LeetCode-in-Java

1498. Number of Subsequences That Satisfy the Given Sum Condition

Medium

You are given an array of integers nums and an integer target.

Return the number of non-empty subsequences of nums such that the sum of the minimum and maximum element on it is less or equal to target. Since the answer may be too large, return it modulo 109 + 7.

Example 1:

Input: nums = [3,5,6,7], target = 9

Output: 4

Explanation: There are 4 subsequences that satisfy the condition.

[3] -> Min value + max value <= target (3 + 3 <= 9)

[3,5] -> (3 + 5 <= 9)

[3,5,6] -> (3 + 6 <= 9)

[3,6] -> (3 + 6 <= 9)

Example 2:

Input: nums = [3,3,6,8], target = 10

Output: 6

Explanation: There are 6 subsequences that satisfy the condition. (nums can have repeated numbers). [3] , [3] , [3,3], [3,6] , [3,6] , [3,3,6]

Example 3:

Input: nums = [2,3,3,4,6,7], target = 12

Output: 61

Explanation: There are 63 non-empty subsequences, two of them do not satisfy the condition ([6,7], [7]). Number of valid subsequences (63 - 2 = 61).

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 106
• 1 <= target <= 106

Solution

import java.util.Arrays;

public class Solution {
    public int numSubseq(int[] nums, int target) {
        // sorted array will be used to perform binary search
        Arrays.sort(nums);
        int mod = 1000_000_007;
        // powOf2[i] means (2^i) % mod
        int[] powOf2 = new int[nums.length];
        powOf2[0] = 1;
        for (int i = 1; i < nums.length; i++) {
            powOf2[i] = (powOf2[i - 1] * 2) % mod;
        }
        int res = 0;
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            if (nums[left] + nums[right] > target) {
                // nums[right] which is macimum is too big so decrease it
                right--;
            } else {
                // every number between right and left be either picked or not picked
                // so that is why pow(2, right - left) essentially
                res = (res + powOf2[right - left]) % mod;
                // increment left to find next set of min and max
                left++;
            }
        }
        return res;
    }
}

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