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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
#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
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