The is a classic search algorithm problem often used in computer science education and competitive programming (most notably on platforms like HackerEarth). It is a practical application of Binary Search . Video Title Cherokee Dass Dr Assjob 3 Upd →
import bisect Download | Euro Truck Simulator 2 Version 1.45
def solve_index_of_monk(arr): n = len(arr) # Arrays to store the counts left_counts = [0] * n right_counts = [0] * n # --- Pass 1: Left to Right (Count Smaller) --- # 'seen' maintains a sorted list of elements encountered so far seen = [] for i in range(n): # bisect_left returns the insertion point which equals # the count of elements smaller than arr[i] pos = bisect.bisect_left(seen, arr[i]) left_counts[i] = pos # Insert current element into the sorted list bisect.insort(seen, arr[i]) # --- Pass 2: Right to Left (Count Larger) --- seen.clear() # Reset the sorted list # Iterate backwards for i in range(n - 1, -1, -1): # bisect_right returns the insertion point. # Elements larger than arr[i] = Total elements - insertion point pos = bisect.bisect_right(seen, arr[i]) right_counts[i] = len(seen) - pos bisect.insort(seen, arr[i]) # --- Final Aggregation --- max_value = -1 max_index = -1 for i in range(n): current_val = left_counts[i] + right_counts[i] if current_val > max_value: max_value = current_val max_index = i return max_index, max_value
$$ \text{Value}_i = \text{countSmallerToLeft}_i + \text{countLargerToRight}_i $$
Here is a useful piece covering the concept, the algorithm, and a practical implementation. The problem typically presents a scenario like this: An array of integers is given. You are asked to calculate a "Score" or "Value" for each element based on the number of elements to its left that are smaller than it and the number of elements to its right that are larger than it.
The goal is usually to find the maximum value in the resulting array or the index of a specific value. A naive approach (nested loops) would compare every element against every other element, resulting in a time complexity of O(N²) . For large datasets (e.g., $N = 10^5$), this is too slow.