some subset of them by 2*B. If you sort them it will be some prefix.

Sort the heights and try every possible split into two halves [0, i-1] and [i, n-1], where you add B to the left half and subtract B from the right half. Then the difference for that split is

`max(a[n-1] - B, a[i-1] + B) - min(a[0] + B, a[i] - B)`

.Can you please explain how you arrived at this approach.

I'm not sure, mostly intuition. But maybe this can help:

So we have to add B to some elements $$$a_{i_1}...a_{i_k}$$$ and subtract B from the rest $$$a_{j_1}...a_{j_l}$$$. Let's say these sets are sorted, so the minimum of the first is $$$a_{i_1} + B$$$. The minimum of the other set is $$$a_{j_1} - B$$$. The overall minimum will then be $$$min(a_{i_1} + B, a_{j_1} - B)$$$.

But if $$$a_{j_1} < a_{i_k}$$$ (i.e. they are not a split into halves of the sorted array), then we could exchange $$$a_{i_k}$$$ and $$$a_{j_1}$$$ and achieve a larger minimum.

Same can be argued for maximum.

Thanks :)

Hello , can you please provide the code based on whatever you understood? It would be helpful.

With this,there is chance that we are making height of tower negative, which is not ideal

I think I have got a working solution which is logical as well and handles non-negative condition also. The concept is the same but only one change, we start checking for min and max elemnts only when arr[i] >= k. The reason for that is if arr[i] < k then, all elements before it are also less than k, so we can't decrease them, so we will have to increase them, also since arr[i] < k, we can't decrease arr[i] to arr[n-1] by k, because then arr[i] will be negative.

In short we will have to increase all elements of the array, and that will not help us as the difference remains same.

So we have to skip all elements less than k, and only when we reach an element >= k, we can perform the operation which is: 1. Increase all elements before, and decrease all elements after arr[i] (including it) by k.

class Solution { public: int smallestRangeII(vector& A, int K) { int N = A.size(); sort(A.begin(),A.end()); int ans = A[N-1] — A[0];

}; the correct and accepted code

Thanks for this. It really helped. Working perfectly fine. :)

what if the heights are negative??

what does K do

Auto comment: topic has been updated by Aafat_Wapas (previous revision, new revision, compare).first find minimum and maximum element of the array and find the mid = (min + max)/2

decrease all height more than mid by B and increase all height more than mid by B.

sort the array.

the diff between min and max of new array will be the answer.

we also have to check for answer in the original array.

How did you came up with the logic? and what to do if element is equal to mid?

Use this for reference only

That would fail for the testcase k = 6, n = 4, arr = [35, 43, 43, 23]. The average would be 36. If you decrease all elements less than 36 by 6 and decrease the other ones then you'll get [41, 37, 37, 29]. Answer would be

12this way.Where as if you decrease the first element (35) by 6. Decrease second and third by 6 and increment the last element by 6, you would get [29, 37, 37, 29]. The answer would be

8is the most optimal solution.Not failing. It's average of min and max.. not an average of all elements.

fails for testcase, I guess

`7 2 [[38, 43]]`

## include<bits/stdc++.h>

## define ll long long

using namespace std;

int main() {

}

https://stackoverflow.com/a/63220955/10015362

This link has explained it very well. Also the gfg practice question and gfg solution is not optimal and hence wrong.

Example testcase:

5 5

7 6 2 1 10

What is wrong in my approach

every state of dp will store a pairdp(i,0) will store that using arr[i]-k and what will be the (min,max) from index 0 to i

dp(i,1) will store that using arr[i]+k and what will be the (min,max) from index 0 to i

ans=min(dp(n-1)(0).second-dp(n-1)(0).first,dp(n-1)(1).second-dp(n-1)(1).first)Transition in statetemp1=min(dp[i-1][1].first,arr[i]-k) //dp[i-1][1].first will store the minimum vale till i-1 index considering arr[i]-k

temp2=max(dp[i-1][1].second,arr[i]-k)

temp3=min(dp[i-1][0].first,arr[i]-k)

temp4=max(dp[i-1][0].second,arr[i]-k)

if(abs(temp1-temp2)>abs(temp3-temp4)) dp[i][0]={temp3,temp4}

else dp[i][0]={temp1,temp2}

//same way we can find dp[i][1]