# Maximize value of (arr[i] – i) – (arr[j] – j) in an array in C++

C++Server Side ProgrammingProgramming

## Problem statement

Given an array, arr[] find the maximum value of (arr[i] – i) – (arr[j] – j) where i is not equal to j. Where i and j vary from 0 to n-1 and n is the size of input array arr[].

If the input array is {7, 5, 10, 2, 3} then we can obtain 9 maximum value as follows−

(element 10 – index 2) - (element 2 – index 3)
(10 – 2) – (2 – 3) = 8 – (-1) = 9

## Algorithm

1. Find maximum value of (arr[i] – i) in whole array.
2. Find minimum value of (arr[i] – i) in whole array.
3. Return difference of above two values

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMaxDiff(int *arr, int n){
if (n < 2) {
cout << "Invalid input" << endl;
exit(1);
}
int minVal = INT_MAX;
int maxVal = INT_MIN;
for (int i = 0; i < n; ++i) {
int result = arr[i] - i;
if (result > maxVal) {
cout << "Max = " << arr[i] << " - " << i << endl;
maxVal = result;
}
if (result < minVal) {
cout << "Min = " << arr[i] << " - " << i << endl;
minVal = result;
}
}
return (maxVal - minVal);
}
int main(){
int arr[] = {7, 5, 10, 2, 3};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Maximum value = " << getMaxDiff(arr, n) << endl;
return 0;
}

## Output

When you compile and execute the above program. It generates the following output −

Maximum value = Max = 7 - 0
Min = 7 - 0
Min = 5 - 1
Max = 10 - 2
Min = 2 - 3 9
Published on 24-Dec-2019 07:18:42