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Find the Index of the Large Integer

Updated on 26 May, 2025
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Problem Statement

In this task, you are presented with an array arr where all elements are identical except for one that is distinctly larger. Direct access to the array is not available; instead, operations on the array are to be performed through a provided API named ArrayReader. This API offers two methods:

  1. compareSub(int l, int r, int x, int y): Compares the sum of elements between indices l to r with another sum of elements from x to y. It returns:
    • 1 if the sum from l to r is greater.
    • 0 if both sums are equal.
    • -1 if the sum from x to y is greater.
  2. length(): Returns the total number of elements in arr.

The challenge lies in identifying the index of the element with the largest value using no more than 20 calls to compareSub(). You can consider the operations of both methods to be constant time, O(1).

Examples

Example 1

Input:

arr = [7,7,7,7,10,7,7,7]

Output:

4

Explanation:

The following calls to the API
reader.compareSub(0, 0, 1, 1) // returns 0 this is a query comparing the sub-array (0, 0) with the sub array (1, 1), (i.e. compares arr[0] with arr[1]).
Thus we know that arr[0] and arr[1] doesn't contain the largest element.
reader.compareSub(2, 2, 3, 3) // returns 0, we can exclude arr[2] and arr[3].
reader.compareSub(4, 4, 5, 5) // returns 1, thus for sure arr[4] is the largest element in the array.
Notice that we made only 3 calls, so the answer is valid.

Example 2

Input:

nums = [6,6,12]

Output:

2

Constraints

  • 2 <= arr.length <= 5 * 105
  • 1 <= arr[i] <= 100
  • All elements of arr are equal except for one element which is larger than all other elements.

Approach and Intuition

The aim is to locate the singularly large element in an array of otherwise identical elements using minimal API calls. Given the nature of the array and the functions available, a strategic approach using comparisons between elements is most effective. Let's outline the methodology:

  1. Initial Setup: Determine the length of the array using length() from ArrayReader.

  2. Binary Search Approach: Implement a binary search method to effectively narrow down the potential index of the largest element.

    • Start with the entire range of the array and progressively reduce the searching range by half based on the output of compareSub().
    • Compare the middle element with its neighbors. Adjust searching boundaries based on which sub-array contains the largest sum.

  3. Efficiency Considerations:

    • Each comparison ideally halves the search space, adhering to the binary search's logarithmic time complexity.
    • Given the maximum constraint, ensuring the method operates within 20 API calls demands that searches must efficiently reduce the potential candidates.

  4. Edge Case Handling:

    • If the array has its extremes as lengths (i.e., near 2 or 500,000), ensure the strategy still performs under the maximum calls limit.
    • Properly handle cases where the largest element might be at boundaries.

This method utilizes the unique properties of compareSub() to effectively search without directly comparing every element, fitting within given constraints and utilizing minimal API calls for a solution.

Solutions

  • Java
  • Python
java 
class Solution {
    public int findOddElementIndex(ArrayReader reader) {
        int start = 0;
        int totalLength = reader.length();
        while (totalLength > 1) {
            totalLength /= 2;
            final int comparisonResult = reader.compareSub(start, start + totalLength - 1, start + totalLength, start + 2 * totalLength - 1);
            if (comparisonResult == 0) {
                return start + 2 * totalLength;
            }
            if (comparisonResult < 0) {
                start += totalLength;
            }
        }
        return start;
    }
}

The given Java code aims to find the index of an odd-element scenario in a list using an ArrayReader interface. The solution employs a bi-partitioning approach similar to binary search, which efficiently reduces the problem size at each step. Here's a breakdown of the implementation process:

  • Start with the initial index set to zero and fetch the total length of the array through the ArrayReader.length() method.
  • Use a binary search mechanism, where the array is virtually divided into two halves in each iteration of the while loop as long as the totalLength is greater than one.
  • Compare the sum or particular condition (undisclosed, but typically could be the sum or other aggregate properties as per ArrayReader.compareSub method) of two halves of the current segment:
    • If the comparison result equals zero (comparisonResult == 0), it means the odd element lies outside the currently considered segments, hence move the index forward by two times the totalLength.
    • If comparison results indicate that the right half is greater (comparisonResult < 0), adjust the start index to move to the right half by adding totalLength to start.
  • This division and comparison continue until the total length of considered elements reduces to one, at which point you've narrowed it down to the segment where the odd element is located.
  • Return the start index as the position of the odd-element or anomaly within the array.

This method ensures a logarithmic reduction in search space, making the solution very efficient for large datasets. The nature and exact application of the ArrayReader.compareSub function need clarification, but it plays a crucial role in comparing segments of the array.

python 
class Solution(object):
    def retrieveIndex(self, data_source):
        start = 0
        size = data_source.length()
        while (size > 1):
            size //= 2;
            comparison = data_source.compareSub(start, start + size - 1, start + size,
                                    start + size + size - 1)
            if comparison == 0:
                return start + size + size
            if comparison < 0:
                start += size
        return start

The provided Python code is designed to find the index of the largest integer within a data collection leveraging the divide-and-conquer strategy, optimized for performance by using binary search principles.

  • The retrieveIndex function initializes by setting start to 0 and determining the size of the data source.
  • The process includes a loop that continuously halves the size of the searchable segment in each iteration, reducing the target area significantly with each step.
  • During each loop iteration, the function compares two halves of the current segment using the compareSub method:
    • If compareSub returns 0, it means both halves are equal, and there's no difference to decide the direction of search, hence ends the search.
    • A negative return indicates the larger value is in the second half, adjusting the start index accordingly.
  • When the size reduces to one, the loop exits, and the position of the largest integer at that point is returned.

This method ensures O(log n) complexity, making it efficient for large datasets by minimizing the number of comparisons needed to locate the largest integer's index.

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