Number of Subsequences That Satisfy the Given Sum Condition

class Solution {
    public static int binarySearchInsertPos(int[] array, int goal) {
        int start = 0, end = array.length - 1;
        while (start <= end) {
            int center = start + (end - start) / 2;
            if (array[center] <= goal) {
                start = center + 1;
            } else {
                end = center - 1;
            }
        }
        return start;
    }
        
    public int countSubsequences(int[] array, int limit) {
        int size = array.length;
        int modulo = 1_000_000_007;
        Arrays.sort(array);
            
        int[] powers = new int[size];
        powers[0] = 1;
        for (int i = 1; i < size; ++i) {
            powers[i] = (powers[i - 1] * 2) % modulo;
        }
            
        int result = 0;
    
        for (int leftIndex = 0; leftIndex < size; leftIndex++) {
            int rightIndex = binarySearchInsertPos(array, limit - array[leftIndex]) - 1;
            if (rightIndex >= leftIndex) {
                result += powers[rightIndex - leftIndex];
                result %= modulo;
            }
        }
            
        return result;
    }
}

Solve the problem of finding the number of subsequences in an array that meet a certain sum condition by applying the provided Java solution. Understand the approach of using binary search for efficiency and modular arithmetic to handle large numbers.

• First, sort the input array to simplify the search for subsequences satisfying the sum condition.
• Calculate power of twos up to the array's length using modular arithmetic. This is to efficiently compute the number of subsequences.
• For each element in the array, use the helper method binarySearchInsertPos to find the appropriate position in the array where the sum condition breaches. This method returns the position to insert an element, which indirectly helps in identifying valid subsequences' upper bound.
• Count valid subsequences from the current element's index to the calculated upper bound using powers of two. This exploits the property that for sorted arrays, every subsequence formed with the current element and elements before the upper bound is valid.
• Handle large result numbers by taking modulo $10^9 + 7$ at each step to prevent integer overflows.

Each iteration effectively counts the number of valid subsequences involving the current element of the array, ensuring the solution is both efficient and straightforward to implement. The combination of sorting, binary search, and modular arithmetic provides a robust framework for solving similar problems in subsequence counting under specific constraints.

class Solution:
    def countSubsequences(self, arr: List[int], limit: int) -> int:
        length = len(arr)
        modulus = 10 ** 9 + 7
        arr.sort()
            
        result = 0
            
        for start in range(length):
            end = bisect.bisect_right(arr, limit - arr[start]) - 1
            if end >= start:
                result += pow(2, end - start, modulus)
            
        return result % modulus

The provided Python code solves the problem of finding the number of non-empty subsequences within a given array such that the sum of the smallest and largest elements in each subsequence is less than or equal to a specified limit. Here’s a breakdown of how the solution works:

• Firstly, make sure the array is sorted. This order helps to easily identify the largest and smallest values in any subsequence.
• Set up a variable result to keep count of the valid subsequences.
• Iterate through each element of the array treating each element as the potential smallest in a subsequence.
• For each starting element, find the end boundary (the largest number) for a valid subsequence using binary search (bisect_right). This optimizes the search by narrowing down the possible largest values under the condition that their sum with the smallest element is within the limit.
• Calculate the number of valid subsequences from the current start to end boundary using the formula 2^(end - start), which leverages the properties of sets where each element can either be included or excluded.
• Use modulo arithmetic to handle large numbers and prevent overflow, ensuring the result fits within typical integer limits of computational environments.
• Return the final count of subsequences modulo 10^9 + 7, which often appears in computational problems to maintain result size manageable.

Make sure to include the standard library function bisect.bisect_right to perform the binary search over the array, optimizing the evaluation of possible subsequences. The solution is efficient, utilizing sorting alongside binary search, ensuring that the operations remain manageable even for larger sizes of input arrays.