Program to find maximum adjacent absolute value sum after single reversal in C++

Suppose we have a list of numbers called nums and we can reverse any sublist in the list at most once. After performing this operation, we have to find the maximum possible value of

$\displaystyle\sum\limits_{i=0}^{n-2}| nums[i+1]-[nums[i]|$

So, if the input is like nums = [2, 4, 6], then the output will be 6, as when we reverse [4, 6] we will get the list as [2, 6, 4] and the value |2 − 6| + |6 − 4| = 6

To solve this, we will follow these steps −

• if size of nums <= 1, then −

  • return 0

• ans := 0

• n := size of nums

• for initialize i := 1, when i < n, update (increase i by 1), do −

  • ans := ans + |nums[i] − nums[i − 1]|

• orig := ans

• for initialize i := 1, when i < n − 1, update (increase i by 1), do −

  • ans := maximum of ans and orig − |(nums[i] − nums[i + 1]| + |nums[0] − nums[i + 1]|

  • ans := maximum of ans and orig − |(nums[i] − nums[i − 1]| + |nums[n − 1] − nums[i − 1]|

• pp := −|nums[1] − nums[0]|

• pm := −|nums[1] − nums[0]|

• mp := −|nums[1] − nums[0]|

• mm := −|nums[1] − nums[0]|

• for initialize j := 2, when j < n − 1, update (increase j by 1), do −

  • jerror := |nums[j + 1] − nums[j]|

  • ans := maximum of ans and (orig + pp − jerror − nums[j] − nums[j + 1])

  • ans := maximum of ans and (orig + pm − jerror − nums[j] + nums[j + 1])

  • ans := maximum of ans and (orig + mp − jerror + nums[j] − nums[j + 1])

  • ans := maximum of ans and (orig + mm − jerror + nums[j] + nums[j + 1])

  • pp := maximum of pp and −|nums[j] − nums[j − 1]|

  • pm := maximum of pm and −|nums[j] − nums[j − 1]|

  • mp := maximum of mp and −|nums[j] − nums[j − 1]|

  • mm := maximum of mm and −|nums[j] − nums[j − 1]|

• return ans

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; int solve(vector<int>& nums) {    if (nums.size() <= 1)    return 0;    int ans = 0;    int n = nums.size();    for (int i = 1; i < n; i++) {       ans += abs(nums[i] − nums[i − 1]);    }    int orig = ans;    for (int i = 1; i < n − 1; i++) {       ans = max(ans, orig − abs(nums[i] − nums[i + 1]) +       abs(nums[0] − nums[i + 1]));       ans = max(ans, orig − abs(nums[i] − nums[i − 1]) + abs(nums[n    − 1] − nums[i − 1]));    }    int pp = −abs(nums[1] − nums[0]) + nums[0] + nums[1];    int pm = −abs(nums[1] − nums[0]) + nums[0] − nums[1];    int mp = −abs(nums[1] − nums[0]) − nums[0] + nums[1];    int mm = −abs(nums[1] − nums[0]) − nums[0] − nums[1];    for (int j = 2; j < n − 1; j++) {       int jerror = abs(nums[j + 1] − nums[j]);       ans = max(ans, orig + pp − jerror − nums[j] − nums[j + 1]);       ans = max(ans, orig + pm − jerror − nums[j] + nums[j + 1]);       ans = max(ans, orig + mp − jerror + nums[j] − nums[j + 1]);       ans = max(ans, orig + mm − jerror + nums[j] + nums[j + 1]);       pp = max(pp, −abs(nums[j] − nums[j − 1]) + nums[j − 1] +       nums[j]);       pm = max(pm, −abs(nums[j] − nums[j − 1]) + nums[j − 1] −       nums[j]);       mp = max(mp, −abs(nums[j] − nums[j − 1]) − nums[j − 1] +       nums[j]);       mm = max(mm, −abs(nums[j] − nums[j − 1]) − nums[j − 1] −       nums[j]);    }    return ans; } int main(){    vector<int> v = {2, 4, 6};    cout << solve(v); }

Input

{2, 4, 6}

Output

6