// The Original Problem Description
// # Write a function, `nearest_larger(arr, i)` which takes an array and an
// # index.  The function should return another index, `j`: this should
// # satisfy:
// #
// # (a) `arr[i] < arr[j]`, AND
// # (b) there is no `j2` closer to `i` than `j` where `arr[i] < arr[j2]`.
// #
// # In case of ties (see example below), choose the earliest (left-most)
// # of the two indices. If no number in `arr` is larger than `arr[i]`,
// # return `nil`.
// #
// # Difficulty: 2/5
//
// def nearest_larger(arr, idx)
//   diff = 1
//   while true
//     left = idx - diff
//     right = idx + diff

//     if (left >= 0) && (arr[left] > arr[idx])
//       return left
//     elsif (right < arr.length) && (arr[right] > arr[idx])
//       return right
//     elsif (left < 0) && (right >= arr.length)
//       return nil
//     end

//     diff += 1
//   end
// end

// What I have found is that
// when a number which is greater than the number at the given index is found,
// the function immediately returns it without considering numbers left in the array
// though there might be numbers which are less than the number found earlier and
// greater than the number at the given value
// For example
// nearest_larger([2,4,3,8], 0) should return 2 because the nearest larger number of
// 2 at the index of 0 is 3 at the index of 2. However the solution function returns 1 where
// 4 is at because the function returns when it finds a number
// which is greater than the number at the given index for the first time
// So I have tried to fix the issue below.
// I made it to keep iterating through the whole array even after a number which is greater than
// the number at the given index is found.

#include <iostream>
#include <vector>
using namespace std;
int nearest_larger(vector<int> &array, int index) {
    int result_index = 0;
    int temp_index = 0;
    bool found_one = false;
    for (int i=0; i<array.size(); i++) {
        if (i != index) {
            if (array.at(index) < array.at(i)) {
                temp_index = i;
                if (!found_one) {
                    // if a number which is greater than the number at the given index
                    // for the first time
                    result_index = temp_index;
                    found_one = true;
                } else {
                    // if at least a number which is greater than the number at the given index
                    // is already found, then need to compare it with upcoming numbers
                    // to determine the number at result_index is really a nearest larger number
                    if (array.at(temp_index) < array.at(result_index)) {
                        result_index = temp_index;
                    }
                }
            }
        }
    }
    return result_index;
}

int main () {
    vector<int> array = {2,4,3,8};
    cout << nearest_larger(array, 0) << endl;
}