# Maximum of Absolute Value Expression in C++

C++Server Side ProgrammingProgramming

Suppose we have two arrays of integers with equal lengths, we have to find the maximum value of: |arr1[i] - arr1[j]| + |arr2[i] - arr2[j]| + |i - j|. Where the maximum value is taken over all 0 <= i, j < arr1.length. So if the given two arrays are [1,2,3,4] and [-1,4,5,6], the output will be 13.

To solve this, we will follow these steps −

• Define a method called getVal, that will take array v

• maxVal := -inf, minVal := inf

• for i in range 0 to size of v

• minVal := min of v[i] and minVal

• maxVal := max of v[i] and maxVal

• return maxVal – minVal

• From the main method, do the following

• make an array ret of size 4

• n := size of arr1

• for i in range 0 to n – 1

• insert arr1[i] – arr2[i] + i into ret[0]

• insert arr1[i] + arr2[i] + i into ret[1]

• insert arr1[i] – arr2[i] - i into ret[2]

• insert arr1[i] + arr2[i] - i into ret[3]

• ans := -inf

• for i in range 0 to 3

• ans := max of ans and getVal(ret[i])

• return ans

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int getVal(vector <int>& v){
int maxVal = INT_MIN;
int minVal = INT_MAX;
for(int i = 0; i < v.size(); i++){
minVal = min(v[i], minVal);
maxVal = max(v[i], maxVal);
}
return maxVal - minVal;
}
int maxAbsValExpr(vector<int>& arr1, vector<int>& arr2) {
vector <int> ret[4];
int n = arr1.size();
for(int i = 0; i < n; i++){
ret[0].push_back(arr1[i] - arr2[i] + i);
ret[1].push_back(arr1[i] + arr2[i] + i);
ret[2].push_back(arr1[i] - arr2[i] - i);
ret[3].push_back(arr1[i] + arr2[i] - i);
}
int ans = INT_MIN;
for(int i = 0; i < 4; i++){
ans = max(ans, getVal(ret[i]));
}
return ans;
}
};
main(){
vector<int> v1 = {1,2,3,4}, v2 = {-1, 4, 5, 6};
Solution ob;
cout << (ob.maxAbsValExpr(v1, v2));
}

### Input

[1,2,3,4]
[-1,4,5,6]

## Output

13
Published on 13-Apr-2020 15:35:18
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