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Number of NGEs to the right in C++
You are given an array and the index of the target element. We have to count the number of elements greater than the given element to its right. Let's see an example.
Input
arr = [2, 3, 5, 1, 4, 2, 6] index = 3
Output
3
The target index element is 1. There are three elements i.e..., 4, 2, 6 that are greater than 1 on its right side.
Algorithm
- Initialise the array and index of target element.
- If the index is greater than or equal to the length of the array, then return -1.
- Write a loop that iterates from the next element of the given index.
- Increment the count if the element is greater than the target element.
- Return the count.
Implementation
Following is the implementation of the above algorithm in C++
#include <bits/stdc++.h>
using namespace std;
int getNextGreaterElementsCount(int arr[], int n, int index) {
if (index >= n) {
return -1;
}
int count = 0;
for (int i = index + 1; i < n; i++) {
if (arr[index] < arr[i]) {
count += 1;
}
}
return count;
}
int main() {
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
int n = 8, index = 1;
cout << getNextGreaterElementsCount(arr, n, index) << endl;
return 0;
}Output
If you run the above code, then you will get the following result.
6
Updated on: 26-Oct-2021
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