Formatted question description: https://leetcode.ca/all/1533.html

# 1533. Find the Index of the Large Integer

Medium

## Description

We have an integer array arr, where all the integers in arr are equal except for one integer which is larger than the rest of the integers. You will not be given direct access to the array, instead, you will have an API ArrayReader which have the following functions:

• int compareSub(int l, int r, int x, int y): where 0 <= l, r, x, y < ArrayReader.length(), l <= r and x <= y. The function compares the sum of sub-array arr[l..r] with the sum of the sub-array arr[x..y] and returns:
• 1 if arr[l]+arr[l+1]+...+arr[r] > arr[x]+arr[x+1]+...+arr[y].
• 0 if arr[l]+arr[l+1]+...+arr[r] == arr[x]+arr[x+1]+...+arr[y].
• -1 if arr[l]+arr[l+1]+...+arr[r] < arr[x]+arr[x+1]+...+arr[y].
• int length(): Returns the size of the array.

You are allowed to call compareSub() 20 times at most. You can assume both functions work in O(1) time.

Return the index of the array arr which has the largest integer.

Follow-up:

• What if there are two numbers in arr that are bigger than all other numbers?
• What if there is one number that is bigger than other numbers and one number that is smaller than other numbers?

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 with arr).

Thus we know that arr and arr doesn’t contain the largest element.

reader.compareSub(2, 2, 3, 3) // returns 0, we can exclude arr and arr.

reader.compareSub(4, 4, 5, 5) // returns 1, thus for sure arr 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 * 10^5
• 1 <= arr[i] <= 100
• All elements of arr are equal except for one element which is larger than all other elements.

## Solution

Initialize rangeStart to 0 and rangeEnd to reader.length() - 1, so [rangeStart, rangeEnd] is the range that must contains the index of the large integer.

Each time calculate rangeLength = rangeEnd - rangeStart + 1, calculate mid = (rangeEnd - rangeStart) / 2 + rangeStart, and calculate l = rangeStart, r = rangeStart + rangeLength / 2 - 1, x = rangeEnd - rangeLength / 2 + 1, y = rangeEnd. Use l, r, x, y to call reader.compareSub. If the result is 0, then the remaining number is at index mid, so return mid. If the result is -1, then the large number must be in the right half of the interval, so set rangeStart = x. If the result is 1, then the large number must be in the left half of the interval, so set rangeEnd = r. Repeat the process until rangeStart and rangeEnd become the same. Finally, return rangeStart.

/**
* // This is ArrayReader's API interface.
* // You should not implement it, or speculate about its implementation
* interface ArrayReader {
*     // Compares the sum of arr[l..r] with the sum of arr[x..y]
*     // return 1 if sum(arr[l..r]) > sum(arr[x..y])
*     // return 0 if sum(arr[l..r]) == sum(arr[x..y])
*     // return -1 if sum(arr[l..r]) < sum(arr[x..y])
*     public int compareSub(int l, int r, int x, int y) {}
*
*     // Returns the length of the array
*     public int length() {}
* }
*/

class Solution {
int rangeStart = 0, rangeEnd = reader.length() - 1;
while (rangeStart < rangeEnd) {
int rangeLength = rangeEnd - rangeStart + 1;
int mid = (rangeEnd - rangeStart) / 2 + rangeStart;
int l = rangeStart, r = rangeStart + rangeLength / 2 - 1, x = rangeEnd - rangeLength / 2 + 1, y = rangeEnd;
int compare = reader.compareSub(l, r, x, y);
if (compare == 0)
return mid;
else if (compare < 0)
rangeStart = x;
else
rangeEnd = r;
}
return rangeStart;
}
}