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Maximum of Absolute Value Expression in C++
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
#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
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