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Minimise the maximum difference between the heights
By Aafat_Wapas, history, 5 years ago, In English

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.

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5 years ago, # |
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5 years ago, # |
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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.

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5 years ago, # |
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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).

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    5 years ago, # ^ |
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    Can you please explain how you arrived at this approach.

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      5 years ago, # ^ |
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      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.

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    4 years ago, # ^ |
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    With this,there is chance that we are making height of tower negative, which is not ideal

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5 years ago, # |
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what does K do

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5 years ago, # |
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5 years ago, # |
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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.

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4 years ago, # |
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What is wrong in my approach

every state of dp will store a pair

dp(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 state

temp1=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]

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